What we will do is we will look at, let me
explain what we are trying to do now right,
so that we get the context right. We know
that the error in the solution en+1 satisfies
this is determined by this iteration equation
is that right. Now what we have is we are
going to, this error corresponds to a function
exy, which is error in the original solution
okay.
So remember that every time that you have
phi h, there is a corresponding phi, right,
there is a corresponding I would say, though
they are set of points, it actually corresponds
to a function, it is a function, because we
can use linear interpolants to find the value
at any given point, am I making sense, okay.
See as I said before in an earlier class you
are supposed to be giving me a solution to
Laplace equation.
And if I say that I have a square and in this
case we have actually gone from, we have taken
an L/L square, for the sake of this discussion
we have gone to an L/L square. If you say
that I have gone to an L/L, if you are saying
that you give me a solution to Laplace equation
in the square, right, that means I have a
right to pick any point and say what is a
function value at that point and you have
to give me a value.
You cannot say no it is not one of my grid
points, I cannot give you a value that is
not an acceptable answer, then you have not
given me a solution, anywhere in this L/L
square if you say that you have solved Laplace
equation satisfying the boundary conditions
that I have prescribed, then you must necessarily
give me a value, right, at any given point,
you understand what I am saying.
Which is why we talked about the HAT functions
and all those ways by which given values at
nodes you are actually able to give me intermediate
points that was the objective, okay, that
is the reason why that we know that there
is an underlying function representation,
though I am giving you nodal values, is that
fine, so exy corresponds to that. Now what
we are asking, the question that we are asking
is, if I go through this sequence en+1 is
phi times en.
If I keep repeating this, I will generate
a sequence, which is e0, which is our first
error that we make when we assume the solution,
then you have e1, e2, e3, these are not e
as an two points on one, this is e as an error,
right, so e4 and so on, so the question that
we have is does the sequence converge? What
we are proposing to do now in today's class
what we are proposing to do is we are going
to expand this using Fourier series, okay.
The original equation, this is the linear
equation. Fourier series is a linear combination
of science and co-science, okay, right, so
if I would substitute the Fourier series,
any mode, any single mode to the Fourier series,
if I ask the question what does this equation
do to any single mode to the Fourier series,
right, if I pick a general mode and ask the
question what does this do the amplitude of
that mode?
Is first wave number, or the second wave number,
or the third wave number, right, I can address
that question individually because the equation
is linear, is that fine, okay, because this
cropped up, so, essentially what I am going
to do is, I am going to write e of x, y as
the Fourier's, but it is in 2 dimension, the
Fourier series in 2 dimensions and I know
that the boundary conditions are 0, because
this is the error.
So the value of the error on the boundaries
is 0, because the boundary conditions are
applied exactly, so it is nonzero only on
the interior, okay, so if you do Fourier series
normally you have to do something called periodic
extension and all that, because your initial
mode if you think about it just in one dimension
if this is x, that is y, just in one dimension
your initial mode sin x will be something
like this, okay.
The initial mode sin x will be something like
this, it is 0 here, it is 0 there, your error
can contain for example your error could just
be this, okay, so you have to able to represent
this, so this goes from 0 to L, what we normally
do is we extend it to –L and there is a
function on that side that we are not going
to worry about because I do not care it, it
is not in my problem domain.
So as a consequence this e of x,y, I am just
saying this so that you can go back and look
up your Fourier series, right, and figure
out where it is, okay. There is a consequence,
I can write e of x,y as double summation because
it is in 2 dimensions al bm exponent i pi
lx/L and 
exponent i pi my/L, is that fine, everyone
and the summation right now, this summation
I will just write l and m.
The summation if you go look up your Fourier
series, the summation actually goes from –n
to +n or something of that sort, okay I will
just write l and m, you go check this out,
okay because there are grid points that go
because I have extended it to –l, there
are grid points that go the other side if
I started numbering this at 1 this will be
-1, -2 and so on, but we do not go and get
into that, I do not want to get into that,
okay.
So I just leave it as m and l, okay, fine.
Now where do I want to, at what points, but
I do not though this is a continuous function,
I planned to substitute it into that iteration
equation, right, which is a discrete equation
and that is going to be evaluated at my grid
points, which are xpq and ypq, so because
it is a uniform grid, I do not need the q
here, nor do I need the p here.
So xp is p times h, h is the grid size and
yq is q times h, right, so this turns out
to be exp, so I will write e of x, p, y, q,
if I write that as epq turns out to be summation
over l, summation over m, al bm exponent i
pi x/l, x is ph/L and i pi m, there is an
L here qh/L, fine, now we have already seen,
so ep+1q what is the relationship between
ep+1q and epq? So the difference will only
be a e power i pi h, right.
You can work that out, so ep+1, so it will
be turned out to be an e power i pi h/L times
let me write exponent, I do not want to write
e power because I am already using e for exponent,
i pi h/L * epq is that fine, right, now because
they are equivalent rules h/L is known, so
I have to have the summation, okay, I do not
want to do this, okay, I am doing this a bit
early okay, let me do this, I have to take
component by component first, okay.
Now this e I want to substitute into en+1
= p times en, okay because this equation is
linear, this is linear, I can swop out the
sums and I can swop out the double summation
and so on and component by component, I can
actually extract out the a's and L's component
by component by appropriately dotting into
the appropriate e power appropriate sign and
co-sign, right.
There are all orthogonal to each other and
so on, so in fact it is enough if I can take
any one wave number and ask the question what
is the effect that this iteration equation
has on that wave number, okay, I have an orthogonal
set essentially, I can ask what happens to
any one of them, is that fine okay, so the
objective of course is we want to find out
which wave number is decaying the slowest.
You want to find out which of these wave numbers
is decaying the slowest, so I can pick an
l and an m, so for an arbitrary l and m, so
I pick l and m arbitrarily and I ask what
happens to that component, so if all the others
was 0, right and that the error was in that
form. There are different ways by which we
can argue this, all the other coefficients
was 0 and only the l and m components were
left.
How would that l, m component grow, okay that
is another way that we can look at it, okay,
how would the l, m, if I had only this, how
would this component grow, is that fine. There
are different ways that you can look at this,
so that is one possibility that you can ask
the question how would that l, m component
grow, so I should in theory also I add a subscript
right, I should in theory also I add right.
So exy, I should also add a subscript saying
that I am going to do the l, m component,
but just so that I do not make it too complicated,
I will just leave the, we know that we are
dealing only with the, I have picked an l,
m component, okay. So what does that do, let
substitute it into our equation and see what
that does, so epq at n+1 is 0.25 times ep+1
q at n + ep-1 q at n+1, we will leave this
at n + epq+1 at n +epq-1 at n, is that fine.
So for the l, m component, for l, m component
what do I have? ep +1q in fact is exponent
i pi lp+1 h/L = exponent i pi l h/L * e power
pq, is that fine, okay, so we get these expressions
ep+1q = this exponent i pi lh/LS N and epq,
ep-1q = exponent –i pi l/n ep-1q and similarly
you get the other ones with +m and –m, the
wave numbers will change, so ep+1q going back
to our iteration equation epq n+1 is 0.25
times, is that fine.
epq at n * 
exponent i pi l/N + exponent – i pi l/N
+ exponent i pi m/N + exponent i pi – i
pi m/N, is that okay, is that fine, I have
replaced h/L by N here okay, h/L is 1/N is
that fine, h/L is 1/N. Now we will use the
fact that e power i theta exponent of i theta,
since I have called the error e, I pay the
price for that right, so exponent of i theta
= cos theta + i sin theta substitute back
here and you will get epq n+1 remember this
is for wave numbers l, m.
We are just asking the question what happens
to the wave numbers l, m, okay, = epq n/4
of course * 2 cos pi l/N + 2 cos pi m/N. So
the gain going from one iteration to another
iteration the gain is basically the ratio
of these amplitudes, okay, so it is epq at
n+1/epq, which is g, this happens to be real,
so we do not have to worry about, right, normally
if g were complex, then we would have to do
gg bar or something of that sort, it happens
to be real.
So we want only the gain and we want the modulus,
right, the absolute value and this is mod
of 2 cos pi l/N + 2 cos pi m/N/4, is that
okay, everybody with me. Now we ask the questions
what is the value, right, when is this maximum,
I want the maximum possible value. Now it
turns out the error, see the error is 0 at
the boundaries, we do not have electrical
engineers, we would say we do not have a DC
component, right.
So, the summation need not start at 0, A0,
we would not have an A0, or a b0 right, so
the summation starts 
the wave numbers that we get 
go from one basically through n-1 that are
of interest to us, okay, as I said the DC
component is not there, the error is 0 on
all the boundaries, so fortunately for us
the expression that we have g as mod 2 cos
pi l/n + 2 cos pi m, I just repeat it here
/N fortunately the expression that we have
mod/4.
We ask the question when does this become
maximum, when does it take it is maximum value
and that occurs fortunately for us it occurs
when at the 2 extremes 1 and n-1, we do not
really have to hunt anywhere in between, am
I making sense, because if you go to 0, if
at 0 was possible then the cos would be 1,
if you went to pi cos would be -1, right,
which are the extreme values that it can take.
We have a modulus so the sign does not matter,
so I substitute 1 and this gives me the max
over l and m, okay, so you see the game that
we have played, we have used the fact that
the equation is linear, you have considered
an arbitrary wave number l and n, we have
got the gain that we can get for that l and
m, and now we are asking the question for
which l and m is it maximum, is that gain
maximum okay that is the key.
So the max/l and m gives me g max, which is
4 cos pi/N/4, which is cos pi/N, is that fine,
okay, which of course I can expand using MacLaurin
series, so MacLaurin series would give me
and I will just use the first 2 terms, MacLaurin
series, this will give me approximately 1-pi/N
square 1/2, other terms I will ignore those
pi/N square, okay, so this is our largest
Eigen value, this is basically the spectral
radius, you understand.
When I put it through the crunch once I iterated
once, right, if I go from an n titrate to
an n+ first titrate, n was chosen arbitrarily
from an n titrate to an n+ first titrate the
gain that I get is of this order, this is
the largest gain corresponds to the largest
Eigen value, right, it corresponds to the
largest Eigen value, what is this value, so
if you take say for example n is 100, what
are you going to get?
We can just estimate it, if you take n as
100 pi squared is like 10, we will do an engineering
approximation, pi squared is like 10, right,
so the denominator gives me 1- it is of the
order of 1-1/2000 that is like 0.999 is there
one more 9 enough, fine something of that
sort, do you understand what I am saying,
okay, so for the first mode if you have an
error I am drawing it only in one dimension
now, we will forget the other.
If you have an error, if your initial guess
is such that the error is like that, that
is the first mode that is going to decay,
this amplitude is going to decay at this rate,
every iteration that you do that amplitude
will be multiplied by this number, it is going
to take forever to converge, am I making sense,
okay, right, so as I said we will do, in the
next class I will do a demo and you will actually
see how bad it can be.
Fine, are there any questions, yes please,
“Professor - student conversation starts”
because exponential is orthogonal (()) (24:16)
exponentials are orthogonal to each other
and equation is linear, right, it is like
the equivalent of saying that sigma f = 0
then you do statics or dynamics, you say sigma
f = 0 then you can say the x component of
force is 0, y component of force is 0, z component
of force is 0, there you are using orthogonality,
okay, “Professor - student conversation
ends”
But you also need that the equation into which
I am substituting, say sigma f = 0 fortunately
happens to be a linear equation, right, so
ijk are orthogonal, so you can do it component
wise, but the equation that you are going
to decompose has to be linear, if the equation
is not linear then you run into difficulty,
right because you can get coupling terms and
all of that kinds, is that right, is that
make sense.
You have to a bit careful because the ijk
argument you have to be a bit careful, right,
that analogy only go so far, right, be a bit
careful, okay so if the equations were nonlinear
in this case, right if you had u, du, dx term
right which we are all familiar with from
our fluid mechanics, if you had u, du, dx
term, if you substitute Fourier series into
it, let us forget exponentials, let us just
stick with signs and co-signs.
If the u you are looking at the sin x term,
du, dx would give you a cos x term right and
suddenly u du dx gave you a sign 2x term,
do you understand because this is the sin
x cos x, okay the u du dx term contributes
to the sin 2x component, so you cannot, when
you say I am going to decompose it, you have
to be a bit careful, okay, if you are going
to decompose it component wise, you have to
be a bit careful, do you understand, right.
So it is important that the iteration equation
is linear, that is very important, fine, okay.
So you can see that if you try to get any
sensible, if you try to get even n=10 right
it is going to give you like 1-1/20, even
for n=10 you are going to end up even for
n=10 g max you are going to be like 1-1/20,
am I making sense, so the convergence rate,
I mean it is quite, so you may be happy with
how fast it runs with for n=10.
But the minute you want to say if you want
to try something larger now you are talking
about let me try at n = 1000 then it gets
really bad, right, if for whatever reason
you want to use higher resolution, then it
gets really bad, okay, there are no other
questions, the other thing that we looked
at was writing A using the fact that A transpose
= A, so we use the fact that A is symmetric,
you remember when I say A what I am talking
about, A is symmetric.
Symmetric A, right, that is this equation
could be transformed to a discrete version,
which was A phi = b, okay, this is what I
meant, so this could be transformed into to
stick to our consistent notation, this could
be transformed into this and I had basically
said that for consistency.
I mean not for consistency, but to make it
look like a traditional standard problem that
you are used to, if you just write it as ax
= ax tilde = b then corresponding to this
we can come up with the function q, which
is the function of x tilde, which is 1/2 x
tilde transpose A x tilde – x tilde transpose
b tilde. I am sticking the tilde underneath
just to indicate that they are vectors.
And I made it a b tilde because I have called
it x tilde that is only reason why we are
doing what we are doing, so that the equation
is consistent and we already saw that the
gradient of Q x tilde gives us because A is
symmetric this gives us Ax tilde –b = 0
for extremer, I think I had a sign error there
earlier, so the one dimensional equivalent
of this or one dimensional 
analog if I just consider one coordinate,
so I can have Q of x, this is just x = 1/2
Ax squared –bx.
So normally when we do flow past anything,
flow past cylinder or something of that sort,
right, I just set up the boundary conditions
and then we will talk about it later, so and
we have Laplace equation for so if you say
the flow is rotational, so I would not go
through the fluid mechanics of all of this,
right, so if this is the solid cylinder and
there is a fluid flow past the cylinder, there
is a flow past the cylinder, okay.
So what is the boundary condition that you
have on the cylinder? They has no penetration
boundary, it is called the no penetration
boundary condition or the solid wall condition,
right, which has the normal velocity is 0,
the normal component of velocity is 0, so
there is no normal component of the velocity,
there is only a tangential component, okay,
that is grad phi doted with m = 0, right.
And from our understanding of directional
derivatives, which will need in a little while
now, this tells us that dou phi/ dou n= 0.
We are not going to do flow past cylinder
here, this is just for motivation, we are
not planning to do flow past cylinder, okay,
so we will restrict ourselves to a box, but
the fact that matters that anywhere on the
box it is possible that your given dou phi/dou
n =0 or dou phi/dou n = something, okay.
If this represents a room and there is some
air coming in through a air condition or whatever
it is then dou phi/dou n may give you some
velocity at that point, which is a speed at
which the air conditioner is injecting air
into, so you may be given the dou phi/dou
n value, what we have done so far is we have
prescribed the phi values on the boundary,
but the question is what happens if we are
not given phi, but you are given dou phi/dou
n, okay.
Fine, are we given these on all the boundary,
are we given it only on one boundary, you
might have studied this is in PDE, I am pretty
sure in partial differential equation course,
you would have seen Dirichlet problems, Neumann
problems, and Robin problems. So Dirichlet
problems you are given the function value
on the boundary, Neumann problem you are given
derivatives on the boundary, and Robin problem
is a mixed problem.
You are given combinations of directives then
function value, okay, so we have to basically
see how we will handle the problems that have
derivative conditions given on the boundary,
right, so that is a very small segment that
we will do that in the next class and we will
also do a few demos, okay. Thank you.
