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PROFESSOR STRANG: So, let's
see, you probably guessed on

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that quiz problem three,
it wasn't what I meant.

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I get a zero for that problem.
 

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But you'll get probably good
numbers, so that's an election

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gift if it comes out that way.
 

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So they're all in the hands
of the TAs to be graded.

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We have a holiday
Monday, I think.

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We come back to Fourier.
 

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Now, so we just have a
concentrated shot at Fourier,

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just about eight or nine
lectures in November.

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So stay with it and that'll
be of course the subject

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of the third quiz.
 

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Which will have no mistakes.
 

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It'll be solved by the TAs in
advance and we'll spot things.

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So, and if we have the quizzes
to return to you by Wednesday

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that will be great.
 

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I hope so, but they
have a big job.

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A little bit, looking far ahead
at the end of Fourier the quiz

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is December 4th, I think
that's a Thursday.

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And that's the end
of the course.

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So December 4th, so we'll
be ending the course

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a little bit early.
 

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Because I'll be in Hong
Kong, to tell the truth.

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And and we've done a lot, and
with the review sessions

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we're really doing well.
 

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So, that's the future.
 

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Fourier, today is an important
day too, finite elements in

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2-D, that's a major part of
computational science

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and engineering.
 

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The finite element idea, the
idea of using polynomials, you

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can find in some early papers
by Courant, a mathematician in

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New York, and by a guy in China
neat guy named Fung Kong But

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those papers were sort of,
you could do it this

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way if you wanted.
 

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It was really the structural
engineers in Berkeley and

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elsewhere who made it
happen ten years later.

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And the whole idea
has just blossomed.

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Continues to grow.
 

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So I had an early book, in the
`70s, actually, about the

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mathematical underpinnings.
 

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The math basis for the
finite element method.

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And many other finite
element books.

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Professor Bathe you
know, teaches a full

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of course on that.
 

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But, I think we can get the
idea of finite elements here.

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We did them in 1-D, and now
there's a MATLAB problem and

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I'd like to just describe
that particular problem

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if I can, as an example.
 

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And, of course, you would use
the code that's printed in the

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book, and that's available on
the website just to download.

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But the problem is not
on a square domain.

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It starts on a circle, so that
the first lines of the code,

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the calling the MATLAB command
square grid, are

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not applicable.
 

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So you have to create,
then, a mesh.

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Well, I have a suggested mesh
so I'll draw that, and then

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from that you want to make a
list of all the node points.

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A list, P, of - so what the
code needs is two lists.

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Well, let me draw a picture
of, well, it's a circle.

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And I'm going to be solving
Poisson's equation.

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The equation will be
-u_xx-u_yy=4, in the circle.

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So it's Poisson but with a
constant right hand side.

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That will mean that all the
integrals of F times v, all the

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right hand side of our discrete
equation will be, the integrals

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are all easy because we
just have a constant there

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times the trial function.
 

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OK, and then on the boundary
is going to be u=0.

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On the boundary.
 

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So it's a classic problem.
 

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And we can say what
the solution is.

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So it's one with a
known solution.

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I think it would be x squared.
 

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No, I guess one, one minus
x squared minus y squared.

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This should all be
on the .086 site.

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I just didn't have a chance
to look this morning

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to be sure it got up.
 

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So you can watch, here.
 

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So that, I hope, does
solve the problem.

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Two x derivatives give us
a two, two y derivatives

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another two, so we get four.
 

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So we know the answer; the
question is, and I'm interested

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in this question, for research
reasons too, is what's the

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error when you go to a polygon?
 

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You go to a, these curved
boundaries don't get

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correctly saved.
 

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You approximate them
by straight lines.

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That would be the first idea.
 

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And with this, all this
symmetry, let's keep the

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problem nice and use
a regular polygon.

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So maybe I'll try to draw
one with about eight

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sides, but, OK.
 

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So we impose u=0
at these nodes.

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So u is zero at those nodes,
and then we have a mesh.

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So we want to create a mesh.
 

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OK, so with all the symmetry
here, the natural idea would be

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to start with eight pieces, or
M pieces if I have, this is a

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regular M side, let's say, and
I'll take M to be eight

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in this picture.
 

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And I think we can work
on just one triangle.

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By rotational symmetry,
all those triangles are

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going to be the same.
 

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So I think our domain is really
this one triangle here.

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That's where we're working.
 

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And in that triangle,
I think we have zero

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boundary conditions.
 

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And across this edge I
think we have natural

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boundary conditions.
 

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Slope zero, if I see
the picture correctly.

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The rotational symmetry
would mean that things

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are not changing.
 

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That every triangle
is the same.

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So I think on these
boundaries it's the Neumann

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condition, dU/dn=0.
 

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And I'm frankly not sure what
to do at the origin, so I'll

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maybe just try both
ways and see.

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OK, so there is a real problem.
 

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Of course, it's artificial
in the sense that

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we know the answer.
 

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But it's a real open question
of what does the error look

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like, from doing that.
 

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So that's the goal and let me
just say the problem I'll ask

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you to do, and it probably is
quite enough to be ready for

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next Friday, is to use
piecewise linear elements.

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Which is what I'm going to do.
 

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What every discussion of
finite elements will begin

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with, linear elements.
 

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Those pyramids that I
spoke about it at the

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end of last time.
 

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So that's what I hope, but
actually I would be highly

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interested if anybody got
into the problem, to

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try quadratic elements.
 

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So I'll just say here,
second-degree quadratic

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polynomials would
be more accurate.

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Would be more accurate.
 

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So, in other words, this is a
first type of finite element

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called P_1, for polynomials
of degree one.

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These guys, I would call P_2,
for polynomials of degree two.

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And I've mentioned here the
possibility of using quads.

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Instead of triangles, if I had
squares for example, the

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simplest element
would be a Q_1.

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So these are, if I manage today
to tell you about how to use

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P_1 and P_2 and Q_1,
you're on your way.

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And for the requirements of
this course, P_1 is the

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first point to understand.
 

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OK.
 

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While I'm speaking about codes
and meshes, let me draw the

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mesh I proposed in the
homework problem.

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So I thought OK, we just have
to have a simple mesh here.

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So let me draw that line in.
 

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I know all these points, right?
 

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This is 0, 0 here.
 

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And that point's on the circle.
 

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And so is this, that point
might be, so what's

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the angle there?
 

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That angle is probably pi/8.
 

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The whole angle would be 2pi/8,
and eight of them would

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go all the way around.
 

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So I think that angle is pi/8.
 

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And so this point
would be cos(pi/8).

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sin(pi/8).
 

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We know where they are.
 

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But that's going to be
what we have to list.

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We have to list where
are the coordinates of

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all the mesh points.
 

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So let me describe the rest of
the mesh points and then see

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what that list would look like.
 

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And once you've created the
list, the code will take over.

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And then you plot the results
and see what's going on.

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So let me suggest a mesh.
 

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It's pretty straightforward.
 

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I just divided this piece into
N, this center thing into N,

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so I called that distance h.
 

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So Nh gets me out to here.
 

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Whatever that is, that's
cos(pi/8), I guess, that's the

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x coordinate of that line.
 

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So those are mesh points and
then let me keep drawing these.

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So these will be mesh
points too, and these

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will be mesh points too.
 

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So at this point, I've got
probably 12 or 13 mesh points.

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But I've got quads, right?
 

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Well, I've got a couple of
triangles here that I'm not

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going to touch, those are fine.
 

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But these are quads and they
could be used with the Q_1

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element, but I'm thinking
let's stay with triangles.

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So I just suggested to put
in triangles, put in these

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diagonals, keeping symmetry.
 

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And so there's the mesh.
 

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There's the mesh.
 

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00:13:15 --> 00:13:19
And then what does
the code ask for?

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So it's got 13 mesh points.
 

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And the code, first of all it
wants a list of the coordinates

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of all those mesh points.
 

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So that the code will, and I
better number them, of course.

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So let me number them one,
shall I number them the center

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guys first, one, two, three,
four, five, and then up here

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six, seven, eight, nine,, now
I don't know if this

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is a good numbering.
 

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Ten, 11, 12, 13.
 

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00:13:48 --> 00:13:57
Why don't we, just to have some
consistency within plans,

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why don't you, don't
have to take N=4.

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00:14:01 --> 00:14:06
I hope you'll take, what did
I take N as four or five?

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00:14:06 --> 00:14:08
Yeah, right.
 

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00:14:08 --> 00:14:09
N is 4.
 

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But you'll want to
try different N's.

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00:14:13 --> 00:14:18
N=4 would be a good crude start
to see what's going on, but

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00:14:18 --> 00:14:23
then I hope you'll go higher
and get better accuracy.

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00:14:23 --> 00:14:27
And you can see how the
accuracy improves, how

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you get closer to that
as n gets bigger.

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OK, but got the nodes numbered.
 

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Oh, I better number
the triangles.

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OK, how shall we
number the triangles?

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Shall we do along
the top, or this?

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I don't know.
 

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What do you want to do for the
numbering of the triangles?

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Maybe run along the top, and
then run along the bottom,

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because then it'll
practically be a copy.

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So I didn't leave myself much
space, but one, two, three,

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four, five, six, seven.
 

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00:15:04 --> 00:15:08
Seven triangles along the top
and seven along the bottom, so

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I have 14 triangles
in this mesh.

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So it's a mesh with 14
triangles and 13 nodes.

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00:15:23 --> 00:15:26
And I know the positions
of everyone, right?

232
00:15:26 --> 00:15:29
I know the x,y coordinates
of every one.

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So what the code will want is
a list of those coordinates.

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So a list, P, P will be
a list of coordinates.

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The first guy will be
of, the nodes so 13

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rows, three columns.
 

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So it's a little 13 by three
matrix that tells you

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where all the nodes are.
 

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So the first one on that
list would be (0,0).

240
00:16:01 --> 00:16:03
that's for node one.
 

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00:16:03 --> 00:16:06
And the second one would be
whatever the coordinates of

242
00:16:06 --> 00:16:11
that are, something zero.
(h,0), I guess it is.

243
00:16:11 --> 00:16:18
The third one will be (2h,0),
and so on, and then complete

244
00:16:18 --> 00:16:21
the list of 13 positions.
 

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So you are then told the code
where all the nodes are.

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00:16:28 --> 00:16:30
What else do you
have to tell it?

247
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Not much.
 

248
00:16:32 --> 00:16:35
You now have to tell it
about the triangles.

249
00:16:35 --> 00:16:39
So now for every, why do
I say three columns?

250
00:16:39 --> 00:16:44
Maybe only two, is it?
 

251
00:16:44 --> 00:16:46
You see the point already.
 

252
00:16:46 --> 00:16:53
I've forgotten, maybe, yeah.
 

253
00:16:53 --> 00:16:58
I don't see why.
 

254
00:16:58 --> 00:17:02
Well for triangles, I'm going
to need three, for nodes

255
00:17:02 --> 00:17:03
maybe it's only got two.
 

256
00:17:03 --> 00:17:07
Maybe it's 13 by two.
 

257
00:17:07 --> 00:17:10
I don't see why I need three.
 

258
00:17:10 --> 00:17:11
But, anyway.
 

259
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Then the other list
is triangles.

260
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So this will be the list,
t, and so it takes

261
00:17:20 --> 00:17:22
triangle number one.
 

262
00:17:22 --> 00:17:24
Which is right here.
 

263
00:17:24 --> 00:17:26
That very first triangle.
 

264
00:17:26 --> 00:17:31
And what does it have to
tell us about triangle one?

265
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The three nodes.
 

266
00:17:32 --> 00:17:37
If it tells us the three node
numbers, and this list, P, gave

267
00:17:37 --> 00:17:40
their positions, we've got it.
 

268
00:17:40 --> 00:17:45
So how many triangles
did that I have?

269
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14?
 

270
00:17:46 --> 00:17:56
So t will be 14 by three, and
so the first guy will be just

271
00:17:56 --> 00:18:00
one, node number one,
node number two, and

272
00:18:00 --> 00:18:02
node number six.
 

273
00:18:02 --> 00:18:06
That will tell us which
is the first triangle.

274
00:18:06 --> 00:18:10
And the second triangle, I
guess I've drawn from two.

275
00:18:10 --> 00:18:14
Two, seven to six, right?
 

276
00:18:14 --> 00:18:19
That's a very skinny triangle
up there but it's the one that

277
00:18:19 --> 00:18:22
started at node two, went up
to seven and back to six.

278
00:18:22 --> 00:18:26
So a list like that.
 

279
00:18:26 --> 00:18:32
And then the code
will do the rest.

280
00:18:32 --> 00:18:35
I hope.
 

281
00:18:35 --> 00:18:38
Almost all the rest.
 

282
00:18:38 --> 00:18:41
The code will create
the matrix K.

283
00:18:41 --> 00:18:47
It'll create the matrix K
with, it'll be singular.

284
00:18:47 --> 00:18:51
Boundary conditions
won't yet be in there.

285
00:18:51 --> 00:18:56
And then a final step after
that K, or maybe we could call

286
00:18:56 --> 00:19:02
it K_0 is created, a final
step will be to fix u.

287
00:19:02 --> 00:19:04
At least at these three points.
 

288
00:19:04 --> 00:19:08
So these three will
be boundary nodes.

289
00:19:08 --> 00:19:12
And as I say, I'm not too sure
about that one, I apologize.

290
00:19:12 --> 00:19:14
At those boundary nodes
I'm going to take the

291
00:19:14 --> 00:19:16
values to be zero.
 

292
00:19:16 --> 00:19:21
So this is going to be zero
along the whole edge, because

293
00:19:21 --> 00:19:24
if it's zero there, zero there,
zero there and zero there, and

294
00:19:24 --> 00:19:27
if it's linear, it's zero.
 

295
00:19:27 --> 00:19:35
So the final, sort of,
subroutine in the code, the

296
00:19:35 --> 00:19:39
final group of commands you
want to impose, zeroes here,

297
00:19:39 --> 00:19:46
that should then make the
matrix K invertible, and then

298
00:19:46 --> 00:19:49
you've got KU=F to solve.
 

299
00:19:49 --> 00:19:54
So what the code is doing
is creating K and F.

300
00:19:54 --> 00:19:57
You see the overall picture?
 

301
00:19:57 --> 00:20:00
I jumped right into this
particular mesh, particular

302
00:20:00 --> 00:20:06
problem, but now I really
should back up to

303
00:20:06 --> 00:20:09
where it starts.
 

304
00:20:09 --> 00:20:13
This this is going to be
the weak form of Laplace,

305
00:20:13 --> 00:20:15
Poisson, maybe I'll make a
 

306
00:20:15 --> 00:20:24
little space to put in
Poisson's name too.

307
00:20:24 --> 00:20:29
You have a picture already of
what this weak form is about,

308
00:20:29 --> 00:20:32
so now I'm really backing
up to the start.

309
00:20:32 --> 00:20:36
I take the equation and
I get its weak form.

310
00:20:36 --> 00:20:39
And remember that's in the
continuous case, as it

311
00:20:39 --> 00:20:41
was in the quiz problem.
 

312
00:20:41 --> 00:20:45
The first step is the
continuous weak form, and then

313
00:20:45 --> 00:20:53
the second step is choose test
functions and, trial - I'm

314
00:20:53 --> 00:20:59
sorry, gosh, I'm in
bad shape here.

315
00:20:59 --> 00:21:02
Because they're the same.
 

316
00:21:02 --> 00:21:04
This isn't my worst
error, today.

317
00:21:04 --> 00:21:11
But those are trial functions,
and these are test functions.

318
00:21:11 --> 00:21:14
OK, questions at this point,
because I've, yeah, thank you.

319
00:21:14 --> 00:21:15
Good.
 

320
00:21:15 --> 00:21:16
Let's look at this picture.
 

321
00:21:16 --> 00:21:22
AUDIENCE: [INAUDIBLE]
 

322
00:21:22 --> 00:21:30
PROFESSOR STRANG:
Because, we did.

323
00:21:30 --> 00:21:30
That's right.
 

324
00:21:30 --> 00:21:40
So because my question is what
is the continuous problem, I

325
00:21:40 --> 00:21:44
would like to solve has
u=0 on the polygon.

326
00:21:44 --> 00:21:50
So in a way you can forget
the circle, where we

327
00:21:50 --> 00:21:52
know the answer now.
 

328
00:21:52 --> 00:21:57
We really are looking on the
polygon, and I would like to

329
00:21:57 --> 00:22:01
know what's the solution
like on that polygon.

330
00:22:01 --> 00:22:06
And then so there are two
steps, the first step was

331
00:22:06 --> 00:22:09
start with a circle,
we have the answer.

332
00:22:09 --> 00:22:14
Second step is go to a polygon,
continuous problem, Poisson's

333
00:22:14 --> 00:22:16
equation in the polygon.
 

334
00:22:16 --> 00:22:18
How different is
that from this?

335
00:22:18 --> 00:22:23
Because this will not satisfy
the polygon boundary

336
00:22:23 --> 00:22:24
conditions.
 

337
00:22:24 --> 00:22:28
So that's the circle answer.
 

338
00:22:28 --> 00:22:32
Then the question is,
what's the polygon answer.

339
00:22:32 --> 00:22:36
And I don't know that.
 

340
00:22:36 --> 00:22:39
You may say a regular
polygon, you can't do that.

341
00:22:39 --> 00:22:41
I didn't think you can.
 

342
00:22:41 --> 00:22:44
It's amazing, but probably
a triangle or a square.

343
00:22:44 --> 00:22:49
So if M is three or four,
probably some formulas

344
00:22:49 --> 00:22:50
would be available.
 

345
00:22:50 --> 00:22:54
But I think once we get higher,
I don't know the answer to the

346
00:22:54 --> 00:22:59
Dirichlet problem, to Poisson's
equation on a polygon.

347
00:22:59 --> 00:23:02
On a regular polygon and that's
what I would really like

348
00:23:02 --> 00:23:03
to know more about.
 

349
00:23:03 --> 00:23:06
And how do I find
out more about it?

350
00:23:06 --> 00:23:08
By finite elements.
 

351
00:23:08 --> 00:23:11
With your help.
 

352
00:23:11 --> 00:23:16
Taking that polygon, breaking
it into a mesh, looking only

353
00:23:16 --> 00:23:22
at one triangle just for
simplicity, and getting

354
00:23:22 --> 00:23:24
u finite elements.
 

355
00:23:24 --> 00:23:26
Well, I should say u_p_1.
 

356
00:23:26 --> 00:23:30
 
 

357
00:23:30 --> 00:23:34
That's the finite element
solution using linear.

358
00:23:34 --> 00:23:40
I would really like to know
u_p_2, the finite element

359
00:23:40 --> 00:23:43
solution which will be better.
 

360
00:23:43 --> 00:23:45
If I use quadratics.
 

361
00:23:45 --> 00:23:47
So now I get the fun of
describing the linear

362
00:23:47 --> 00:23:51
elements, the quadratic
elements, the quads.

363
00:23:51 --> 00:23:53
But did I answer
that question OK?

364
00:23:53 --> 00:23:54
Yeah.
 

365
00:23:54 --> 00:23:59
So this is the problem I would
like to know the answer to.

366
00:23:59 --> 00:24:04
If I have this equation, zero
boundary conditions on a

367
00:24:04 --> 00:24:09
regular polygon with M
sides, what's the answer?

368
00:24:09 --> 00:24:13
And it's going to be close to
this, but it won't be the same.

369
00:24:13 --> 00:24:17
Because this does not vanish
on the polygon edges.

370
00:24:17 --> 00:24:22
And I would like to
compare the slopes, too.

371
00:24:22 --> 00:24:25
So the homework problem asked
you not only to compare

372
00:24:25 --> 00:24:32
u circle with u_p_1,
but also the slopes.

373
00:24:32 --> 00:24:37
The slopes here are easy,
slopes here are easy because

374
00:24:37 --> 00:24:39
it's a bunch of flat functions.
 

375
00:24:39 --> 00:24:43
So the slopes are just
constant in each triangle.

376
00:24:43 --> 00:24:48
OK, I'm guessing that the error
gets smaller as you go in.

377
00:24:48 --> 00:24:52
I think that if you plot the
error, it'll be largest out

378
00:24:52 --> 00:24:53
here and get small there.
 

379
00:24:53 --> 00:24:56
But remains to be seen.
 

380
00:24:56 --> 00:24:57
So I hope you enjoy
 

381
00:24:57 --> 00:24:59
- yeah, good.
 

382
00:24:59 --> 00:25:05
AUDIENCE: [INAUDIBLE]
 

383
00:25:05 --> 00:25:07
PROFESSOR STRANG:
Rather than seven.

384
00:25:07 --> 00:25:13
AUDIENCE: [INAUDIBLE]
 

385
00:25:13 --> 00:25:17
PROFESSOR STRANG:
No, the middle.

386
00:25:17 --> 00:25:21
There's nothing magic about
any particular mesh.

387
00:25:21 --> 00:25:31
I just chose this mesh as
pretty good, and actually,

388
00:25:31 --> 00:25:34
I'm imagining M could
get pretty big.

389
00:25:34 --> 00:25:35
That would be interesting.
 

390
00:25:35 --> 00:25:40
M=8 would be interesting,
M=16, M=1,024, now then

391
00:25:40 --> 00:25:42
I'd really get interested.
 

392
00:25:42 --> 00:25:50
OK, but so if M is 1,024, then
this side would be very small.

393
00:25:50 --> 00:25:52
Right?
 

394
00:25:52 --> 00:25:55
And I just wanted
more triangles.

395
00:25:55 --> 00:25:58
Actually, I would like
more than I've got.

396
00:25:58 --> 00:26:05
I'd like, if M was really
big, then probably N should

397
00:26:05 --> 00:26:07
be at least that big.
 

398
00:26:07 --> 00:26:09
So I should have a thousand
this way, if this

399
00:26:09 --> 00:26:11
is just a tiny bit.
 

400
00:26:11 --> 00:26:14
I just want little tiny
h, and then, yeah.

401
00:26:14 --> 00:26:18
Actually, that might
not be too bad.

402
00:26:18 --> 00:26:25
If m anM N were roughly
comparable, then that length

403
00:26:25 --> 00:26:28
would be roughly comparable
to these lengths.

404
00:26:28 --> 00:26:33
And the triangles would
be pretty good shape.

405
00:26:33 --> 00:26:36
And that's what
you're looking for.

406
00:26:36 --> 00:26:43
I think there's a lot of
experiments to be done here.

407
00:26:43 --> 00:26:45
So, I'm thinking
then of M and N.

408
00:26:45 --> 00:26:48
Here I took N to be just four.
 

409
00:26:48 --> 00:26:50
When M was eight.
 

410
00:26:50 --> 00:26:51
That's fine.
 

411
00:26:51 --> 00:26:56
But if you keep M and N roughly
the same size, then you've got

412
00:26:56 --> 00:27:00
triangles that are not
too long and skinny.

413
00:27:00 --> 00:27:04
I'll tell you when
you might want.

414
00:27:04 --> 00:27:07
So generally you want
nice shaped triangles.

415
00:27:07 --> 00:27:13
You don't want angles very
small or very large, usually.

416
00:27:13 --> 00:27:20
But there would be, anybody in
Course 16 can imagine that if I

417
00:27:20 --> 00:27:29
am computing the flow field
past a wing, that long, thin

418
00:27:29 --> 00:27:33
triangles in the direction
of the wing are natural.

419
00:27:33 --> 00:27:41
I mean, somehow a problem
like true aerodynamics is

420
00:27:41 --> 00:27:43
by no means isotropic.
 

421
00:27:43 --> 00:27:46
I mean, the direction of the
wing is kind of critical to

422
00:27:46 --> 00:27:49
whether the plane flies, right?
 

423
00:27:49 --> 00:27:54
So don't make the
wing vertical.

424
00:27:54 --> 00:27:57
And if you want accuracy, then
you have long, thin triangles

425
00:27:57 --> 00:27:59
in the direction of the flow.
 

426
00:27:59 --> 00:28:02
But here we're not
doing a flow problem.

427
00:28:02 --> 00:28:06
We haven't got shocks, or
trailing edges, and other

428
00:28:06 --> 00:28:10
horrible stuff that
makes planes fly.

429
00:28:10 --> 00:28:12
We just got Poisson's equation.
 

430
00:28:12 --> 00:28:13
OK.
 

431
00:28:13 --> 00:28:15
Thanks for those good
questions, another one.

432
00:28:15 --> 00:28:19
AUDIENCE: [INAUDIBLE]
 

433
00:28:19 --> 00:28:20
PROFESSOR STRANG: What would
the dimension look like?

434
00:28:20 --> 00:28:24
Ah, would you like me to show
you something about quadratics?

435
00:28:24 --> 00:28:27
Yeah.
 

436
00:28:27 --> 00:28:30
Shall I jump into quadratics,
it's kind of fun.

437
00:28:30 --> 00:28:37
Quadratics, so let me just do,
so I'll come back to the weak

438
00:28:37 --> 00:28:41
form, right it's totally,
oh I'll do it now.

439
00:28:41 --> 00:28:44
It's so simple I don't
want to forget it.

440
00:28:44 --> 00:28:48
The weak form, so I write the
equation down, -u_xx-u_yy=f.

441
00:28:48 --> 00:28:51
 
 

442
00:28:51 --> 00:28:57
This is the continuous weak
form equal f(x,y), OK?

443
00:28:57 --> 00:29:00
So that's the strong form.
 

444
00:29:00 --> 00:29:04
And I've made it the Laplace
in here to keep it simple,

445
00:29:04 --> 00:29:06
on any right hand side.
 

446
00:29:06 --> 00:29:09
OK, how do I get
to the weak form?

447
00:29:09 --> 00:29:14
Just remind me, I multiply
both sides by any

448
00:29:14 --> 00:29:15
test function v(x,y).
 

449
00:29:17 --> 00:29:23
Multiply by v(x,y), and
then what do I do?

450
00:29:23 --> 00:29:28
I integrate over
the whole region.

451
00:29:28 --> 00:29:35
So that's the weak form,
dxdy, this is for all v,

452
00:29:35 --> 00:29:40
all v(x,y), all, I'll say
all admissible v(x,y).

453
00:29:42 --> 00:29:45
So that's the weak form.
 

454
00:29:45 --> 00:29:51
If this holds for all this
great family of v's, the idea

455
00:29:51 --> 00:29:55
behind it is, that if this
holds for all these trial

456
00:29:55 --> 00:29:59
functions, test functions,
v(x,y), the only way that can

457
00:29:59 --> 00:30:03
happen is for this to
actually equal that.

458
00:30:03 --> 00:30:12
That's a fundamental lemma in
this part of math, and of

459
00:30:12 --> 00:30:17
course it has to be spelled out
more than I'm doing in words.

460
00:30:17 --> 00:30:22
But the idea is that if these
hold for such a large class of

461
00:30:22 --> 00:30:25
v(x,y), then the only way that
can happen is for the

462
00:30:25 --> 00:30:26
strong form to hold.
 

463
00:30:26 --> 00:30:29
For this to actually
match this.

464
00:30:29 --> 00:30:31
OK, so that's the start.
 

465
00:30:31 --> 00:30:35
But then what's the next
step in the weak form?

466
00:30:35 --> 00:30:38
I I like the right hand
side but I'm not so crazy

467
00:30:38 --> 00:30:39
about the left hand side.
 

468
00:30:39 --> 00:30:43
I'm not crazy about it because
this says second derivatives

469
00:30:43 --> 00:30:49
of u, and my little roof
functions, pyramid functions,

470
00:30:49 --> 00:30:51
haven't got second derivatives.
 

471
00:30:51 --> 00:30:55
So I would be dead in the water
without doing the natural step

472
00:30:55 --> 00:30:58
that makes everything
beautiful, which is?

473
00:30:58 --> 00:31:00
Integration by parts.
 

474
00:31:00 --> 00:31:01
Integrate by parts.
 

475
00:31:01 --> 00:31:05
Move derivatives
of of u, on to v.

476
00:31:05 --> 00:31:10
One derivative onto v, off
of u, so then u and v

477
00:31:10 --> 00:31:12
each have one derivative.
 

478
00:31:12 --> 00:31:17
I can use my piecewise linear,
piecewise quadratics, all my

479
00:31:17 --> 00:31:20
finite elements are
going to go fine.

480
00:31:20 --> 00:31:22
So I integrate by parts.
 

481
00:31:22 --> 00:31:28
So integrate by parts, and
what is that mean in 2-D?

482
00:31:28 --> 00:31:30
Of course I have a
double integral here.

483
00:31:30 --> 00:31:37
So integrate by parts, that
mean you the Green's formula.

484
00:31:37 --> 00:31:43
That was the key point
of this Green, or

485
00:31:43 --> 00:31:47
Gauss-Green's, formula.
 

486
00:31:47 --> 00:32:01
Can I do it first in, this is
-div(grad u), times vdxdy, we

487
00:32:01 --> 00:32:03
can write out all the terms.
 

488
00:32:03 --> 00:32:05
We can use vector notation.
 

489
00:32:05 --> 00:32:09
I could use that nabla,
that upside down triangle

490
00:32:09 --> 00:32:10
notation, or whatever.
 

491
00:32:10 --> 00:32:13
But maybe good to see it
a few different ways.

492
00:32:13 --> 00:32:15
So what's the point?
 

493
00:32:15 --> 00:32:20
When I integrate by parts, that
minus disappears to a plus, I

494
00:32:20 --> 00:32:26
have a double integral then,
and these derivatives move off

495
00:32:26 --> 00:32:32
of, I'm taking one derivative
off of here, the divergence

496
00:32:32 --> 00:32:35
moves over there, but when
the divergence moves

497
00:32:35 --> 00:32:38
onto v it becomes?
 

498
00:32:38 --> 00:32:39
The transpose.
 

499
00:32:39 --> 00:32:40
It becomes gradient.
 

500
00:32:40 --> 00:32:52
And so this is gradient
view, gradient of v. dxdy,

501
00:32:52 --> 00:32:54
plus boundary terms.
 

502
00:32:54 --> 00:32:58
The integral of,
what is, let's see.

503
00:32:58 --> 00:33:02
 
 

504
00:33:02 --> 00:33:11
What do I have in this
integral, I have grad u dot n,

505
00:33:11 --> 00:33:14
times v around the boundary.
 

506
00:33:14 --> 00:33:18
And that's with my boundary
conditions that's going

507
00:33:18 --> 00:33:20
to be gone, so I can
come back to that.

508
00:33:20 --> 00:33:26
Now, you all looked a little
uncertain when I wrote

509
00:33:26 --> 00:33:29
Green's formula this way.
 

510
00:33:29 --> 00:33:33
For this problem I can
write it more easily.

511
00:33:33 --> 00:33:36
This is my left side.
 

512
00:33:36 --> 00:33:40
I want to write the answer, I
just want to write this weak

513
00:33:40 --> 00:33:44
form in a much simpler form.
 

514
00:33:44 --> 00:33:49
So let say, what
have I got here.

515
00:33:49 --> 00:33:54
Well, all I've got is one
derivative is moving

516
00:33:54 --> 00:33:56
off of u and on to v.
 

517
00:33:56 --> 00:33:58
And the minus sign is
disappearing, so I have

518
00:33:58 --> 00:34:02
du/dx times dv/dx.
 

519
00:34:02 --> 00:34:03
Right?
 

520
00:34:03 --> 00:34:06
One off of u, onto v.
 

521
00:34:06 --> 00:34:10
The other term, one y
derivative, moving off

522
00:34:10 --> 00:34:12
of this and onto v.
 

523
00:34:12 --> 00:34:15
Minus sign again going to
a plus. du/dy, dv/dy.

524
00:34:15 --> 00:34:19
 
 

525
00:34:19 --> 00:34:20
That's the integral.
 

526
00:34:20 --> 00:34:22
That's it, that's cool.
 

527
00:34:22 --> 00:34:24
Easy to do.
 

528
00:34:24 --> 00:34:29
And on the right hand side
of course I have no change.

529
00:34:29 --> 00:34:31
The integral of
f(x,y)*v(x,y)*dy.

530
00:34:31 --> 00:34:34
 
 

531
00:34:34 --> 00:34:36
Now, that's the
weak form, dx/dy.

532
00:34:36 --> 00:34:44
 
 

533
00:34:44 --> 00:34:52
Here it is, weak form.
 

534
00:34:52 --> 00:34:55
That's pretty nice.
 

535
00:34:55 --> 00:34:58
Beautifully symmetric, though
the matrix that comes up when

536
00:34:58 --> 00:35:04
we plug in finite specific
trial functions and test

537
00:35:04 --> 00:35:07
functions is going to be
a symmetric matrix K.

538
00:35:07 --> 00:35:13
And the integrals of first
derivatives, so as long as our

539
00:35:13 --> 00:35:19
functions, our trial functions
and test functions are

540
00:35:19 --> 00:35:24
continuous, that is,
they shouldn't jump.

541
00:35:24 --> 00:35:28
If the trial functions or test
functions jump, then if I have

542
00:35:28 --> 00:35:31
a jump, then the derivative
would be a delta.

543
00:35:31 --> 00:35:34
I'd have another delta here,
I'd have an integral delta, a

544
00:35:34 --> 00:35:37
delta times delta, and
I don't want that.

545
00:35:37 --> 00:35:39
That's infinite.
 

546
00:35:39 --> 00:35:45
Those discontinuous elements
would not be conforming, and

547
00:35:45 --> 00:35:48
that's a whole new world of
discontinuous Galerkin.

548
00:35:48 --> 00:35:52
I'd have to impose penalty
stuff, and Professor

549
00:35:52 --> 00:35:54
Peraire I mentioned.
 

550
00:35:54 --> 00:36:00
And others, Professor Darmofal
in aero are experts on this.

551
00:36:00 --> 00:36:02
We're doing continuous form.
 

552
00:36:02 --> 00:36:03
CG.
 

553
00:36:04 --> 00:36:07
Our piecewise linear,
piecewise quadratic,

554
00:36:07 --> 00:36:08
they'll be continuous.
 

555
00:36:08 --> 00:36:11
All I have to do is
these derivatives.

556
00:36:11 --> 00:36:14
Integrate those things and
that's what the code will do.

557
00:36:14 --> 00:36:19
OK, I've got to the weak form.
 

558
00:36:19 --> 00:36:22
That's the weak form.
 

559
00:36:22 --> 00:36:25
Now comes the finite
element idea.

560
00:36:25 --> 00:36:28
So there is our weak
form, now ready for the

561
00:36:28 --> 00:36:31
finite element idea.
 

562
00:36:31 --> 00:36:34
OK, so what was that idea?
 

563
00:36:34 --> 00:36:36
That's the continuous problem.
 

564
00:36:36 --> 00:36:47
Now, the finite element idea
is, plug in U as a combination.

565
00:36:47 --> 00:36:52
Let me write out the terms.
 

566
00:36:52 --> 00:36:55
You know what's coming here.
 

567
00:36:55 --> 00:36:58
If I'm using finite elements,
I'm going to choose nice

568
00:36:58 --> 00:37:07
polynomials, phi,
say, N of them.

569
00:37:07 --> 00:37:10
That would be like, one for
every node, so I would

570
00:37:10 --> 00:37:14
have 13 functions here.
 

571
00:37:14 --> 00:37:18
I'm going to choose the v's
to be the same as the phis.

572
00:37:18 --> 00:37:23
And then, I'm working then
in 13 dimensions instead

573
00:37:23 --> 00:37:26
of infinite dimensions.
 

574
00:37:26 --> 00:37:29
So what do I do?
 

575
00:37:29 --> 00:37:33
For this limited subspace, this
finite element subspace, this

576
00:37:33 --> 00:37:38
piecewise polynomial, piecewise
linear subspace, I plug that

577
00:37:38 --> 00:37:44
into the weak form and I test
it against 13 V's,

578
00:37:44 --> 00:37:45
which are phis.
 

579
00:37:45 --> 00:37:53
So I plug that in,
so now what is K?

580
00:37:53 --> 00:38:00
Now let me just say, so I now
have the integral of, yeah I

581
00:38:00 --> 00:38:03
guess I'd better plug it in.
 

582
00:38:03 --> 00:38:07
K_ij would then
be the integral.

583
00:38:07 --> 00:38:09
I'm just copying
the weak form in.

584
00:38:09 --> 00:38:22
Of dU/dx, no, sorry I'd better
just plug it in first.

585
00:38:22 --> 00:38:32
dU/dx*dV/dx plus dU/dy*dV/dy,
those are the integrals

586
00:38:32 --> 00:38:33
I have to do.
 

587
00:38:33 --> 00:38:37
And on the right hand side
I have to do the fV.

588
00:38:37 --> 00:38:40
 
 

589
00:38:40 --> 00:38:44
OK, plug that in.
 

590
00:38:44 --> 00:38:48
That's the integral
over the whole domain.

591
00:38:48 --> 00:38:54
When I plug it in this U
is a combination of known

592
00:38:54 --> 00:39:05
functions and the V's
will be the same guys.

593
00:39:05 --> 00:39:08
So what am I going to get here?
 

594
00:39:08 --> 00:39:10
It's just as in 1-D.
 

595
00:39:10 --> 00:39:14
So no new ideas entering here.
 

596
00:39:14 --> 00:39:20
The new idea's going to enter
when I construct these phis.

597
00:39:20 --> 00:39:23
Let me just say,
though, one thing.

598
00:39:23 --> 00:39:29
In 1-D, we've pretty much
had a choice of, when

599
00:39:29 --> 00:39:31
it was one dimension.
 

600
00:39:31 --> 00:39:34
Just remember that.
 

601
00:39:34 --> 00:39:40
In one dimension, when I had
these hat functions, when I had

602
00:39:40 --> 00:39:44
these guys, integrated against
these guys, I pretty much had a

603
00:39:44 --> 00:39:48
choice of did I want to think
about integrating that hat

604
00:39:48 --> 00:39:50
function against that one.
 

605
00:39:50 --> 00:39:53
Or actually it was
their derivatives.

606
00:39:53 --> 00:39:58
It was the integral of U, yeah.
 

607
00:39:58 --> 00:40:02
Of phi, what I needed
was all the integrals

608
00:40:02 --> 00:40:04
of phi_i', phi_j'.
 

609
00:40:06 --> 00:40:12
Those are what I needed,
these go into K.

610
00:40:12 --> 00:40:13
Into the matrix K.
 

611
00:40:13 --> 00:40:16
In fact, that's what
equals K_ij, the

612
00:40:16 --> 00:40:18
integral of phi prime.
 

613
00:40:18 --> 00:40:20
In 1-D.
 

614
00:40:20 --> 00:40:21
OK.
 

615
00:40:21 --> 00:40:26
Now, what I was going
to say, I could do it

616
00:40:26 --> 00:40:28
this way if I wanted.
 

617
00:40:28 --> 00:40:30
But you remember the
other way to do it?

618
00:40:30 --> 00:40:33
Was elements at a time.
 

619
00:40:33 --> 00:40:36
So this was one method here.
 

620
00:40:36 --> 00:40:43
That found the entries of
K separately, one by one.

621
00:40:43 --> 00:40:47
The other way was take the
elements, one by one.

622
00:40:47 --> 00:40:52
So the other way was take an
element like this element.

623
00:40:52 --> 00:40:56
It's got two functions,
two trial functions

624
00:40:56 --> 00:40:57
are involved there.
 

625
00:40:57 --> 00:41:02
There's a little two by
two, so this is four.

626
00:41:02 --> 00:41:06
Two by two element matrices.
 

627
00:41:06 --> 00:41:08
K equals.
 

628
00:41:08 --> 00:41:12
And the quiz
recalled that part.

629
00:41:12 --> 00:41:13
That approach.
 

630
00:41:13 --> 00:41:16
So what I want to say is
that's the right way to

631
00:41:16 --> 00:41:18
do it in two dimensions.
 

632
00:41:18 --> 00:41:20
A triangle at a time.
 

633
00:41:20 --> 00:41:22
That's the way the
code will do it.

634
00:41:22 --> 00:41:26
It creates these little element
matrices, and then it stamps

635
00:41:26 --> 00:41:30
them into the big matrix K.
 

636
00:41:30 --> 00:41:32
Alright.
 

637
00:41:32 --> 00:41:42
So I want to do this integral
one triangle at a time.

638
00:41:42 --> 00:41:45
Is the good way.
 

639
00:41:45 --> 00:41:47
OK, and that's what
the code will do.

640
00:41:47 --> 00:41:52
Actually, I think that the
best way to learn these

641
00:41:52 --> 00:41:55
steps is just to read
the lines of the code.

642
00:41:55 --> 00:41:59
You can read them in the
book, Page 303 or something.

643
00:41:59 --> 00:42:04
And you'll see it just
doing all the steps

644
00:42:04 --> 00:42:05
that need to be done.
 

645
00:42:05 --> 00:42:07
One triangle at a time.
 

646
00:42:07 --> 00:42:11
So, now.
 

647
00:42:11 --> 00:42:12
Now comes the fun.
 

648
00:42:12 --> 00:42:15
I get to answer what do
these piecewise linear

649
00:42:15 --> 00:42:16
elements look like.
 

650
00:42:16 --> 00:42:18
What do the quadratic
elements look like.

651
00:42:18 --> 00:42:24
What do the Q_1 quad
elements look like?

652
00:42:24 --> 00:42:28
This was the golden age of
finite elements, when people

653
00:42:28 --> 00:42:35
invented these ways to create
piecewise polynomials.

654
00:42:35 --> 00:42:37
And it continues.
 

655
00:42:37 --> 00:42:41
People are still inventing,
I had a email this week,

656
00:42:41 --> 00:42:44
somebody says I've got
spectral elements.

657
00:42:44 --> 00:42:47
People are going higher
and higher degrees.

658
00:42:47 --> 00:42:51
You, know sixth degree,
eighth degree.

659
00:42:51 --> 00:42:53
In order to get more accuracy.
 

660
00:42:53 --> 00:42:55
OK, let's start with P_1.
 

661
00:42:57 --> 00:43:02
How do I describe a P_1
element inside a triangle?

662
00:43:02 --> 00:43:11
So in a triangle, the unknowns
will be the value, this has a

663
00:43:11 --> 00:43:14
height U_1, this has a
height U_2, and a height

664
00:43:14 --> 00:43:17
U_3 at those nodes.
 

665
00:43:17 --> 00:43:24
Inside the triangle, the
function U is linear. a+bx+cy.

666
00:43:24 --> 00:43:31
 
 

667
00:43:31 --> 00:43:37
Then, you see that if I know
these three values, then I

668
00:43:37 --> 00:43:39
know these three numbers.
 

669
00:43:39 --> 00:43:40
And vice versa.
 

670
00:43:40 --> 00:43:43
There's a three by
three matrix, right?

671
00:43:43 --> 00:43:44
There has to be a three by
 

672
00:43:44 --> 00:43:47
- any time you see pictures
like this, this is like the

673
00:43:47 --> 00:43:51
good part of 18.085 is to
realize that if I have three

674
00:43:51 --> 00:43:55
numbers here, three values and
I've got three coefficients,

675
00:43:55 --> 00:43:58
that there's some three by
three matrix that connects them

676
00:43:58 --> 00:43:59
that you're going to need.
 

677
00:43:59 --> 00:44:04
That's like a meta-message
of this course.

678
00:44:04 --> 00:44:13
Is, you've got to translate
between the node values

679
00:44:13 --> 00:44:15
and the coefficients.
 

680
00:44:15 --> 00:44:19
Because the node values
are the unknowns, right?

681
00:44:19 --> 00:44:22
These are the guys that are
multiplying the pyramid

682
00:44:22 --> 00:44:25
function, this is multiplying
a pyramid function

683
00:44:25 --> 00:44:26
with height one.
 

684
00:44:26 --> 00:44:31
At that point, going down to
zero, so this one will be a

685
00:44:31 --> 00:44:36
pyramid function of height U_2
times one, going down to zero.

686
00:44:36 --> 00:44:37
And U_3.
 

687
00:44:38 --> 00:44:42
So we've got a flat
function in here.

688
00:44:42 --> 00:44:44
And it looks exactly like that.
 

689
00:44:44 --> 00:44:46
OK?
 

690
00:44:46 --> 00:44:50
So what do I want to say?
 

691
00:44:50 --> 00:44:54
When we know the positions
of these three nodes

692
00:44:54 --> 00:44:57
from our list, P, right?
 

693
00:44:57 --> 00:45:00
These were the crucial
things we did.

694
00:45:00 --> 00:45:05
The positions of all the nodes,
we know where they are.

695
00:45:05 --> 00:45:10
Then there has to be a three by
three matrix that will now

696
00:45:10 --> 00:45:13
connect to the coefficients.
 

697
00:45:13 --> 00:45:15
Why do we want the
coefficients?

698
00:45:15 --> 00:45:18
Because those are what we
do when we integrate.

699
00:45:18 --> 00:45:21
The coefficients are what we
need, we need to integrate

700
00:45:21 --> 00:45:21
dU/dx, dU/dy, dU/dz.
 

701
00:45:23 --> 00:45:27
Sorry, dU/dx, dU/dy.
 

702
00:45:27 --> 00:45:31
Are you visualizing
this overall solution

703
00:45:31 --> 00:45:36
capital U, yeah.
 

704
00:45:36 --> 00:45:40
So what the overall solution
capital U, you should visualize

705
00:45:40 --> 00:45:44
with a combination of all the
little u's, is zero here and

706
00:45:44 --> 00:45:48
then it's going to go up and
these triangles and bend around

707
00:45:48 --> 00:45:51
and, I don't know,
maybe down again.

708
00:45:51 --> 00:45:53
Or maybe, no, maybe
it keeps going up.

709
00:45:53 --> 00:45:58
This is probably the largest
value, because it's the largest

710
00:45:58 --> 00:46:00
value and the correct solution.
 

711
00:46:00 --> 00:46:11
So is this is probably going to
be the highest point of this.

712
00:46:11 --> 00:46:14
What's the Forbidden
City, right?

713
00:46:14 --> 00:46:20
In China, is in Beijing
is like, or a single,

714
00:46:20 --> 00:46:23
do pagodas have flat?
 

715
00:46:23 --> 00:46:24
No.
 

716
00:46:24 --> 00:46:29
We we will meet
pagoda functions.

717
00:46:29 --> 00:46:34
But this would be just an
ordinary western roof, I guess.

718
00:46:34 --> 00:46:35
Just flat pieces.
 

719
00:46:35 --> 00:46:36
Yeah.
 

720
00:46:36 --> 00:46:40
OK, see, you've got to see
the whole thing and then

721
00:46:40 --> 00:46:42
you look at each piece.
 

722
00:46:42 --> 00:46:47
Each piece looks like that,
and the integrals are doable.

723
00:46:47 --> 00:46:54
OK, so while I'm going here,
I want to do quadratics.

724
00:46:54 --> 00:46:55
You'll get the idea right away.
 

725
00:46:55 --> 00:46:59
So, same triangle, now I'm
going to have quadratics.

726
00:46:59 --> 00:47:02
So I'm now going to have, so
this won't be the arrow, this

727
00:47:02 --> 00:47:04
arrow will now go this way.
 

728
00:47:04 --> 00:47:10
I'm going to have dx squared,
exy, and f y squared.

729
00:47:10 --> 00:47:12
So now how many
coefficients have I got

730
00:47:12 --> 00:47:15
to determine a quadratic?
 

731
00:47:15 --> 00:47:18
Six, right? a, b, c, d, e, f.
 

732
00:47:18 --> 00:47:20
How many nodes do I need?
 

733
00:47:20 --> 00:47:21
Six.
 

734
00:47:21 --> 00:47:22
Where are they?
 

735
00:47:22 --> 00:47:29
Well, the natural positions are
those guys in the mid-point.

736
00:47:29 --> 00:47:33
So now, those are
all nodes now.

737
00:47:33 --> 00:47:38
Some nodes are at vertices
of triangles, some

738
00:47:38 --> 00:47:39
nodes are at midpoint.
 

739
00:47:39 --> 00:47:44
But remember, we've got other
triangles hooking on here, many

740
00:47:44 --> 00:47:48
other triangles, all with
their own six nodes.

741
00:47:48 --> 00:47:51
Well, not their own,
because they share.

742
00:47:51 --> 00:47:53
That's a big point.
 

743
00:47:53 --> 00:47:59
So there's a grid of triangles,
with nodes for quadratic.

744
00:47:59 --> 00:48:02
And we've got one, two, three,
four, five, six, seven, eight,

745
00:48:02 --> 00:48:09
nine, ten, 11, 12, 13, 14,
15, 16 nodes, I think.

746
00:48:09 --> 00:48:13
And within each triangle,
this is what we've got.

747
00:48:13 --> 00:48:17
So there's a six by six
matrix for each triangle.

748
00:48:17 --> 00:48:21
A six by six matrix which will
connect the values U_1, U_2,

749
00:48:21 --> 00:48:29
U_3, U_4, U_5, U_6
for this triangle.

750
00:48:29 --> 00:48:33
Connect those six heights
with these six numbers.

751
00:48:33 --> 00:48:42
And what will the roof look
like within that triangle?

752
00:48:42 --> 00:48:44
Well, sort of curved.
 

753
00:48:44 --> 00:48:45
A parabola, right?
 

754
00:48:45 --> 00:48:49
A parabola somehow in 2-D,
it'll look like this, yeah.

755
00:48:49 --> 00:48:49
Yeah.
 

756
00:48:49 --> 00:48:52
And here's the key question.
 

757
00:48:52 --> 00:48:59
Will that roof, that curvy
roof, fit the one over there?

758
00:48:59 --> 00:49:01
Because if it didn't
fit, we're in trouble.

759
00:49:01 --> 00:49:04
This derivative would have a
delta function, and we've got

760
00:49:04 --> 00:49:07
delta functions, and integral
squaring them would

761
00:49:07 --> 00:49:09
give infinite.
 

762
00:49:09 --> 00:49:10
So here's the question.
 

763
00:49:10 --> 00:49:16
Why does this roof, using these
six points, fit on to the roof

764
00:49:16 --> 00:49:23
that uses U_7, U_8, U_9,
and U_3 U_4 and U_6?

765
00:49:23 --> 00:49:26
Why do those two
roofs fit together?

766
00:49:26 --> 00:49:30
This one piecewise polynomials?
 

767
00:49:30 --> 00:49:33
Of course, the
slope will change.

768
00:49:33 --> 00:49:35
But the roof won't have a gap.
 

769
00:49:35 --> 00:49:37
Water won't go through it.
 

770
00:49:37 --> 00:49:38
Why's that?
 

771
00:49:38 --> 00:49:41
Do you see why?
 

772
00:49:41 --> 00:49:43
Because what do they
share, what do those

773
00:49:43 --> 00:49:46
two curvy roofs share?
 

774
00:49:46 --> 00:49:48
They share a side.
 

775
00:49:48 --> 00:49:53
They share the same
values along the side.

776
00:49:53 --> 00:49:57
And are those three values that
are shared along the side

777
00:49:57 --> 00:50:03
sufficient to make it
match all along the side?

778
00:50:03 --> 00:50:03
Yes.
 

779
00:50:03 --> 00:50:05
That's the important question.
 

780
00:50:05 --> 00:50:08
Finite elements lives or
dies on that question.

781
00:50:08 --> 00:50:14
The answer is yes, because
along that side, if I just

782
00:50:14 --> 00:50:20
focus on that side, where these
three values are shared on both

783
00:50:20 --> 00:50:23
sides, by the triangle
on both sides.

784
00:50:23 --> 00:50:28
Along that edge, what kind
of a function have I got?

785
00:50:28 --> 00:50:31
It's second degree.
 

786
00:50:31 --> 00:50:34
This is whatever, when I
restrict this to just run along

787
00:50:34 --> 00:50:37
a line, it's a parabola.
 

788
00:50:37 --> 00:50:40
And the parabola is determined
by those three values.

789
00:50:40 --> 00:50:42
So having it right at three
points means I have it

790
00:50:42 --> 00:50:44
right the whole way.
 

791
00:50:44 --> 00:50:44
Yeah.
 

792
00:50:44 --> 00:50:49
So there you see what quadratic
elements would look like, and

793
00:50:49 --> 00:50:55
you could extend the code in
the book and on the CSE site to

794
00:50:55 --> 00:50:57
work for quadratic elements.
 

795
00:50:57 --> 00:51:01
And you want to just guess what
cubic elements could look like?

796
00:51:01 --> 00:51:03
I'm sorry, we've run five
minutes over, but maybe

797
00:51:03 --> 00:51:06
finite elements is worth it.
 

798
00:51:06 --> 00:51:13
So if I had cubic elements,
any idea how many?

799
00:51:13 --> 00:51:18
So I'm now going up to, I'm
adding g x cubed, h, i,

800
00:51:18 --> 00:51:25
j, any idea how many
coefficients I now have?

801
00:51:25 --> 00:51:28
Four new ones plus
these six is ten.

802
00:51:28 --> 00:51:29
I need ten nodes.
 

803
00:51:29 --> 00:51:33
Where I am I going to put
ten nodes in this triangle?

804
00:51:33 --> 00:51:36
I want to put them, I'd like
to have some on the edges.

805
00:51:36 --> 00:51:39
Because the edges help me make
triangles match each other.

806
00:51:39 --> 00:51:42
They'll just be like
bowling balls.

807
00:51:42 --> 00:51:52
So here's six, oops, that
wouldn't be believable.

808
00:51:52 --> 00:51:54
Is that right?
 

809
00:51:54 --> 00:51:55
Four, three, two, and one.
 

810
00:51:55 --> 00:51:56
Yeah.
 

811
00:51:56 --> 00:51:57
Yeah.
 

812
00:51:57 --> 00:51:58
OK.
 

813
00:51:58 --> 00:52:06
So, now I've got a bubble node
inside and I've got four nodes

814
00:52:06 --> 00:52:13
of vertices and two points, at
two 1/3 points, and that

815
00:52:13 --> 00:52:16
will then match the
triangle next to it.

816
00:52:16 --> 00:52:19
Because four points
determine a cubic.

817
00:52:19 --> 00:52:23
There you go, I hope you
have fun, I hope you

818
00:52:23 --> 00:52:24
have a great holiday.
 

819
00:52:24 --> 00:52:30
I'll see you Wednesday for
Fourier and always open for

820
00:52:30 --> 00:52:32
questions on the MATLAB.