All right, so that's word making as a remark.
And that remark is that
the local definition
of basis functions.
Leads to global
basis functions.
Okay?
Global basis functions associated
With each node
Say xe, all right?
So, if we look at the global basis function associated with node e,
that global basis function has compact support.
Right?
In elements omega e-1 and omega e, sorry,
not omega e+1 it should be just make a e in this case.
Okay?
Now the fact that this global basis function is associated with the elements,
adjoining a given node will be observed in higher dimensions as well.
Okay?
Because in higher dimensions, it won't be just to the right and left.
There would be neighbors in other directions as well.
Okay?
All right.
Fine.
So this is another property of the types of basic functions we are defining,
so now with all of this in place we have a clean way to write out our trial
solution as well as our waiting function, okay.
And it's useful now to go back and recall from the types of
integrals that we need to evaluate in the Galerkin weak form.
Just what we need to do with these finite dimension functions.
Okay?
So let's recall.
So the kinds of integrals we need to assemble come from the fact that we have
an integral, sorry,
we have a sum over e of an integral over omega e of a function of this type.
Right? The gradient of our waiting
function x sigma h Adx equals
integral, sorry, sum over e,
integral over omega e, wh f.
It's worthwhile remembering that f is a function of as defined with respect to x,
which turned can be parametrized by c, right, using our bi unit dimension.
Okay, this x Adx +
wh L tA, I'm sorry,
there is no dx there.
Okay?
So these are the integrals that we need to compute.
Using these finite dimension functions.
Now, one more thing I need to state as we go on, okay.
And that is the following, an important note to make is this, okay?
Consider element one and omega e,
for e=1.
Okay.
If we have that as our domain and
let's suppose that this is omega 1, okay?
The very first element.
We know that this point here is x1, which in this case,
because of the way we are setting up the problem, is the point 0.
Right? And this is element,
this is global node number x2.
Okay.
Recall that here we have a Dirichlet boundary condition.
Right, at x=0.
So out here we know that uh=0.
Right?
It's equal to u nought but then actually let's just leave it as u nought.
I'm sorry, I'm getting ahead of myself by seeing that 0.
That's u nought.
Okay?
And yes we know that if we're thinking of this bar,
we're thinking of it as an elastic bar.
And if it is fixed on the left that u nought will indeed be 0.
But for now let's just leave it as u nought.
Okay.
So that part is straightforward.
What about wh there?
What should happen with the waiting function at that point?
Because it corresponds to the Dirichlet boundary condition.
Right, there's something special that happens to it.
Think about it.
Yeah, wh goes to 0, okay?
Always.
Now it does not matter what the actual value of u naught is.
I know I got ahead of myself a couple of minutes ago and
said that u nought is equal to 0.
It often is.
Even if it were nought equal to 0, the fact that you have a Dirichlet
bonded condition there implies that wh has to be equal to 0.
Okay, that's how we've constructed our rating function space.
What this tells us is that as far as the field wh is concerned within
this element, we only truly need that basis function.
Okay.
Because we are already going to make sure thus in fact,
choosing just a single basis function allows us to
ensure that wh does indeed vanish at x=0, okay.
And in particular we choose only N2 to make the basis function there.
Okay?
So, what this says is that now if I
just pull out that element, okay,
now we go on because we know that at wh at 0,
and x=0, has to be equal to 0.
What it tells us is that the only basis function we need in that element is N2.
Okay, N2 of x is only basis function,
fn, short for function for
representing the waiting function, okay?
It's not the case for the trial solution.
Only for the waiting function, right?
Do we need a single basis function,
so N2 of x is the only basis
function needed for wh of x for e=1.
Okay?
That's something to observe here.
All right?
However, for any other element,
e=2, and so on, right up to
number of elements in the problem.
We have properly wh of x sub e
equals sum A going from 1 to number
of nodes in the element, NA (x) dAe.
Okay, for wh of x where e=1,
we don't need a sum, right,
we have a single basis function,
we just have that, and
I realized that I use d here.
We are using c to write out the degrees of freedom,
corresponding to the waiting function.
Okay?
So there's a difference in the way we choose to interpolate, or
to represent our waiting function, depending upon the elements.
Okay?
As something to think about,
what if we had another Dirichlet boundary condition in our problem,
what if we had a Dirichlet boundary condition on the right end, as well.
Then for that element also we would do the same sort of thing right,
we would use only a single basis function for that one.
Okay, so that's just something to keep in mind,
I won't develop that idea right now, we come back to it when we need it.
Keep it, that's just something to keep in mind.
Okay, what I want to do as well is look at
what we need for our finite dimensional functions, right?
Let me just do this for you.
Just go back to this previous slide and recall that when we look at this
finite dimensional weak form, that's written just below the word recall.
You observe that we have
To compute a gradient of our shape function.
Furthermore, you observe that
sigma h is defined as e times x.
Okay?
So we see the need to compute gradients of our functions.
Here, however, we do not need anything.
We do not need the compute gradients, right?
And there, we just need to evaluate the function at L, okay?
At any rate, what this tells us is that we need to compute.
Compute or calculate wh,
x for the weak form.
And we also need to compute x and
in particular x is needed,
for computing the stress sigma h.
Okay?
So, how do we go about doing this?
Easy enough.
Let's just recall that, again, taking advantage of the fact that we
have a local representation, right?
So, we can look at our trial solution in element e, in a general element e.
And, so I will write it as, sorry, not a summation over e, but a summation over a.
Summation over a, 1 to a number
of nodes in the element, NA,
I will now write it as it is a function of c, okay?
We know that we originally defined it as a function of x, but
then, by going to our bi unit domain,
we observed that it's more convenient to write it as parametrized by c.
Okay, we have this, dAe, and we also have wh of e.
Function of c is also = to sum over E.
NAc, ce.
Right.
Well, we also need to compute those components.
Right.
And the way we can do that is just by computing NA,
x in both those expressions.
Right, because d, the dAe's and
cAe's are just degrees of freedom, right?
They do not have any dependence upon the position, right?
So we observe that in order to compute these gradients of the trial solution and
of the waiting function, we just go to our representation for
the corresponding functions.
We essentially need compute gradients of our basis functions, okay?
So how do we do that?
Well, we just invoke the change rule,
right, so sum over A Right.
So now we write.
NA, x as NA,c,
c,x dAe,
okay?
And this quantity is done in exactly the same way sum over A,
NA,c, which is easy to compute,
because we are indeed actually defining
the Na's as being parametrized by c, okay?
This time, c,x,
cAe, okay?
Now this is useful, right, because it tells us how
we can compute our gradients of required fields,
except for the fact that we may say well,
how do I go around actually computing that quantity?
Hardware, write the derivative of my position in the bi-unit domain
with respect to my physical coordinate x, okay?
Think about it.
We'll come to it, but we will do it in the next segment.