All right, we'll continue.
So what we did the previous segment was use this mapping,
between the parent and physical element sub domains to observe
that we have actually a convenient way of writing out our gradients.
Okay.
We'll continue now from we, from there and
consider explicitly the three integrals that go into our weak form.
Okay.
So, with that background now, we are getting into,
the assembling the integrals in the finite dimensional weak form.
Okay.
The integrals.
In.
The finite dimensional.
Weak form.
Okay, so, let's consider the first of our integrals, right, so.
We are going to consider this
one integral over omega e, w,
h comma i, j, h, i, d, v.
Okay?
We first invoke our constitutive relation to write this as integral over omega e.
W h comma i.
We pick up a minus sign here,
minus kappa i j, u h comma j.
D, v.
Okay, that minus sign just came from the fact that the flux is written as minus
the conductivity tensile.
Multiplying the, gradient of the field of interest.
Right?
The final solution field.
Okay, we'll take care of that minus sign later on.
Okay.
So we have this and now, essentially, what we are going to do is use what
we've just discovered about writing out the gradients of w h and u h.
Right?
Okay.
So, over element e this lets us write it out as,
integral over element, over omega e.
Right.
Of w h comma i, we will write as sum over a.
I'm not going to write the limits on a.
Right. We know the limits,
they go from one to number of nodes in the element.
Well okay, let me write them as always for the very first time and
then I'll abandon writing that.
Okay.
Sum over the nodes
in the element,
n a comma i c a e.
I put parentheses on that.
Kappa i j, okay.
Open another set of parentheses for
a sum over b, going from one to number
of nodes in the element, okay?
N v.
Comma j, d,
a, e, d, b.
And, we have a minus sign.
Okay?
And as we did in the one d case, perhaps it is useful just
to remind ourselves that, that is w h comma i, and that is u h comma j.
Okay, and now we, use our
discovery of how to write those
gradients of basis functions.
Noting that those gradients as expressed here are readings with respect to physical
coordinates.
Right?
Okay.
So this is now integral over omega e minus integral over omega e,
open the first set of parentheses.
Sum over A.
And I'm going to abandon writing the limits from here on.
Now, n a comma capital i.
Okay and recall that when I write n a comma capital i,
we are talking of taking derivatives with respect to, what?
Right.
With respect to the, coordinates in the parent sub domain.
Right.
This times c I
comma, little i.
Okay?
Multiplying c,
a, e, okay?
Capital I, j.
Open parentheses again,
sum over b, abandoning limits,
n b comma capital J.
C J comma little j d,
b, e, d, v.
And then note now that I was not quite as careful when I wrote that.
That's not a D, A, E.
That's a D, B, E.
Okay.
Right, note that what I'm doing
here is just, partial of c i,
with respect to x, little i.
Okay.
Same thing here.
Okay.
Partial of c capital J with
respect to x little j, okay?
And this is.
Partial of n a, the a basis function.
The basis function for node a, or the basis function which is one at node a.
With respect to c capital I, and this is
Okay.
Right?
Fine. So
we know how to compute these quantities now.
Right?
Okay.
So.
Let's move on, and actually let's make our expression here a little simpler
by observing that the degrees of freedom of course are independent of
position, and therefore they can be pulled out of the integral.
Okay.
So we take the step that we have taken before in the one d problem.
This is equal to minus,
now, sum over a and b.
Okay.
We have here c, a, e.
Integral over omega e,
n a comma i, c,
capital I comma
little i kappa i j
n b comma capital
J c capital J little j.
Okay.
All of this integral d v.
And just for complete clarity I'm going to put parentheses on this.
Right, on the integral.
And note that the whole thing is then multiplied by d b e.
Okay, some things to note.
Though, perhaps one way to think about this is that we have an integral,
right, and we have a complicated looking integrand here.
Okay.
Each integrand, right?
Has indices a and b, right?
And these indices a and b come from the particular
degrees of freedom over the element that we are talking about.
All right, so this would be, so the picture here would be,
let me draw it here.
The picture here would be
In the physical domain, we have our element and we have a node a.
All right?
Or b.
Okay?
The index a or b runs over those nodes, right?
It runs over the degrees of freedom that correspond to the nodes, right?
And over the others of course.
Right? It runs over all of those.
Okay.
So having chosen a particular degree,
a particular combination of degrees of freedom for a and b, right?
Okay.
For that particular combination of a and b we have the functions n a and n b, okay?
Now, inside of the, inside of our integrand.
Observe that there is another comp well there is another set of
multiplications happening and I'm trying to look for a different color here.
There we go.
Okay.
Using green.
Okay. Inside the integrand there is Einstein
summation convention applied all over the place okay so
there's a Einstein summation convention here.
Okay?
Here here, As well as here.
Okay?
Again, if you've studied continuum physics and done this in three dimensions,
you've probably seen this sort of thing.
Often.
Okay?
So, whereas, we have Einstein summation convention for summing over indices
that correspond to either physical or parent domain coordinates.
That's what the little i j and the capital i j refer to, right?
We have used an explicit sum.
There, to run over the degrees of freedom of the element.
Okay? I just want to make this little bit clear.
Okay, so if that part is clear,
we'll proceed and in order to proceed, we observe that.
The way I've written it out so far,
even though we've actually moved over to writing our you know computing
derivatives with respect to the barren sub domain coordinates.
Our integral is still over the physical element, right?
And that elemental volume here really defers to the physical element.
Okay? The next step we are going to take is to
convert this also into an integral over the parent sub domain.
Okay, so now we are properly changing variables into the parent sub domain.
Okay?
So we are really completing
change of variables.
To c.
All right?
And what that means is that we need to essentially convert the elemental volume.
Okay.
Now, here is our map.
Okay.
Okay. That's omega e.
And this is constructed of course from omega c.
Okay.
And we have our map.
X of c, right?
So for an arbitrary point, physical point little x here,
and an arbitrary point c in the parent sub domain.
Okay, what you want to think now about is how do we represent the elemental
volumes, right?
And in order to do that consider the fact that we have here.
An elemental volume.
Okay, that's my D, capital V.
And here, we have an elemental
volume, that I'm going to
represent as d, v, c.
Okay?
D v sup c, just telling us that that's an elemental volume in the parent sub domain.
Now, a another result from the fact that we have this mapping and
that we also have the tangent map, J of c.
The result that follows from this is the following.
One can show that d
v equals determinant of J.
D v c.
Okay?
Again this is a standard result.
If you've ever played around with these sorts of configurations, can be proven and
it's conventionally done if you're studying kinematics of deformable bodies.
Okay.
But we are just going to use this result.
Okay.
We are essentially going to use this result in our, integral, right,
that we're working with.
This one, right?
We are going to try to replace that, elemental volume with this expression.
Right.
That is however, perhaps best done in a separate segment.
So we'll stop the segment here.