Okay.
So we're now shaping up to get to the main
result related to finite element analysis,
right?
An, an analysis of
the conversions of the method.
To get there, to,
to actually get to that result,
there is one more class
of results that we need.
So the title of this
segment is going to be
Sobolev estimates.
Sobolev of course, refers to the type
of space that we are working with and
that I'll tell you what we're
trying to estimate here.
In order to say what we
are trying to estimate
let us sort of reuse our symbol U,
capital U of h that we introduced before.
It continues to represent
a general member of the space Sh,
which means it satisfies the,
the rationally boundary condition and
it lives in H1.
Okay?
For, for our particular problem,
it, it lives in H1.
But in the context of Sobolev estimates,
which are applicable to spaces
with which, which are applicable
in general to Hn spaces.
Let me just redefine this for now.
Okay.
Let me say that Belongs to Sh,
which now consists of all functions.
That for our, for our purposes now,
which actually hold in
more general settings.
Okay?
Consist of Hn functions.
And additionally, of course, they need to
satisfy the Dirichlet boundary condition.
Right?
And so, if we were doing the,
the Dirichlet-Neumann problem,
we would have At 0 equals 0.
Okay.
So we're re-invoking this,
this same function capital.
Remember, U, capital Does not need to
be the, the finite element solution.
It could be, right?
But it does not need to be, right?
So let me state that.
Does not necessarily
represent little
Right?
Which is the finite element solution.
It is any other solution in Sh, right?
Where Sh now has been defined to
include all functions that belong,
that live in Hn.
Okay?
All right.
I want now to consider a very special
member a member of this set of functions.
Okay?
Consider
Such that.
At any nodal point
let me say, this as X
let me just say, x sub A.
Okay?
So At x sub A equals little at x sub A.
Right?
Okay.
But if A here just denotes our global
numbering of degrees of freedom,
then this remember,
using our finite element
vector notation is just
the degree of freedom, d sub A.
Okay?
All right.
All right.
So what this, what this says is that A is
a global degree of freedom, right?
And this, this then implies that
xA is a globally numbered node and
dA is a globally numbered
Trial solution degree of freedom.
Right.
Or finite element solution degree
of freedom.
Okay?
All of that should be clear.
So what we're saying is that we want to
consider a special member of this class,
where Is equal to the finite
element solution at the node,
the degrees of freedom.
Right?
At the nodes, okay?
But you know that if we do this,
we would have capital Actually
equals to little everywhere, right?
Because of the fact of the finite
element representation.
Okay?
Right?
So this is not,
not exactly the solution we want.
Not exactly the the function we want.
Consider another function,
U tilde h which is such, that
U tilde h.
Okay?
At xA, now equals the exact
solution evaluated at that point.
Okay?
The idea is that if you had the exact
solution and you could evaluated
it at a nodal point, right?
Take that value and
let's suppose that u tilde h is such that
it hits the exact solution
right there on the nodes.
So what we're seeing here
is that u tilde h, right?
Expressed now as a function of
parametrize by little x, right?
By position.
U tilde h is what is said
to be nodally exact, okay?
I'm going to sketch it out for
us, suppose this is our domain.
And for simplicity, let's just suppose
we have linear basis functions.
Okay?
And suppose our exact solution is
the following.
Let me go to a different color here.
Our exact solution I'm going
to plot in green, okay?
Let's suppose this is our exact solution,
okay?
All right.
This is our exact solution and what we're
talking of doing here is following.
We are talking of we're talking
of looking at these nodal points.
And identifying the value of
the exact solution at the nodes.
Okay?
And then we're talking of so, so this, so
this is u, the exact solution, okay?
All right, and what we are,
what we want to do now is
identify the u tilde h.
Okay?
U tilde h, remember, is a member of S h,
right, which means it
inherits the representation
that we get from our basis functions.
If we are working with linears here,
u tilde h can at,
can at best be linear for each element.
However ,we want it to be such that
actually hits the exact solution
at the nodes.
Okay?
For many of the other elements it's,
it's not going to be very different from
the exact solution because perhaps
the exact solution is not too different
a linear, sorry, it's not too different
from a linear solution elsewhere.
Okay.
You see that there is a difference, right?
See, this is our u tilde h.
And this is what we mean by saying that
u tilde h is nodally exact, right,
it's exactly the nodes.
It hits the exact solution in the nodes,
but it is a member of S h, so
over an element in this particular case it
is as, at, at, it is, it is, it is linear.
Okay?
And let me go back to my red here, okay.
So this, this is just plotting
up a solution up here.
All right?
Okay, so this is what we
mean by saying u tilde h is nodally exact.
Such a function, u tilde h,
is what is often called
an interpolate of the exact solution.
Because what we are doing is taking
the exact solution at the nodes, and
then using the basis functions to
essentially interpolate between those
exact nodal values to get u tilde h, okay?
U tilde h is nodally exact and
is also called the interpolate.
Okay?
All right.
Okay, and just remember that u
tilde h belongs to S h, right, and
that's re, reflected in the fact
that it is linear over elements.
In this case we are considering linears,
okay?
So we're considering, in this particular
example only linear basis functions.
All right, okay.
So this is what we have.
Now, here is a result.
The theory of Sobolev spaces, right,
or, or mathematics of Sobolev spaces,
functional analysis on Sobolev spaces,
gives us the following so
called interpolation Error estimate.
In Sobolev spaces.
Okay, the interpolation error estimate
is for the following quantity.
It holds for the m norm,
the h m norm of the difference
between this interpolate and
the exact solution.
Okay?
It's important to note why this is
called the interpolation error, right?
So the term on the left is
my interpolation error.
Interpolation error is this, okay?
U tilde h minus u.
The reason it's called interpolation
error should be pretty obvious, right?
And what is that?
Think about it.
What it is saying is that, suppose you
had the exact solution, right, and
at the nodes you were to
hit the exact solution,
which is what U tilde h does, right?
But then because of your choice of a
finite dimensional basis over the element,
you are now interpolating
from that exact solution.
But since you are interpolating
with a finite dimensional basis,
you fail to hit the exact solution over
an element, right, most prominently here.
Therefore, this difference that
we are looking at is indeed
the interpolation error, okay?
All right.
So the Sobolev, estimate here, and,
and I'm, I'm just going to state it,
we're not going to prove it,
is the following.
It is that this interpolation error
estimate is bounded from above
by c as a constant.
h e is our element size.
Okay, and in these error estimates
the we assume, or the error estimates
have been derived for the case in which
we have a uniform element size, okay?
So the c h e to the power
of a number alpha,
which I will define in a little bit,
okay, c h e to the power alpha
times the h r norm of u, okay?
So what do we have here?
U tilde h minus u is
the interpolation error.
It's the error incurred by the fact
that we're using finite dimensional
basis functions even if we had
the exact solution at the nodes, okay.
H e, as before, is the element size.
c is a constant.
r is what is called the regularity
of the exact solution.
Okay?
So note that when we're talking
about u r the r norm of u,
we're seeing that yes, if we were to
square integrate the exact solution and
also square integrate its derivatives up
to r, right, up to its rth derivative,
we would get a quantity which is bounded,
right, and that is the r norm.
Okay?
All right, so
let's just recall that this quantity,
the r notm of our u Is
a measure of smoothness,
of regularity, really.
r comes from regularity.
Right, which is really
a measure of smoothness
Of u, all right?
If you have the r norm of u, it means that
you can take up to r derivatives of u and
still have that quantity bounded,
which gives you some sense
of how small the u is, okay?
All right, the next thing I need to
tell you is about what alpha is.
Right, alpha is the exponent in to which
h e is raised, right, the power of h e.
Alpha satisfies it,
it's an exponent, obviously.
And it satisfies.
The following condition.
Alpha is equal to the minimum of
right, k plus 1 minus m,
and r minus m.
Well, what is k now?
k, k is the order of the polynomial
order of our finite dimensional basis.
k is the polynomial order of
the finite element, sorry,
finite dimensional basis.
Okay?
All right.
Let me do just well, let, let,
let me rewrite the result here, okay.
So that is h minus u interpolation error.
We are computing the m norm of it, right?
And this is lesser than or
equal to c h e to the power alpha.
Okay, so what this is telling us
is that supposing in most cases,
let us suppose that we have
a solution just to fix ideas.
Let's suppose that our exact solution
is such that we can actually take
derivatives of it up to a very high
order and you know, we're, we're able
to take derivatives of it to a very high
order, which means it's very smooth, okay?
What it means is that in this definition
of alpha, r is going to be very big, okay?
So even though we want to take, so, so,
we, we are going to take m derivatives of
the solution and we are interested
in knowing what the m norm of it is.
If r is very big,
what it says is that the order of our of
our exponent here,
the size of our exponent is controlled by
the polynomial order of the basis
functions that we have.
Okay?
Now why does this matter?
The question is, what happens with our
interpolation error as we refine the mesh?
Okay, if r is big enough,
then alpha essentially
reduces to k plus 1 minus m, right?
So if r is large, okay, which means
the exact solution is very smooth.
Right, we have a nice,
well-behaved problem.
Okay?
In this setting what we will
see is that alpha being
the minimum of those two quantities,
is indeed going to come down
to k plus 1 minus m, okay?
And therefore,
what this thing is saying, then,
is that our result is that the,
the interpolation error.
The m norm of the interpolation
error is lesser than or
equal to c h e to the power k plus
1 minus m times this r norm of u.
But then if u is very smooth, we expect
that the r norm is not very big, right?
We won't get a very large number
when we integrate the functions,
to integrate the function and
its r derivatives up to r, okay?
But that tells us then that
now as as h e tends to 0,
right, if our quantity k plus
1 minus m is greater than 0,
what happens with the interpolation error?
What happens with the norm
of the interpolation error?
It vanishes, right?
Okay?
It vanishes at the rate k plus 1 minus m,
right?
So it tells us that any nor, that,
that this h, sorry, this m norm that
we want to compute of the interpolation
error also tends to 0 at
the rate k plus 1 minus m.
Okay?
So as we refine the mesh,
as we make h e smaller, right,
remember h e going to 0 means we
are looking at mesh refinement.
We are going to smaller and
smaller elements.
Okay?
So what the Sobolev estimate tells us is
that yeah, as you refine the mesh eh,
if you were to look at this
interpolation error it will vanish,
provided, provided what?
Provided that number is greater than 0.
How can we make that
number greater than 0?
We can make the number greater than 0 by
taking a higher polynomial order, right,
or making sure that we are not taking too
many derivatives in computing the norm.
So we don't, if we are not looking for
very high m norms, right,
we're not looking for taking many
derivatives when we compute the m norm,
right, then this holds.
Right, and
this is an important thing to know.
This, this property, note we should
note that this property comes to us
directly from the Sobolev space.
Right, we're not saying anything yet
about the finite element solution.
This result is a property
of our Sobolev space
S h.
Okay?
How is it a property of
our Sobolev space S h?
Well, that's what determines
the polynomial order k.
Right?
That is determined by,
by the space we have picked for
S h, right?
So it says as long as you pick that to be
high enough, and as long as you're not
looking for too many derivatives in this
interpolation error it does converge.
Okay?
No talk yet
about the finite element solution.
Okay?
That will come next.
And that will come in the next segment.
