What I'd like to do in this segment is talk about how we actually carry out
integration of the different terms that show up in our weak forms.
And I'm referring specifically to the fact that when we carried out the formulation
for linear and quadratic basis functions.
We explicitly carried out certain integrals which we could do by
the way because we assumed that our coefficients of the problem,
our, our coefficient e if we were talking about elasticity,
or it would be like, it would be ended something like a conductivity,
if we were talking about some other problem.
That coefficient was uniform over the domain.
We also did the same thing for the forcing function, right?
The distributed forcing function f we say was uniform over the domain.
It made our integrations fairly easy, okay.
However, that's not always the case.
So we will need to develop the ability
to carry out numerical integration.
Okay.
It's needed if the coefficient in our problem,
such as E our forcing function f.
And other quantities, maybe the area a, and
so on are complec or let,
let me see, complicated functions, right, we're not really talking about getting
into the comple, into a complex space here, but, but,
but we need them, however, of e and f and so on, are complicated functions.
All right, functions of position x.
Okay?
All right we could also need them if
the sort of basis functions we need are very high order.
Right. And sometimes integrating them or,
or if all of those basis functions are something more complicated than simpler,
simple sort of polynomials we're using here.
Okay, all right
also a complicated
Basis functions
are used, okay?
Now, so develop numerical integration for this.
In particular,
the approach that we will take is what is called Gaussian quadrature.
Okay? We will take, we will consider Gaussian
quadrature, which can be shown to be optimal for polynomials.
Okay by optimal we mean that it's possible to integrate certain types of polynomials.
Polynomials exert an order exactly, okay?
[INAUDIBLE] There's a systematic rule about how we go about doing this.
All right, so let's see what we need.
If you look at to any of our integrals either, either the integrals going to
the stiffness matrix or into the for, into the distributed forcing function,
you will see that we always are faced with the task of integrating over minus 1 and
1, a function of the form g of z dz.
Okay.
The whole business with any sort of numerical integration,
any sort of quadrature is the following.
We replace it with a sum, okay?
Over L.
Going from 1 to mint, where n int simply means number of integration points, okay?
We replace it with the sum where every of term in that sum,
in that series if you like is in that finite series, is g
evaluated at a certain value of c indexed by L.
Multiply it by what we call a weight for L.
Okay?
This is the general form of a quadrature rule.
Alternately and numerical integration rule.
Let me say a few things here.
N sub int is the number
of integration.
Points.
Okay?
See L is an integration point.
Okay.
Wl is the corresponding weight.
Weight ascribed.
To the integration point.
Okay here of course
L equals 1 to an int.
And here, too.
Okay?
Now, essentially the way to understand this perhaps is to say that,
well, if this is what we're trying to do, if this is c,
and this is our function g, and maybe g looks like that and
we are here between minus 1 and 1, right?
What we're trying to do here is pick certain points z,
that is zL, one of those zL's, we look at the value of g, right?
And that value of z, right?
So this is g at z sub L.
Okay?
And we are basically forming that sum,
like I wrote on the previous slide.
Okay. We're giving some certain,
certain weight to the value of G, X, E, L.
All right?
Okay.
And then there are just rules, right?
Before we get to those rules let's just see one more thing.
The weight wL are such that some over L going
from 1 to n int, wL has to be equal to 2.
Can you think of why this might be the case?
Why do we have this requirement?
It's because
If g of z is a constant,
we know that integral minus 1 to 1 gz,
dz, is equal to twice of
that constant, right?
Right?
Well, this is satisfied automatically
by saying that what, when we sum over L
g of zL, wL, if that is a constant, and
this rule for wL is satisfied, right?
For the wL to satisfied we get back twice of the constant.
Okay, so we ensure that at the very least we can integrate constants exactly.
All right, so
Okay.
All right, here are the integration rules, okay.
So n int equals 1, okay?
For this rule, z1 equals 0, right?
And w1 equals 2.
For n int equals 2.
We have z1 equals
minus 1 over root 3.
W1 equals 1.
Z2 equals 1 divided by root 3.
W2 equals 1.
Going on, n int equals 3 as z1 equals,
minus 3 over 5 square root.
W1 equals 5 over 9,
z2 equals 0.
W2 equals 8 over 9.
Z3 equals 3 over 5,
square root w3 equals 5 over 9.
You can check that for
all three rules I put down the sum of the w's is always equal to 2.
Okay, these are all rules from Gaussian quadrature.
Right, and it goes on, right.
We can right out rules for any order of n int.
Okay, I mentioned that Gaussian quadrature is optimal.
It is optimal in the sense that a Gaussian quadrature rule.
With n int points, right,
and n integration points.
Exactly integrates.
All polynomials of order.
Order lesser than or
equal to twice n int minus 1.
Okay.
So, if we go back and look at
n int equals 1 integrates linears exactly.
N int equals 2 integrates octo cubics exactly.
N int equals three integrates octo pentics.
All right. And so on.
All right.
That's what we need to save for numerical integration and
we'll end this segment here.