Alright, we'll continue with our development of
basis functions in multiple dimensions.
What we did in the previous segment was to, observe that the stensor product
of idea allows us to very conveniently construct, the higher order,
the functions in, in higher dimensions, using the functions in one D.
Okay, in order to continue and, and provide us with,
ourselves with everything we need to complete formulations in two and
three dimensions, there is one thing we haven't yet done.
Even for our 3D formulation of the linear elliptic PDE for
the scaler variable, and that thing is integration.
All right, we haven't talked about how we do numerical integration in
multiple dimensions, all right, that's what we will aim to do in this segment.
So the topic of this segment is numerical integration.
N again.
One through three dimensions.
Well, let me write out the word.
The approach we will use is conceptually similar to what we did in previous
segment.
Right? Side
of forming the extensive product formulas.
Okay, and to start out let's go back to one d.
Okay, in one d this sort of situation we are confronted with is the following.
We need to integrate from minus one to one.
Some function,
say G of C1, D, C1.
Okay. Every single integral that
we need to evaluate in our finite element formation reduces to this.
Okay.
And of course this is the form that's applicable to every component of,
of a matrix.
Whether it's our stiffness matrix, or conductivity matrix, or, or
our forcing function.
And so on.
Okay? And you recall how we did this in one d?
Right? How did we go
about doing numerical integration in one d?
Remember?
Yes we did it with numerical quadrature, right?
Which basically says that okay let me just do a sum instead of the integral.
Okay?
I'm going to do a sum with l going from
one to number of integration points Okay?
I am going to evaluate g xi 1
at all these different values of l,
all right?
And once I do that.
I am going to multiply each of those function evaluations with the,
a weight, w L, okay?
So, what we have here are quadrature points.
P-T is short for point and that is a weight.
Okay, that's the general approach for numerical quadrature,
of course we use Gaussian quadrature.
Okay?
And Garrison Quarters said that well if you choose
to do it with a single integration point.
Okay?
You would pick c l equals zero and
w l equals two.
Right.
Okay and the reason for that weight of w l equals two is something that we understood
from how we would integrate constants.
All right?
And this went on, right?
So, for instance, if you went as far as n int equals 3.
What we would see, it.
What we would observe is that we would get sorry.
I should have paid attention to my
notation here,
this will be c one, here right?
Okay. So if we were doing an n equals three,
then c one, one at n equals one
would be minus square root of three over five.
W one would be, five over nine.
C one two would, by symmetry, be the point zero.
W two would be eight over nine.
And, c one three would be the point,
root three over five.
And the weight, you may remember, or
you may figure it out from symmetry, would, again, be five over nine.
Okay? And we saw that you know,
Gaussian quadrature we said that a rule of
that n int rule, right,
or n int point rule integrates.
[SOUND] A polynomial of
order two, n, int,
minus one exactly.
Okay? So, we'll recall all this.
What we're going to see is, that, when we go to multiple dimensions,
it is essentially a tens of product idea.
Right? And already,
tens of product rule that calls on this basic idea.
All right.
So let's go ahead and see how that works out.
Okay.
All right, so in 2D.
In 2D as you may imagine you need to
integrate terms of the following type.
Need to integrate, okay?
Minus 1 to 1.
G, generally a function of c1 and c2.
d c 1 d c 2. Okay, and these limits, let's say,
for c 2, and this is for c 1.
All right.
It's really very straightforward.
Let's suppose we first do the integral over c 1, okay,
and we're going to do numerical quadrature, okay?
So, right, so I have numerical
integration or quadrature.
Right?
So what numerical integration says is that well I can just break this up.
I can write this as an integral
c2 equals minus 1 to 1 and now for the inner integral.
Right?
Over c1 I will first use quadrature.
Okay? So what this says is that I'm now
going to talk of doing a sum let's say L1 equals
to number of integration points along that dimension.
Okay?
All right, so we have L1 to number of integration points.
We have G, C-1, L-1.
Alright, you pick quadricant points only, along the C-1 coordinate.
And, you leave the C-2 as
the continuous coordinate.
All right. So, you, you retain, therefore,
the elemental d x e 2, okay?
Actually, let me write that.
We're first doing the integration for expectancy 1, right?
So then, we just multiply this by w and 1, okay?
So, what I have here in parentheses is the result of having it integrated over C 1.
Right, yah we did numerical integration over at c 1,
but nevertheless there we have it.
Okay, d c 2.
Right.
And then we come back and do the same thing over at c 2.
All right.
So numerical integration would give us this, and
then as another step we would get now Sum
l 2 equals 1 to n int.
Sum l 1 equals
1 to n int.
G c l one.
Sorry.
C 1 l 1,
c 2 l 2, w l 1.
Wl2, and we're done.
Okay?
Now, of course,
this is, based on an assumption of symmetry in c1 and in c2, right?
Symmetry in the sense that we are, considering the case where we have
a polynomial of, perhaps, the same order in c1 and c2.
All right, that's why we've chosen the same number of integration points along
the two directions.
Okay, of course, this does not have to be the case, right.
We are always free to choose different number of integration points in
the different directions.
Right, if we know something about our polygon, right?
So I, I'm going to make this more particular by changing this to n one int.
Right, and this to n 2 right, it just says that we can use different number of
integration points in the different directions, right, so the remark here
Can use different number of integration points.
Along c1 and c2.
Right, if you know something about the polynomials
that we are working with, okay.
And of course we would be,
we could use Gaussian quadrature in each direction, right?
And everything would work out.
The, the integration points along c 1 and c 2 would be the same as
the Gaussian quadrature points and so would the weights, all right?
All right okay, and here, what you will see here, of course, I'm not,
I'm not going to write it down, it's something that you can conclude easily by
going back and looking at the formulas.
What you will see is that, in,
in the two d case, the weights add up to what?
Consider what would happen if you had to integrate a constant.
Okay, so what happens here is that I just write it out as a sentence, here.
Sum of.
Weights.
Equals 4.
Okay and this comes simply because we're integrating over bi-unit domain, right?
And the area of the bi-unit square in 2D is of course 4.
Okay, so that's where that's from.
Alright, now we can extend this to 3D, right?
And this would be completely clear how to do it in 3D.
Right?
In 3D we're trying to integrate something of this form, right?
We're trying integrate c 3 equals minus 1 to 1.
C 2 equals minus 1 to 1.
C1 equal minus
one to one G of
C1 C2 C3,
D c1, D c 2 d c 3.
All right?
I'm straight away going to just write out the formula.
Our numerical quadrature formula is sum l three equals one to n three int, right?
Number of integration points in that c 3 direction.
L-2 equals one to number
of integration points in the, C-2 direction.
L-1 equals one to number of integration
points in the C-1 direction.
G, C-1.
l 1 xi 2 l,
sorry, xi 2 l 2,
xi 3 l 3, right?
Times. wL1,
wL2, wL3.
Okay?
And again, we could be using Gaussian quadrature points and
weights where the points are again simply determined as the For
the points as chosen in the one d case.
So this is how we would construct our, we would,
we would actually evaluate all our, our integrals.
Alright, so, so this is for the generally case, sorry, sorry, not the generally case
but the case where we are looking at either a bi unit domain in one d.
Two d or in 3 d.
Okay, and that bi unitness is reflected in the limits on these integrals.
All right.
We'll end this segment here.
When we come back, we will look at a, different type of basis function.