So, we'll proceed.
What did in the last
segment was look at high,
the behavior of high order modes and
also introduce this notion of
consistency of the finite difference
approach to time discretization.
I ought to mention that proof of that,
those results for consistency,
and order of accuracy are obtained
by essentially looking at
the time exact solution,
carrying out Taylor series expansions,
and finite Taylor series expansions,
and then, and then manipulating them.
It's a, it's a fairly straightforward
exercise but somewhat tedious, okay?
So, in this segment we are going
to essentially move to the end,
end game for this particular topic.
We are going to look at convergence.
Okay, so
we are going to look at convergence.
Of.
Time discrete.
Solution.
All right.
In order to talk of convergence,
as you may recall from our
early treatment of error analysis for
the finite element
method itself, we need first
to identify what our error is?
Okay.
So we define the error, and
as we've done before in the context of
the finite element solution itself we
will define the error as the time,
as the discrete solution, right?
But in this case, we look at
the discrete solution at n plus 1,
minus the time exact solution at n plus 1.
Okay?
All right.
Now this is what we
define as e at n plus 1.
Okay?
Now, e n plus 1 is of course,
is just another vector in the same space.
Right?
Right, just as d and t are sorry,
your d n plus 1 n d at tn plus 1
are vectors and this rndf space.
But what that means is that because we
have a basis our orthonormal basis,
we can expand e as well in that basis,
right?
So we can essentially carry
out the modal decomposition.
Of e at n plus 1.
Right.
And that model of
decomposition essentially
is that e at n plus 1 equals sum over m,
e m for the mode at n plus 1, right,
those are our scalar modal
coefficients times psi m.
All right, just as for any other vector.
Okay, now
how do we pose the question
of convergence, right?
We say that something has converged,
right?
Or we say our solution has converged,
have converged.
Convergence Right,
is the requirement that
as limit as the limit of
n plus 1, tending to infinity, right,
as we take more and more steps, right.
What we expect to see
is that e n plus 1,
dotted with M e n plus 1 equals 0.
Right, that limit is 0.
Okay.
Can you tell me why it is okay to,
or can you think,
why it is okay to put M inside there?
Normally one would say that well,
the error has to vanish, right?
But why is it okay to put M in there?
It's because M is positive definite.
Right?
All right, so let me just state this here.
Since.
M is positive definite,
all right.
What we see is that e n plus 1
dotted with M e n plus 1,
right, equals 0,
if and only if e n plus
1 itself equals 0.
Okay?
Only if,
it's only when that vector
itself is equal to 0 that,
that quadratic product is equal to 0,
since M is positive definite.
And so, and this is why it means that
well, yes, if, if this particular limit is
tending to 0, it has to be that e itself,
e n plus 1 itself is tending to 0.
Okay, so
this is why it is okay to consider this,
as our convergence criterion.
All right.
But a more importantly,
we are not going to work in this form.
Right, we are not going to work,
with e dot with M e.
We're going to work in terms
of modal coefficients,
because our entire analysis has been
in terms of modal coefficients, right.
But it's okay,
because I'm going show that for
this quantity tending to 0.
Right.
This quantity that I put a brace on,
okay, that quantity tends to zero,
only modal coefficients
themselves go to zero.
Okay.
So let me,
I've told you what I'm going to show you.
But in order to write it out, let me say
the following, but not the following.
Okay, e n plus 1,
dotted with M,
e n plus 1 equals
sum over m, e m,
n plus 1 psi m,
dotted with M
sum over l,
e l n plus 1.
Psi.
Right, and
we're going to put parentheses here.
Okay.
Each of those sums is for one of those
e's right, in the quadratic product.
All right?
Okay.
Well, we know how this works now.
This is equal to,
let me see how I want to write it, right.
It is the sum
over m and l,
right, of,
e n, n plus one,
psi m dot M psi l.
All right, and for
no particular reason I'm going to put
parentheses around that, and
here I have e l n plus 1.
All right?
However, we know that this is delta m l.
Because of the order normality
of the psis with respect to m.
Right?
Well, if that is the case,
we know that,
this product that we started
out with, e, n plus one,
dotted with m e, n plus one,
equals sum over m, e m n plus one.
E m n plus one.
All right, essentially what has
happened there is the chronicle delta,
has been used to, to turn that
l index into an m index, right?
So this is basically just
the Euclidean norm if,
with respect to the psi basis,
right, so this is sum
over m, e m n plus one, the whole square.
Okay, so now it's clear that if this
quantity is standing to zero it
means that the modal coefficients,
that sum itself has to tend to zero.
Right, which means the modal coefficients
themselves have to go to zero.
Okay?
So what this implies, finally,
is that, limit, n plus one,
tends to infinity.
E, n plus one, dotted with M e,
n plus one, equals, essentially
limit as n plus one tends to zero,
sorry, tends to infinity.
Of the sum.
Okay?
All right?
So clearly, if this, has to be equal to
zero, if this limit is equal to zero,
it just means that, each, every single
modal coefficient, tends to zero.
Okay?
All right, so limit n plus
1 tends to infinity.
E n plus 1.
Dotted with M e n plus 1, equals zero.
If and only if, each E m n plus 1,
equals zero in the limit.
All right?
So, it means that it's okay to just
look at what happens with the modal
coefficients.
Okay?
All right, so.
Study convergence.
Of modal coefficients,
right, E
n plus one.
Right and now dropping the explicit,
mention of the modal, index.
Right I've just dropped n.
Okay?
All right,
well that's what we're going to do.
Okay, so, how do we set it up?
We set it up by going back and
writing out our time-discrete,
discrete equation,
our modal time-discrete equation in
the form that we set up in order to
pose the question of consistency.
All right, so here is what I say.
Consider.
The time discrete equation
in the following form.
D n plus 1 minus A d n minus
L n equals zero, okay?
And consider the, the exact equation
also forced into this form.
We know, however, that the exact
equation does not cooperate, and.
Give us a zero right hand side,
instead we get delta t, tau at t n.
Okay, right, what I'm going to
do now is subtract the second
equation from the first, okay,
so I'm going to change signs.
Okay, and
now when I add those two together,
I get on the left hand side,
right, d n plus one,
minus d t n plus one,
is essentially our e, at n plus one.
Right?
Okay?
Right?
It is the, error in the corresponding
modal coefficient, right?
And here I get minus A e at n.
All right, the Lns cancel out,
and I'm left with minus
delta t tau at t n.
I'm going to rewrite this
as an expression that
will allow me to write out
a recursive formula for the error.
Error at n plus 1 equals A times
error at e minus delta t.
Tao at tn, okay?
I'm going to use recursion now, and
write from here e at n equals A e
at n minus 1 minus delta
t tau t n minus 1, okay.
What that implies is
that on substituting e as
written here in that expression, right?
I get e at n plus 1 equals
A square e at n minus 1,
minus A to the power
0 delta t, tau at
t n minus 1, okay.
Right.
Minus A to the power 1
delta t tau tn minus 1,
and I just realized that
here it is just tn.
All right, I can take yet
another step, okay.
What, if I were to now rewrite
an equation for e n minus 1.
All right, I would get A e
n minus 2, minus delta t tau
at t n minus 2, okay?
And making the substitution, right?
You see how this is going, right?
What I would get is that e, sorry.
E at n plus 1 equals
A cube e n minus 2,
minus A to the 0 delta t tau tn,
minus A to the 1 delta
t tau tn minus 1,
minus A squared,
right.
Minus A squared delta
t tau at tn minus 2, okay.
Now one can go on with this, and
I'm sure you've all done it at one stage
or the other with similar expressions.
When you go all the way back to,
the zeroth step, right, so
the initial time, we get e at n plus
1 equals A, to the power n plus 1.
Okay, right.
A to the power n plus 1, e at 0, okay.
Minus, sum i
going from 0 to n,
A to the power i,
delta t, tau at tn
minus i, right?
You can check that this is what
the recursive formula reduces to, okay?
All right, let's stare at this.
What is e at 0?
What is the error at time t equal to 0?
It's equal to 0, right?
Because we've made sure that
our initial condition is
actually obtained from the time
exact solution, right?
I mean, at the initial time
there is no error, right,
because we fixed the initial condition.
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