Okay, so we are going to continue with our development of the Galerkin weak form.
[COUGH] For this [COUGH] one-dimensional linear elliptic PDE.
Let's begin by recalling where we were.
So we've got as far as here.
If this is our domain of interest, or this is the physical problem we're looking at,
we have this
bar we have our x-axis,
positions L, 0, 'kay?
And this is our domain omega.
Without those endpoints.
What we did already is to partition this by the use of
points that we're calling nodes.
We've partitioned it into subdomains,
each of which is denoted as omega sub omega e, 'kay, indicating an element.
The nodes that we have used are labeled x1, x2 and so on.
In general, this would be the point xe and that could be the point xe plus 1.
The point with x-coordinate,
x equals L is node, N number of nodes,
N nodes and that indicating number of nodes.
And we also observe that the number of nodes
is equal to our number of elements plus 1 in one dimension.
Okay, in one dimension, also, for
the very specific manner in which we are partitioning omega for
this simplest of finite element examples, 'kay, simplest of finite element cases.
Okay, this is as far as we we've got,
and what this allows us to do is to write the Galerkin weak form.
As follows.
We can now write it as a sum over e,
going from 1 to a number of
elements open integral over
omega e of Wh,x sigma hAdx
equals sum over e going from
1 to number of elements,
integral over omega e WhfAdx plus Wh at L,
which of course is our
last node, times t,
which is the traction times area.
So this is our Galerkin weak form.
Now observe that each integral really is
redefined in this expression as an integral
over the relevant element subdomain.
What we need to do to proceed is to talk of how we're going to write out
e representation for our trial solution.
Right, our trial solution of x and the waiting function.
Okay?
This is what we need to do.
Now, in doing this observe that because of the fact that
we're able to use this decomposition of our integrals over
these element subdomains, we really only need to worry about how
to carry out this representation over an element subdomain.
Okay?
So, let me state that.
Need to represent and wh over omega e,
right, where, of course,
e goes from 1 to number of elements, right?
So we have the advantage of
needing to focus only upon what often gets called a local representation here, right,
local being local to the element, okay, all right?
Okay, so let's look at how we can go ahead and do this now.
So let's then focus upon
a single element.
That is xe, that is xe plus 1.
And omega e.
Okay, all right, so it's all with this subdomain that we need to write out our
trial solution and waiting function.
The way we can do this now
is to define what we will
call local basis functions.
We define local basis
functions on omega e, okay?
All right.
And what you are going to observe is as you may expect
there is going to be a finite number of these local basis functions, okay?
So we define local basis functions on omega e.
And let me also state that we
will define a finite number
of basis functions.
[SOUND] Because we have the advantage of working just locally over the element's
subdomain, we're going to focus on this element's subdomain,
define a finite number of basis functions over that subdomain.
But then, if there is a finite number of basis functions over a single subdomain,
and there is a finite number of subdomains, what we are going to observe,
of course,
is that there is a finite number of basis functions over the domain omega.
Okay? So, we will have a finite number of basis
functions over omega e and therefore.
Over omega itself.
So, we have a finite number of basis functions.
Naturally, of course, our weak form is going to be finite dimensional, right, so
that, that's the connection, right?
So when it comes down to defining a finite number of basis functions we, we
are assured of having a finite dimensional weak form, the Galerkin weak form, 'kay?
This is how we want to think about.
All right, so let's talk about what these basis functions are going to be.
We are going to take in this initial entry-level,
very first presentation of a finite element method, we are going to take
the simplest possible basis functions, 'kay?
We have an element, and it is by choice, by the way,
that I have chosen an element to have only two nodes on it.
We are going to use those two nodes to write out to expand fields
of interest over the element, in terms of just two basis functions, okay?
Okay, so what we are going to do is the following.
We will define two basis functions.
Over omega e, 'kay?
These basis functions are going to be polynomials.
Okay?
And since we are in one dimension,
if we have two basis functions and the basis is complete,
it follows that we are talking of representing
functions as what type of polynomials?
Think about that.
If you have a complete basis, which you're going to use just two basis functions in
one dimension with polynomials, what kind of polynomials are we able to represent?
'Kay, let me just add on that bit about complete basis.
Well, you probably got the answer.
What this implies is that we are essentially going to be representing
functions here as linear polynomials.
On omega e, okay?
This is the simplest sort of polynomial basis function we can a,
adapt, in the most standard type of finite element method.
There are finite element method in which we can even use constant polynomials they
need a little more work than we need to do, than we should be doing at this early
stage of developing the finite method, so, so we go with linear polynomials, 'kay?
All right let's actually write, write these down now.
So the way this works is, is the follows graphically, this what we will do.
We have omega e and I'm going
to write the noted coordinates here.
Our basis functions are the following.
They're linears, right?
That one and that one.
This is N1 for obvious reasons,
that we're going to call N2, okay?
So, N1 is a function of position and
N2 is also a function of position, okay?
With these, what we will be able to do now is to write
out our finite dimensional trial solution, okay?
And for the very first, just, just for this case, I'm going to specify that
this is the finite dimensional trial solution over element e.
'Kay, restricted to element e, and obviously it is restricted to element e,
because it is over element e that we've defined our basis functions.
Okay?
So, the way we write it out in this case is
sum A going from 1 to
number of nodes in the element.
N sub n sub e, is for number of, the capital N is for number,
the little n is for nodes and e is elements, okay.
So we have two nodes in the element, so n and e is equal to 2, 'kay?
So the way that we write this is then NA functional
position times some degree of freedom that has been interpolated.
And we will write that degree
of freedom as d A sub e.
Okay, which in detail is simply N1,
functional position times a degree of freedom,
sorry, that degree of freedom should be 1.
Degree of freedom 1 corresponding to element e,
plus the basis function 2, function of position times
degree of freedom 2 corresponding to element e, okay?
So in this what we
have is that NA of x is
the basis function,
all right, and
dA of sub e is the degree
of freedom, 'kay?
'kay.
Also this is NA is basis function A, where A equals 1,2.
And dA sub e is degree of freedom
A equals 1,2 on element.
E, okay?