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Till now, we have looked at the one-dimensional
finite element problem using linear basis
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function. So, if you had this domain, what
we have done is, we had partitioned the domain
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into elements, we had corresponding nodes
and then called the element, we went ahead
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and constructed approximation using these
functions phi i, which are linear. Also the
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question - is that all we can do or can we
do more than checking such linear approximation?
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The answer is yes. So, what today we are going
to look at, is higher order approximation
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for the finite element solution.
So, what is the notation? Let us first introduce
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when it is a linear approximation, then we
will represent it by P equal to 1, where P
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we call has the order of approximation. Now,
what we have implicitly assumed, I will assume
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it throughout this course, unless it is specified,
specifically, that this order of approximation
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is the same in all the elements, that throughout
the domain, we are going to use the same order
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of approximation; one can always use different
order of approximation, in different regions
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of auto mode, that case we will not handle
now.
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So, let us now see, how this order of approximation
can be increased. So, the next example that
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would we have to do is, P equal 2; that is
the order of approximation is quadratic. For
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this, let us see, what is it that we have
to do, as far as the approximation is concerned.
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So, for this, you take the same domain that
we had taken earlier, we send n load p, this
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is x equal to l, this is x equal to 0 and
some distributor forces f x. For f x, let
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us also takes EA is equal to constant in the
domain, that need not be true. So, using this,
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given this data for the problem, let us now
go ahead and do the approximation. So, how
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are we going to do this? What we need to do
now, is to first break the domain into element
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as we have doing till now.
May be I will take 4 elements here. So, these
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partitions that I have drawn, using these
end vertical lines, these are our elements.
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So, this is element 1, this is element 2,
this is element 3 and this element 4. So,
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in this element, in each of this element,
what I am going to do is add an extra point.
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So, in each of the element, I am going to
add an extra point. This point is located
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at the center of the element. So, now we are
going to do this, the locations of this point.
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So, now, instead of having 1, 2, 3, 4, 5 points
for the 4 elements that we have drawn, now
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we have 9 points. So, we will give them name.
So, this is point x 1, this is x 2, this is
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point x 3, x 4, x 5, x 6, x 7, x 8, x 9.
What you should note, is that these extremities
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of the element are given by the point x 1,
x 3; so, these are the extremities of element
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1; extremities of the element 2 are the point
x 3 and x 5; extremities of the elements 3
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are the points x 5 and x 7; extremities of
the element 4 are the point x 7 and x 9. These
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points x 2, x 4, x 6, x 8 are generally called
mid time nodes or mid edge nodes. So, we will
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give a name to them; these are
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mid side nodes and the points x 1, x 3, x
5, x 7, x 9 are called these n vertices or
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simply the extremities of the elements. So,
this point becomes end nodes or elements nodes
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or extremities of the elements. Now, what
do we do next?
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So, what will do is - over this point checked
that we have created, we are going to use
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it somehow to create our basic function, which
are quadratic in nature. So, I have this point
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set, I am going to construct the basis function,
which now should be quadratic in the elements.
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So, first, I am going to draw; then, we will
talk about how to construct mathematically.
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So, while drawing, what had we said about
the linear function, what we wanted these
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functions to do, what should have a value
1 at one of the nodes and 0 at all other nodes;
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the same principle we are going to use as
far as construction of the quadratic functions.
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So, let us say the function phi 1 that is
corresponding to the nodes x 1 or the point
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x 1. The phi 1 will be 1 at the point x 1
and 0 at all other points; so, if I want to
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make that phi 1 as this function. Similarly,
I would like to make phi 2; phi 2 will be
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a function, which is 1 at this point and the
point x 2, and 0 at all other points. So,
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what we see is that we have defined it in
a local region; that is, phi 2 is only defined
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in the first element.
So, if I can draw that element here, this
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is element I 1, this is element I 2, this
is element I 3 and this is element I 4, then
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I can draw phi 3. According, phi 3 should
be 1 at this point and 0 everywhere else.
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So, by the same token that I have done phi
3 like this, then phi 4. phi 4 should be 1
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here, 0 at all other points phi 4. Similarly,
phi phi phi phi will be 1 here; 0 will be
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where are... Similarly, if I want to use,
phi 6 will be this, phi 7 will be this, phi
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8 will be this and phi 9 will be this.
So, what we got is phi 2 is here, so naming
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wise, this is phi 3, this is my phi 4, this
my phi 5 phi; sorry, I made a mistake here.
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So, I have to go back and change my naming
convention. So, before phi 2, this is going
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to be phi 3, this column is going to be phi
4, this one is phi 5, this one is phi 6, phi
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7, phi 8 and then phi 9.
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So, here, I have the 9 phi, which I have constructed;
they look quadratic, we have to also give
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an expression to them mathematically; why
are they quadratic? Because they have a value
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1 at one point, and 0 at two other points,
and then we extend it as 0 everywhere. So,
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they should be quadratic; that means they
look like that. So, in some will this function,
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how am I going to write my 500 ml solution.
So, u F E of x, we will use our earlier notation,
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is it is going to be, u is the 9-term solution
u 9 of x. This is equal to sigma i is equal
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to 1 to 9 u i phi i.
So, this is a pixel optical. So, now, what
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is this, that the function has to satisfy
in order to be admissible basis function?
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The first thing that we have done till now
is that, this function phi i should be, it
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should be linearly independent; second thing
that they have to satisfy is that they should
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be complete; and third thing that we have
done by construction is they should local
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support, that is, they are only defined in
one or two elements and in the rest of the
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element there, 0.
So, by construction, we see that this property
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has satisfied linear independence; we do not
know this, we do not know. So, let us first
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make this function in such a way that they
are quadratic, linearly independent and complete.
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So, how are we going to do that? So to do
that, let us now consult this function element
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by element. So, if I look at element - a generic
element I k - if you look at the generic elements
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I k, what are the nodes that this elements
has or the point it will be? x 2 k minus 1
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x 2 k x 2 k plus 1. Similarly, which is the
phi, which is non-zero here? So, phi which
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are non-zero, here are
phi 2 k minus 1 phi 2 k plus 1.
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So, let us now go and transform this to our
element notation that we had introduced earlier.
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So, from the element notation point of view,
in the element, I will call the first node
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as x 1 of an element k; the second one, x
2 of an element k; and the third one is x
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3 of element k.
Similarly, the phi 2 k minus 1, in this element
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could become, would be called N 1 of element
k and phi 2 k
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will be called N 2 of k. Similarly, phi 2
k plus 1 will be called n 3 of k. And like
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in the case of linear approximation, in this
we want this function N 1 of k to be to have
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a value 1 at the point x 1 of k and 0 at all
other points. N 2 of k should have a value
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1 at the point x 2 of k 0; at all other points,
N 3 of k should have a value 1 at the point
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x 3 k and 0 at all other points. So, what
we can write in short is, N i k as the point
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x i k is equal to 1 and at all other points
j it is equal to 0. Set property is what we
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want to have for this function. So, if we
want to construct N 1 k with this property
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such that N 1 k is 0 at the point x 2 k and
x 3 k.
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So, how will I form the function N 1 k? This
should minus at the point x 2 k, certainly,
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this representation should have this form,
something into is there, then at x 2 k this
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expression goes to 0, it should also vanish
at the point x 3 k. So, it should also have
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this part sitting there. You see that this
expression vanishes at the point 2 and the
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point 3 of the element; now what else is required
at the point 1? This expression should have
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a value 1. So, what I will do is simply divide
by, what I have done in the bottom part, I
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have replaced x with x 1 k; you see what has
happened, this is a quadratic expression,
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it vanishes at the point where you wanted
to vanish and it has the value 1 at the point
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where we want it to have a value 1.
So, by the same token, can we define N 2 k.
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We are defining this thing only in the element
phi k. This will be equal to what? Very simple;
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it will be x minus, it should vanish at the
point 1, and 2 is going to be x minus x 1
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k x minus not 1 and 2, but 1 and 3 x 3 k,
and it should have a value 1 at the point
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x 2 k. So, it will be x 2 k minus x 1 k x
2 k minus x 3 k.
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Similarly, N 3 k by the same token, can be
written as x minus x 1 k into x minus x 2
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k divided by x 3 k minus x 1 k into x 3 k
minus x 2 k. So, we have constructed the three
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functions at the element level; this satisfy
the three requirements we have imposed as
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far as the values are concerned. Now, what
do we note? That if I look at the function
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phi 2 k minus 1 of x, where is this functions,
and this is for we have to differentiate between
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the two things, this function 2 k minus 1
corresponds to an edge vertex, that is a node
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which is shared by two elements; that is,
this corresponds to if k is 1, it is 1. So,
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it is 1, 3, 5, 7, 9 this function has to be
non-zero in the two elements sharing that
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node. So, this will be equal to N 3 in the
element k minus 1 for it will be N 1 in element
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k; hence, it will be 0 everywhere else.
So, what you said is this function, if I talk
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of this node 2 k minus 1; this is element
I k minus 1; this is element I k. So this,
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the function is given as N 3 of k minus 1,
which we now know how to construct and here
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N 1 in k.
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So, this we can construct, all the global
of basis function, odd ones using these N
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I k(s) defining the element; what about the
phi 2 k that is the even one? So, the even
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one which you see are defined only in the
element. So, the even one - the phi 2 k k
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is equal to 1 2, the number of elements, these
are sometimes called internal bubble functions
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and they are only defined over element I k,
that is, this is my element I k. So, this
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is by the 2 k node and phi 2 k is non-zero
in this element and 0 everywhere else. So,
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what we know now that phi 2 k is equal to
by definition N 2 of element k in element
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I k and 0 everywhere else.
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So, these, now you see that when we went,
we came from the linear to the quadratic.
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The construction has been become a little
involved; nevertheless, the steps that we
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are going to follow will be the same, irrespective
whether it is linear, quadratic, and as we
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will see cubic forth order or fifth order,
any order that we would like to have. Now,
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the question is that - are these functions
linearly independent? So, once I have defined
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these functions phi i, then if you remember,
we had written u 9 of x is equal to sigma
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i is equal to 1 to 9 u i phi i of x. How do
we check linear independence? If you remember,
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linear independence means that if this function
is identically equal to 0, then the coefficient
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u i should come out to be equal to 0, how
do I show that the coefficient c y r 0? So,
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forgot, let us see a very important property
of this function; what are these phi i? If
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you remember this phi i has a value 1 at the
node with respect to all the point, with respect
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to they have been defined, and at all other
points that we have put in the domain, these
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functions are 0.
So, if this function u 9 of x has to be 0
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everywhere, it has to be 0 at the point x
i also, is also equal to 0, which is equal
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to from the summation, what do we get? All
other phi i(s) will vanish, only the u i will
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remain. This will be equal to u i. So, if
I do this for all the 9 nodes, we are having
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the domain, what do I get that from this condition?
That u 9 is equal to 0 everywhere, we get
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that all the coefficient u i are 0, which
means, that the phi i, are... this will be
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called piecewise quadratic; that is, in every
element they are quadratic and linearly independent.
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What about completeness? So, for the completeness
- if we go, and what do we like, want to do?
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We would like to see, whether a function of
this type, if the function f of x, this function
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is given as the generic quadratic a 0 plus
a 1 x plus a 2 x square; if I have this function,
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then I should be able to represent it as the
linear combination of a phi i; that is, I
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will call it as alpha i phi i. Now, what is
this? Because, we are saying these approximating
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functions are piecewise quadratic for completeness,
we want the series represented by these functions
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to exactly represent at least a quadratic
polynomial, not more than that; so, either
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I have, if you see, this is, these are arbitrary
coefficient.
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So, this series has to exactly represent the
constant represented by a 0 or a linear a
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0 plus a 1 x or a quadratic, which is a 0
plus a 1 x plus a 2 x square. So, if it has
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to do it in the whole domain, then it should
also be able to represent a quadratic in an
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element, so in the element if I write... So,
instead of doing it in the whole domain, let
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us do it in the element. So, in the element,
if I want to write f x it is equal to, I will
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call it alpha 1 of k N 1 of k plus alpha 2
of k N 2 of k plus alpha 3 of k N 3 of k,
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and this, let say that the quadratic polynomial
that we have taken in the element, it is of
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this form beta 0 plus beta 1 x plus beta 2
x square.
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Now, by the various functions have been defined,
what will be the alpha 1 of k be? Alpha 1
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of k, where the equal to f evaluated at the
point x 1 of k; alpha 2 of k will be equal
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to f evaluated at the point x 2 of k; alpha
3 of an element k will be equal to f, evaluated
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at the point x 3 of k. So, with this representation
for the alpha, one should go and check that
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indeed, what we have done in the previous
stage is really... I do not know, whether
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this is going to be like this. So, the equalities
is question mark and what we can now check
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is that is not a question mark, but indeed
this is an equality.
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This is a simple algebraic exercise, and one
can show that, yes, indeed just linear combination
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of these three functions is able to exactly
use a quadratic polynomial defined in the
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elements and thus in the whole domain, which
means that this function phi i, the set of
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function phi i represented in the complete
set. So, this in the construction of this
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phi i is... one has to be very careful, one
cannot arbitrarily define this function phi
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i as we wish.
Now, if you see this point x, another point
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we have to note is that the point x 2 of k
that we have taken, x point x 2 of k, we have
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said it is at the mid side, it is not been,
it could be anywhere; but as the convention,
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it is taken at the middle of the element always,
but one should remember it need not be the
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00:31:56,869 --> 00:32:03,619
case, specially when we have solve element,
which we will be dealing with later on.
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00:32:03,619 --> 00:32:10,619
Now, that we have defined the phi i, what
next? So, once we have defined the phi i,
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00:32:12,009 --> 00:32:19,009
the next thing that we have to do is to go
ahead and find the coefficient u i, in order
183
00:32:20,629 --> 00:32:27,629
to form the solution to the problem. So, to
form the coefficient where do we start from?
184
00:32:31,179 --> 00:32:38,179
We start from the weak formulation, and for
the problem that we have taken, the weak form
185
00:32:40,220 --> 00:32:47,220
is very easy, it is x is equal to 0 to L,
it will be EA d u dx f v dx plus p into v
186
00:33:03,340 --> 00:33:10,340
evaluated at L.
Now, in weak form what do we do? We go and
187
00:33:14,899 --> 00:33:21,899
substitute, instead of u 9. So, this will
be replaced by u 9 and to form the i th equation,
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00:33:31,940 --> 00:33:38,940
what we do? We replace v with phi i. So, to
make the notations a little bit compact, this
189
00:33:46,429 --> 00:33:52,710
part as we have done a few lectures ago, they
are going to represent as the bi-linear form,
190
00:33:52,710 --> 00:33:58,499
as I said it is linear in u and v. So, it
will be called as the bi-linear form given
191
00:33:58,499 --> 00:34:05,499
by this compact notation b u v; and this side,
this we are going to call by F of u, which
192
00:34:06,320 --> 00:34:13,320
is the linear function and given by the name
F of u, because here as the number of functions
193
00:34:16,600 --> 00:34:21,909
increase, the algebra get tedious. So, we
would like to be compact only for that reason,
194
00:34:21,909 --> 00:34:28,909
we are introducing these things. So, the i
th equation is given as B of u 9 phi i is
195
00:34:42,460 --> 00:34:46,200
equal to F of phi i.
196
00:34:46,200 --> 00:34:53,200
So, now if I have this, now I can expand the
u 9. So, what will that be equal to? I can
197
00:35:00,760 --> 00:35:07,760
write it as B of sigma i is equal to 1 to
9 u i u not i - we will call it j - u j, phi
198
00:35:23,730 --> 00:35:30,730
i is equal to F of phi i. Now, since this
is bi-linear, by the definition of bi-linear
199
00:35:33,710 --> 00:35:38,410
functions this summation over this coefficient
I can take out. So, this is equivalent to
200
00:35:38,410 --> 00:35:45,410
writing form of j equal to 1 to 9 b of phi
j phi i whole thing multiplied by u j is equal
201
00:35:59,410 --> 00:36:06,410
to F of phi i.
If you see for the i th equation, this becomes,
202
00:36:09,980 --> 00:36:16,980
this term becomes the coefficient of the j
th unknown. So, what is it called? If you
203
00:36:26,270 --> 00:36:33,270
want to go back to what we have to done earlier,
this term will be equal to k for the i th
204
00:36:34,470 --> 00:36:41,470
equation j th call. So, this is k i j. Now,
our job is... So, if this is k i j, then we
205
00:36:42,230 --> 00:36:48,700
can write the whole problem in the form of
the global stiffness matrix k into the unknown
206
00:36:48,700 --> 00:36:55,339
displacement vector u, where now this is k
is 9 by 9, u is the vector of phi 9 this is
207
00:36:55,339 --> 00:36:59,280
equal to the vector F.
208
00:36:59,280 --> 00:37:06,280
So, our job is to find the entries of this
phase and of this vector. So, if we can find
209
00:37:15,569 --> 00:37:21,490
anything, then we have assembled to make this
problem and all that we have to do is invert
210
00:37:21,490 --> 00:37:25,880
this matrix; first, we have to apply one-way
condition, invert the matrix and solution
211
00:37:25,880 --> 00:37:32,880
is there. So, to do this, what do we do? We
have to again go back to the element calculation;
212
00:37:34,329 --> 00:37:41,329
so, this is element level calculation.
Now, let me redraw the whole figures again
213
00:37:53,280 --> 00:38:00,280
for the element; this is x 1 of an element
k, x 2 of an element k, x 3 of an element
214
00:38:03,559 --> 00:38:10,559
k and if you remember, so this was N 1, N
2, N 3; so, this was N 1 of an element k,
215
00:38:14,809 --> 00:38:21,809
which is equivalent to phi 2 k minus 1. This
is N 2 of element k, which is equivalent to
216
00:38:31,660 --> 00:38:38,660
phi 2 k, and this one is N 3 of element k,
which is equivalent to phi 2 k plus 1 in the
217
00:38:46,690 --> 00:38:53,440
element k. Just like we have done for the
linear problem for the phi equal to 1 approximation,
218
00:38:53,440 --> 00:38:59,299
here what do we know, that the phi is which
are non-zero in the element are these three
219
00:38:59,299 --> 00:39:06,299
phi(s), phi 2 k minus 1, phi 2 k and phi 2
k plus 1 equal to 1, 2, 3, 4 that is a number
220
00:39:06,390 --> 00:39:13,390
of up to the number of element.
So, what element is going to contribute to,
221
00:39:13,849 --> 00:39:20,849
is the 2 k minus 1 th equation, 2 k th equation,
and 2 k plus 1 th equation, and which rows
222
00:39:21,549 --> 00:39:27,319
of these equation are going to be non-zero,
which column of these rows are going to be
223
00:39:27,319 --> 00:39:34,319
non-zero? The columns which are going to be
non-zero for each of these rows are the 2
224
00:39:35,069 --> 00:39:41,720
k minus 1, 2 k and 2 k plus 1. So, these columns
only, the element is going to contribute,
225
00:39:41,720 --> 00:39:48,720
and none of the other column it is going to
contribute. So, if I take that b b, that I
226
00:39:51,640 --> 00:39:58,640
had written phi j phi i, which is nothing
but as we have said k i j.
227
00:40:01,569 --> 00:40:07,440
Which of the i and which of the j will get
contribution from the element? As we have
228
00:40:07,440 --> 00:40:13,690
said, i equal to 2 k minus 1, i equal to 2
k and i equal to 2 k plus 1. Similarly, j
229
00:40:13,690 --> 00:40:18,789
is equal to 2 k minus 1, j equal to 2 k and
j is equal to 2 k plus 1. So, let us find
230
00:40:18,789 --> 00:40:24,130
the contribution, how we had found these contributions
earlier.
231
00:40:24,130 --> 00:40:29,339
So, what we are going to define now, it is
for this element k. The element stiffness
232
00:40:29,339 --> 00:40:35,579
matrix - now what will be the size of this
element stiffness matrix? Because there are
233
00:40:35,579 --> 00:40:42,579
now 3 n(s) or 3 phi(s), which are non-zero
in the element. So, it will be a 3 by 3 matrix.
234
00:40:43,640 --> 00:40:50,640
So, what will the entries be? I will write
only of few entries of this and then it will
235
00:40:54,180 --> 00:41:01,180
be integral form x 1 k
236
00:41:06,650 --> 00:41:13,650
to x 3 k EA d N 1 k dx; this I will check.
So, for the first row of this, I will EA d
237
00:41:24,119 --> 00:41:31,119
N 1 k dx into d N1 k dx will be the first
column entry EA d N 1 k dx into d N 2 k dx,
238
00:41:33,049 --> 00:41:40,049
as the second column entry and the third will
be d N 1 k dx into d N 3 k dx is the third
239
00:41:42,400 --> 00:41:45,210
column entry.
And similarly, if I come here to the second
240
00:41:45,210 --> 00:41:52,210
row, so it will be x 3 k EA d N 2 k dx d N
1 k dx and so on; and the third will have
241
00:42:11,930 --> 00:42:18,930
integer x 1 k to x 3 k; you see that limits
of the integration are from one x exterminate
242
00:42:19,630 --> 00:42:24,940
of the element to next exterminate of the
element. So, that is should be kept in mind
243
00:42:24,940 --> 00:42:31,940
d N 3 k dx d N 1 k dx d x. Can I write the
k i j in the element k? This will be equal
244
00:42:46,000 --> 00:42:53,000
to integral x 1 of k to x 3 of k of EA d N
j k dx d N i k dx d x.
245
00:43:12,000 --> 00:43:19,000
Similarly, F i in the element k will be equal
to nothing but x 1 of k to x 3 of k f N i
246
00:43:26,809 --> 00:43:33,109
k dx; remember, that here we are not going
to add the part due to the p that is going
247
00:43:33,109 --> 00:43:38,789
to be added, when we are applying the or the
force found the condition; this is all that
248
00:43:38,789 --> 00:43:45,789
the F i k from the element will have. So,
once we have this, then we have obtained the
249
00:43:46,750 --> 00:43:52,549
element level stiffness matrix and the element
level load vector. So, this will go to the
250
00:43:52,549 --> 00:43:59,549
element load vector.
So, once I have these, now what we do? We
251
00:44:10,890 --> 00:44:16,619
have to find the correspondence between these
element level entries and the global entries;
252
00:44:16,619 --> 00:44:23,619
that is, where in the global stiffness matrix
should be this k i j go globally? It is which
253
00:44:27,780 --> 00:44:34,780
m n it should go to, that is where I have
to assemble these things. Similarly, where
254
00:44:36,349 --> 00:44:43,349
will be the F i at the element go to globally,
which F this equation has this information,
255
00:44:49,799 --> 00:44:50,960
we have to help.
256
00:44:50,960 --> 00:44:56,130
So, what has been done earlier, as far as
this information is concerned, we had created
257
00:44:56,130 --> 00:45:03,130
a local to global numbering or 90 times it
is called the connectivity information. So,
258
00:45:22,930 --> 00:45:29,930
this connectivity information has to be constructed.
So, let us now do it for the elements. So,
259
00:45:42,190 --> 00:45:49,079
what I will do is, here I write the global
number, and here, I will write the corresponding
260
00:45:49,079 --> 00:45:56,079
local number for the element k. So, this is
the local number K in the element; I have
261
00:46:01,309 --> 00:46:06,450
how many? 1, 2, 3; this global number, does
it correspond to 1? Corresponds to which global
262
00:46:06,450 --> 00:46:11,950
basis function? N 1 k corresponds to phi 2
k minus 1. So, it should correspond to globally.
263
00:46:11,950 --> 00:46:18,950
I will re-write it in another form, 2 k minus
1 plus 1 which is nothing but 2 k minus 1.
264
00:46:21,960 --> 00:46:28,960
This one will be N 2 corresponds to phi 2
k globally; so, it will be as... rewrite as
265
00:46:33,609 --> 00:46:40,609
this plus 2; similarly, N 3 k will corresponds
to phi 2 k plus 1. So, remember what we have
266
00:46:48,309 --> 00:46:53,720
done. So, before we go and do the element
calculation, we will stretch the entries of
267
00:46:53,720 --> 00:46:58,680
the global stiffness matrix and the global
load vector equal to 0.
268
00:46:58,680 --> 00:47:04,260
The process of the assembly that we had talked
about in great detail, in the previous lecture,
269
00:47:04,260 --> 00:47:11,260
essentially is adding the elemental contributions
to this global stiffness matrix and the global
270
00:47:11,339 --> 00:47:17,280
load vector, in order to complete the global
equation, because what we have said, is integration
271
00:47:17,280 --> 00:47:23,390
over the full domain can be partitioned into
the integration over the elements, and when
272
00:47:23,390 --> 00:47:27,730
we add these elemental integrals together,
we get the global representation.
273
00:47:27,730 --> 00:47:34,730
So, the global entry which is going to get
contribution from element are going to be:
274
00:47:35,670 --> 00:47:42,670
k 2 k minus 1 plus i, 2 k minus 1 plus j is
equal to... what we are going to do? It is
275
00:47:53,619 --> 00:48:00,619
to the original number, which was sitting
there, I am going to add k i j. So, what has
276
00:48:11,079 --> 00:48:18,079
happened to the global stiffness entry k 2
k minus 1 plus i k 2 k minus 1 plus j? I am
277
00:48:21,099 --> 00:48:28,099
going to add the elemental i j th. So, this
is the part, as the assembly part, as far
278
00:48:28,460 --> 00:48:35,460
as the stiffness is concerned and this for
i, j is equal to 1, 2, 3. Similarly, the F
279
00:48:40,549 --> 00:48:47,549
2 k minus 1 plus i will be equal to F 2 k
minus plus i 1, to this I am going to add
280
00:48:55,210 --> 00:49:02,210
F k i. So, this is the assembly procedure.
So, we have obtained the element calculation,
281
00:49:17,160 --> 00:49:22,010
using the function which we have defined the
elemental level, and assemble them to get
282
00:49:22,010 --> 00:49:26,849
the global equation, in terms of the basis
function that we are after, the coefficient
283
00:49:26,849 --> 00:49:33,849
for the global basis function. One should
remember, that in any finite element approximation,
284
00:49:34,430 --> 00:49:40,180
we are really looking for the coefficients
of the global basis function, and the element
285
00:49:40,180 --> 00:49:47,180
calculation, is a simple way of doing the
integration, relevant that have to be done,
286
00:49:49,890 --> 00:49:54,930
to obtain the entries of the global stiffness
matrix and the global load vector by doing
287
00:49:54,930 --> 00:49:59,210
at the element level, adding it up to get
the global equation.
288
00:49:59,210 --> 00:50:06,210
One question that should come to mind now
is - earlier, we have seen for P equal to
289
00:50:06,970 --> 00:50:13,970
1, that the i th equation had only non-zero
entries corresponding to the i minus 1 column,
290
00:50:15,710 --> 00:50:22,710
the i th column and the i plus 1 th column;
in this case, what are the non-zero entries
291
00:50:23,700 --> 00:50:26,160
for the i th row?
292
00:50:26,160 --> 00:50:33,160
So, let us check; now take the i th row; I
will take this one as the 2 k minus 1 th point.
293
00:50:42,039 --> 00:50:49,039
So, this is 2 k minus 3. Here we have to differentiate
between this nodal function and the internal
294
00:50:56,730 --> 00:51:03,520
bubble function. So, let us take the nodal
function that we have. So, if you see that
295
00:51:03,520 --> 00:51:10,520
I am talking on the equation corresponding
to this one. So, I am corresponding to the
296
00:51:14,770 --> 00:51:21,770
equation corresponding to this function which
is phi 2 k minus 1.
297
00:51:24,799 --> 00:51:31,799
This is the phi that we are going to contribute
to the equation corresponding to this. Certainly,
298
00:51:32,240 --> 00:51:39,240
it is going to be this phi, it is going to
be this phi and this phi. So, this is going
299
00:51:48,250 --> 00:51:55,250
to be phi 2 k minus 3, this is phi 2 k minus
2, this is phi 2 k and this is phi 2 k plus
300
00:52:05,839 --> 00:52:12,839
1. What can you see? That for the 2 k minus
1 th equation will have non-zero entries corresponding
301
00:52:16,720 --> 00:52:23,720
to the 2 k minus 3 column 2 k minus 2 column
2 k minus 1 column 2 k column and 2 k plus
302
00:52:26,079 --> 00:52:33,079
1 column. So, how many columns will be non-zero?
1, 2, 3, 4, 5; so five columns will have at
303
00:52:34,950 --> 00:52:41,950
least, not at least, at most five columns
will have non-zero entries; for all other
304
00:52:43,210 --> 00:52:47,720
phi(s) the column entries are going to be
0, and because all other phi(s) are going
305
00:52:47,720 --> 00:52:53,880
to disappear or become 0 in the region, where
this phi 2 k minus 1 is defined to be.
306
00:52:53,880 --> 00:53:00,880
You see that if right global stiffness matrix,
the number of non-zero entries has increased
307
00:53:05,200 --> 00:53:11,760
in a row. This is the another property of
these approximation that we have going to
308
00:53:11,760 --> 00:53:18,339
see and how they reflect in terms of the part
city of the global stiffness matrix and that
309
00:53:18,339 --> 00:53:23,619
also this part city, we are talking about,
we have talk about solver. So, this we will
310
00:53:23,619 --> 00:53:30,619
tackle at a later date. See, there is the
very curious thing about this function phi
311
00:53:31,750 --> 00:53:38,750
k; this function phi k, if you add them up
anywhere, you see their value is equal to
312
00:53:54,640 --> 00:53:58,240
1; why is it so?
313
00:53:58,240 --> 00:54:05,240
Remember, we had said that these functions
phi k should be able to exactly represent
314
00:54:07,589 --> 00:54:14,589
any polynomial up to a quadratic. So, a constant
is also global polynomial, which should be
315
00:54:15,099 --> 00:54:20,720
represented exactly, and in constant I will
take my function f x, which I want to represent
316
00:54:20,720 --> 00:54:27,720
to be 1. This will be equal to sigma u i phi
i. Now, is this function is equal to 1 everywhere
317
00:54:34,970 --> 00:54:41,770
in the domain? It has to be 1 at the points
x 1, x 2, x 3, x 9. So, if I go to the point
318
00:54:41,770 --> 00:54:48,770
x i, this is still equal to 1, but then this
will be equal to u i, which tells me that
319
00:54:53,670 --> 00:55:00,670
the u i here on the right hand side are all
equal to 1 is the u i are all 1, then what
320
00:55:02,029 --> 00:55:05,539
we get is some of the phi i is equal to 1
and this is also true in the element; same
321
00:55:05,539 --> 00:55:12,539
thing applies in the element. So, in the element,
some of the shape function will be i is equal
322
00:55:13,769 --> 00:55:20,769
to 1 to 3 is equal to 1; this is a property
is that we should keep in mind, and what it
323
00:55:25,140 --> 00:55:32,140
also tells us that it implies, that if I do
this summation, this has to be equal to 0,
324
00:55:37,579 --> 00:55:42,980
because this sum is equal to constant that
delivered the constant is 0; these are certain
325
00:55:42,980 --> 00:55:48,289
things, which we should have in mind. So that
when we go and write the computer program
326
00:55:48,289 --> 00:55:52,769
to do the finite element analysis, we can
check all function, these basis functions
327
00:55:52,769 --> 00:55:57,490
of the element shape functions that we have
called, which have N i is are nothing but
328
00:55:57,490 --> 00:56:04,490
the element shape function, we can check whether
they have been correctly program or not.
329
00:56:07,390 --> 00:56:14,390
The definition of this element shape function
that we have used, this definition is called
330
00:56:19,940 --> 00:56:26,940
Lagrangian shape or basis function; we will
see how to generalize these, to go for a keys
331
00:56:44,079 --> 00:56:45,880
order approximation in future
332
00:56:45,880 --> 00:56:52,880
So, this Lagrangian basis functions that we
have constructed, are note, the only one that
333
00:56:54,089 --> 00:57:01,089
we can have. So, those who are interested,
there is also a very popular family of shape
334
00:57:03,369 --> 00:57:10,369
functions that are basis function, that can
be used is Legendre function, but we are not
335
00:57:11,039 --> 00:57:18,039
going to discuss it, here in this analysis.
In the next class, what we are going to do
336
00:57:20,789 --> 00:57:27,319
is, we are going to look at the cubic approximation,
the P equal to 3 approximation, also construct
337
00:57:27,319 --> 00:57:34,319
basis function, how to do the elemental calculation,
how to go for the load vector and from there
338
00:57:34,369 --> 00:57:39,089
we will generalize to the P for the approximation.
And once you have done the P for the approximation,
339
00:57:39,089 --> 00:57:44,950
then we can do a finite element computation
using any order of approximation in the domain.
340
00:57:44,950 --> 00:57:51,069
After that, we will see how they are going
to benefit us; just doing them is not enough,
341
00:57:51,069 --> 00:57:58,069
why we should do them, how they are going
to benefit us is going to be
in the future lecture.Thank you.
342