Okay, so what this gives us then is that.
This tells us then that e
at n plus 1 equals minus
sum i going from 0 to n
A to the power delta t tao t at n minus i.
Okay.
All right.
Let's work with this.
And we're going to work now,
by invoking some inequality.
Okay.
In particular what we can say,
first of all, is that the first con,
the first step that we will
take is not an inequality.
It is to say that
the magnitude of e n plus 1,
right, is equal to essentially
magnitude of sum i going from 0 to n,
A to the power i delta t tao t n minus i,
okay?
Right, we're taking the magnitude of that
sum, the absolute value of that sum.
Now, is where things
get really interesting.
We, now's where the inequalities come up,
okay?
And, and
remember when we write the inequalities,
I'm going to start out by writing this.
And you recall when we looked at the,
our error analysis for
the finite element method I made the point
that when I write such an inequality,
what I mean is that the previous
right-hand side, this one is bounded
from above by what I'm about to write
now as the new right-hand side, okay?
So, what I get is that,
that is lesser than or
equal to i going from 0 to
n the magnitude of the,
the absolute value of
each one of those terms,
and that's A to the power i,
A to the power of i delta
t tau add t n minus i.
Okay?
All right the absolute value of a sum is
bounded from above by the sum
of the absolute values.
And this result is a very
standard step in analysis.
It is called the triangle inequality.
Okay?
But there is more.
We can say further that that,
that our most recent
right-hand side itself is bounded
from above by this expression.
Okay?
Essentially what we're seeing here that
is that the that any absolute value,
which is a product, right?
So the product,
the absolute value of a product is
boundary from above by the product
of the absolute values.
Right, and this result is called
the Cauchy–Schwarz inequality.
Okay?
But now we have more.
We have stability, right?
If our method is stable, what we are able
to say is that sum i going from 0,
to n of 1 times magnitude of delta
t times tau tn minus i, bounds from
above our previous right-hand side.
Okay?
Essentially,
what I've done is replace,
A i with 1, right.
And why, why am I able to do that?
It's a property of the methods
we are looking at.
What property are we applying here?
It's stability, right?
Because we know that the magnitude of
A has to be lesser than or equal to 1.
The absolute value of A has to be less
than or equal to one for stability,
therefore, A to the power i has
an absolute value also lesser than or
equal to 1.
Okay, so
if you re-substitute that with one,
we get a bounding from above of what we
had as our previous right-hand side, okay.
But now the story goes on,
now we know that delta t is
a positive quantity, right.
So what we can do here is,
we know that delta t's a positive quantity
and, furthermore, we know that for
tau, the condition of consistency
allowed us to say something about it.
Okay, so in one step I'm going to do this.
I'm going to pull the delta t out.
All right.
And I have inside my sum.
C delta t to the power of k.
Okay.
Why am I able to say this?
It's because we said that
we have consistency, okay?
All right, we have consistency.
Now one could introduce another step
inside here, which would be to first say
well, if all the tau's over all the n
minus i's are take maximum of them.
Okay and then I get to this point, right.
So let me say that one can see
inside of here, when they first, if
one wanted to be really careful about this
then I guess one should be really careful.
So one would say, that first of all you
would have a previous step inside here,
which is to say that the,
that this right-hand side, right?
Using that, one could say that sorry,
the entire right-hand side, not just that.
One could say that this
entire right-hand side,
first of all would be
bounded from above by
sum i equals 0 to n delta t times
the maximum over all i, t n minus i.
Right.
One could take that step, right?
Which is to say,
that well since each of the tau,
you know you have a different
value of tau for each time step.
Let's consider the maximum over all i,
okay.
But then we know that that maximum
has to be bounded from above by C
delta t to the power of k,
because we have consistency.
Right, so
then we get to this step that I have here.
Right.
Okay, all right.
Okay now well, what do we get here?
We have delta t times sorry,
the sum here goes from 0 to n.
Sorry, right.
Okay, we have delta t times a sum of n
plus 1 steps, because i is going from
0 to n, of that quantity c delta t
times c delta t to the power of k.
'Kay, so we can very well say that
it is just like multiplying delta t
essentially n plus 1 times, right.
So this thing is lesser than or
equal to t to the power n plus 1.
Sorry, t at n plus 1, not t to the power
n plus 1, but t at n plus 1 times
c plus delta t, to the power of k.
Okay.
All right,
just summing over those n plus 1 steps.
Okay, right right.
Now and now from here, we see however,
we see that as delta t tends to 0, right,
because k is greater than 0, right?
Then as delta d turns to 0,
we see that the righ- hand
side also turns to zero, okay?
All right, but.
Limit delta t tends to 0 of
c delta t to the power k is equal to 0 for
k, greater than 0.
Right, and where do we get this?
Why are we allowed to see
a case greater than zero?
Once again it is
the result of consistency.
Okay, so what we see finally,
is that the absolute
value of our modal
coefficient of the error is
bounded from about by, essentially by 0.
Right.
Which means it, it,
it still tends at 0, right, and
the limit as delta t tends to 0, right.
What we are seeing is that, that
quantity is lesser than or equal to 0.
Okay, so indeed we have conversions.
Right, and that's the end of our proof.
Okay?
Make a quick remark here,
which is that what we've
seen here is a demonstration
of the use of consistency.
And stability.
Implies convergence.
Okay, which is a very standard it's
actually a tier in numerical analysis.
Okay it's called
the Lax theorem.
Okay?
All right we're actually done with
this entire topic of methods for
parabolic problems.
And what and, and the approach we've taken
if you just to summarize is to carry out
a standard spacial discretization
using the finite element method.
But to the time discretization using
finite difference methods, and
since here we're looking
at first order of problems.
We looked at the Euler family.
We analyzed its stability understood
the behavior of its higher order
modes looked at consistency, and
looked at how stability and
consistency give us convergence.
Let me make one more remark
now about about the be,
about the use of the different algorithms,
okay.
In particular, let me do this, okay.
Let me say, let me say that
here I have alpha, okay, and
let's look at alpha equals 0,
alpha equals one-half, and alpha equals 1.
Okay, and just for
connection to other things.
Let's give this,
this algorithms their names, right?
So this is forward Euler.
Right, this is the mid-point rule,
or the Crank Nicolson method.
And this is backward Euler.
Okay let's look here at the stability.
Let's look at order of accuracy.
And let's draw a straight line.
Okay better, and let's look finally
at high order modes, right.
Okay?
Stability of forward Euler is conditional.
Right, and we've seen the cons,
the stability condition.
Midpoint rule and
backward Euler are unconditionally stable.
Okay?
Order of accuracy,
forward Euler has order of accuracy 1,
midpoint Euler has order of accuracy 2,
backward Euler has order of accuracy 1.
For high order modes,
limit as lambda h delta
t tends to infinity.
Limit of A in the tends to infinity not 0.
Okay forward Euler is let me just see
it tends to minus infinity right,
because nothing can be equal
to minus infinity, right.
Midpoint rule.
Limit lender each delta t,
tends to infinity is equal to minus 1.
Okay?
Oscillatory behavior.
For backward Euler we
see that limit lender h
delta t tends to infinity.
Oh, sorry, limit of A here.
Okay, here too, limit of A equals 0.
Okay, so it damps out high order modes,
dissipates them away.
So this is the, broadly speaking,
the behavior of the three main
are the three most commonly
used members of this family.
Depending upon the problem
you choose your method, okay.
All right, we're done with this
segment and with this unit here
when we return we will take up
the problem of Elastodynamics.
