So if we were working in three dimensions, in all three, d equals 3.
And then, the measure of our domain, okay.
Our domain may be in gentle, be referred to as Omega, okay?
Omega to the power of 1 over d would be measure
of Omega to the power of 1 over 3.
And, this would be a.
Length.
Okay.
So this is how we would extend this sort of idea to multiple dimensions.
Okay, but then let's get back to our H1 functions.
So, v belongs to H1 on 0 comma L,
if the integral of the function square plus
now let me write it differently.
Measure of 0 comma L square v comma x square,
the whole thing, dx less than infinity.
So we require that the function itself, and its first derivative squared
integrated be bounded, and H1 there simply refers to the fact that we attempting
here to control not only the function itself, but also its first derivative.
Right. If we go on,
and want to control also the second and higher order derivatives,
then we would accordingly go to higher orders of H.
Right, we go H2, H3, and so on.
Okay, so now you observe that when we say that.
Now, when we see that belongs to Sh,
which where sh is equal to
functions belonging to H1.
When we say this, what we are saying is that is such that we expect that
is such that the function itself be bounded, okay?
Because it is square integrable,
the function itself as well as its derivatives, okay?
And as you can think, now as you can imagine now, we're saying something about,
about the nature of the displacement field that we're expecting to find to,
to find as our, our approximate solution.
Okay?
So, what we are saying here is
that and comma x are.
Square integrable.
Okay, let's look at functions that are H1.
Okay.
Okay, so examples.
Okay examples of H1 functions.
Okay, so are the pcv
of x equals constant,
no problem, okay?
If we say v of x equals
sum k equals 0 to N,
a sub k, x to the power k.
Okay?
This is also all right.
There is a third type of function that I want to consider and I could
write it out in form, int as it in, in the form that I've written out
the first two functions here, but let me instead draw it.
Okay? Let me instead sketch it, because it's,
it, it, will still convey the same idea.
Where as defining it would, would require us to define
things that we will need only later.
Okay, so let me suppose that v is of the following form linear,
maybe quadratic, linear again,
some other polynomial,
linear again, constant, okay?
This sort of function also belongs to H1
Okay.
It belongs to each one because clearly, the function itself, if you square it and
integrate it, is bounded.
At points such as this, you observe that v comma x, okay?
Is v comma x suffers a jump, okay.
Those double brackets there indicate a jump in a function, okay?
We know that there is a jump in the function there.
Because if we look at the derivatives, right, right?
We know that the derivatives are going to look like that or
the derivatives are going to be,
are essentially the tangents to the two sides of the curve, right?
Those tangents are not the same right?
And therefore we, we know that the function itself suffers a jump.
However, these continuous functions can still be squared and integrated.
Okay, in fact,
we consider in exactly such a case in, in the case of L2 functions, right?
We did consider the, the Discontinuous function itself, the Heaviside function.
And, observe that, yes, to belongs to L2.
So, Heaviside function, Discontinuous functions can be squared and integrated.
That's what happens with b comma x when it comes to defining a 2v in H1, okay?
Right.
However, what about a function, which has this form?
All right, let's do,
let's do something that looks pretty much like what we have there.
What about this function?
Right, as we saw in the case above we see that yes here,
the tangents are different, right?
So there is a jump in v comma x.
But out here, what we are seeing is that, v itself suffers a jump.
Okay, and therefore, what happens when we consider v comma x?
V comma x, we know goes as the a delta function,
which would be called at this point x not, right.
We know that d comma x is, is like e, delta of x minus x not.
Okay? And that's where the square root of integ,
integrability of the derivative of three breaks down.
Okay?
So in this case, we have v does not belong to H1.
Okay.
All right this is probably a good place to end this segment.
We've essentially gone over the definition of our very basic function spaces,
introduced these ideas that we, that we need here.
Ideas of square integrability of the functions, and of their derivatives, and
this idea that when we specify these functions to belong to certain spaces,
these Hilbert spaces L2 and H1 by the way are Hilbert spaces.
We gain control over the functions themselves, and their derivatives.