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PROFESSOR: Let's get started.
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Could I start first with
an announcement of a talk
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this afternoon.
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I know your schedules
are full, but --
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the abstract for the talk I
think will go up on the top
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of our web page.
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It's a whole range
of other applications
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that I would hope to
get to, by an expert.
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Professor Rannacher,
his applications
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include optimal control,
which is certainly
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a big area of optimization.
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Actually, MathWorks, we think
of them as doing linear algebra,
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but their number one customer
is the control theory world,
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and it totally connects
with everything we're doing.
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He's also interested in adaptive
meshing for finite element
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or other methods -- how to
refine the mesh where it pays
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off in arrow problems,
all sorts of problems.
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The math behind it is this
same saddle point structure,
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same KK -- when he
says KKT equations,
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that's our two equations.
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I've been calling them,
sometimes, Kuhn-Tucker,
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but after Kuhn
and Tucker, it was
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noticed that a graduate
student named Karush, also a K,
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had written a Masters thesis in
which these important equations
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appeared.
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So now, they're often
called KKT equations.
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I think Rannacher will do that.
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Anyway, I don't
know if you're free,
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but it'll be a full talk in this
program of computational design
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and optimization.
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I don't know if you know
MIT's new masters degree
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program in CDO.
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So it's mostly engineering,
a little optimization
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that's down in
Operations Research,
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and a couple of guys in math.
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So that's a talk
this afternoon, which
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will be right on
target for this area
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and bring up applications
that are highly important.
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Well, so I thought
today, my job is going
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to be pretty straightforward.
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I want to do now the
continuous problem.
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So I have functions as unknowns.
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I have integrals
as inner products.
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But I still have a
minimization problem.
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So I have a minimization
problem and I'll call,
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it's often a
potential energy, so
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let me use P. Our
unknown function is u,
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so that's a function u of
x; u of x, y; u of x, y, z.
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And instead of inner
products we have integrals,
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so there's a c of x.
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This is going to be
a pure quadratic,
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so it's going to lead
me to a linear equation.
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In between comes
something important --
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what people now call the
weak form of the equation.
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So, this is would
be the simplest
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example I could put forward
of the calculus of variations.
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So that's what
we're talking about.
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Calculus of variations.
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What's the derivative
of P with respect to u?
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Somehow, that's what
we have to find,
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and we're going to set
it to zero to minimize.
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Then why do we know
it's a minimum?
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Well, that's always
the second --
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the quadratic terms here
are going to be positive.
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So we have a positive
definite problem.
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Positive definite means
things go this way, convex,
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and you locate the
bottom, the minimum
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is where the derivative is zero.
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But what's the derivative?
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That's the question.
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It's not even called
derivative in this subject,
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it's called the first variation.
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Instead of saying
first derivative,
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I'll say first variation.
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And instead of writing
dP/du, I'll write --
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where can I write it?
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So this is key word,
so this is going
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to be the first
variation, and I'm
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going to write it with a
different d, a Greek delta,
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dP/du.
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It's just sort of a
reminder that we're
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dealing with functions
and integrals of functions
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and so on.
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So it's just change the
notation in the name
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a little as a trigger
to the memory.
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But this board
really has a lot --
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not all the details
are here, of course.
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This is a summary of what the
calculus of variations does.
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So it takes a
minimization problem.
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So I'm looking for
a function u of x.
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And always since we're
in continuous problems,
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we have boundary conditions.
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So as always, those could be --
let me imagine that those are
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the boundary conditions.
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So I would call those
essential conditions.
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Every function that's
allowed into the minimum
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has to satisfy the
boundary condition.
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So it's a minimum over all u
with the boundary conditions,
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with those boundary conditions.
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There are two kinds of
boundary conditions,
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and maybe I'll postpone thinking
about boundary conditions
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till I get the equations,
the differential
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equation inside the interval.
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So all these intervals
go from zero to one.
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I won't put that.
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We're in 1D.
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So, how do you find
the function, u of x?
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that stands for du/dx.
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So the given data
for the problem
00:07:10.340 --> 00:07:12.985
is some load, some
source term, f
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of x, which is going to show
up on the right-hand side,
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and some coefficient
c of x, which is going
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to show up in the equation.
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They depend on x, in general.
00:07:25.570 --> 00:07:30.330
Many, many, many physical
problems look like this.
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Sort of steady-state
problems, I would say.
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I'm not talking about
Navier-Stokes fluid flow
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convection, I'm talking about
static problems, first of all.
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So here's the general idea of
the calculus of variations.
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It's the same as the
general idea of calculus.
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How do you identify a
minimum in calculus?
00:08:03.860 --> 00:08:09.440
If the minimum is at u,
you perturb it a little,
00:08:09.440 --> 00:08:15.720
by some delta u, that I'm going
to call v to have just one
00:08:15.720 --> 00:08:17.970
letter instead of two here.
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You say OK, if I look at that
neighboring point, u plus delta
00:08:22.260 --> 00:08:29.490
u, my quantity is bigger.
00:08:29.490 --> 00:08:31.460
The minimum is at u.
00:08:35.220 --> 00:08:39.370
So we're remembering calculus --
I guess I'm saying I've written
00:08:39.370 --> 00:08:39.980
that here.
00:08:39.980 --> 00:08:42.470
Compare u with u plus
v, which you could
00:08:42.470 --> 00:08:44.920
think of as u plus delta u.
00:08:44.920 --> 00:08:50.790
It's like -- v you might think
of as a small movement away
00:08:50.790 --> 00:08:53.390
from the best function.
00:08:53.390 --> 00:08:55.740
In calculus it's
a small movement
00:08:55.740 --> 00:08:57.650
away from the best point.
00:08:57.650 --> 00:09:01.300
So let me draw the calculus.
00:09:01.300 --> 00:09:04.950
If you think of this blackboard
as being function space instead
00:09:04.950 --> 00:09:08.730
of just a blackboard, then I'm
doing calculus of variations.
00:09:08.730 --> 00:09:10.540
But let me just
do calculus here.
00:09:10.540 --> 00:09:13.330
So there's the minimum, at u.
00:09:13.330 --> 00:09:16.930
And here is u plus v near it.
00:09:16.930 --> 00:09:20.740
Could be on this side or
it could be on this side.
00:09:20.740 --> 00:09:24.020
Those are both u plus
v. Well, you maybe
00:09:24.020 --> 00:09:26.100
want me to call one
of them u minus v.
00:09:26.100 --> 00:09:31.870
But the point is v
could have either sign.
00:09:31.870 --> 00:09:36.230
I'm looking at minimum
sort of inside, where I can
00:09:36.230 --> 00:09:39.410
go to the right or the left.
00:09:39.410 --> 00:09:41.240
So what's the deal?
00:09:41.240 --> 00:09:46.090
Well, that point is then that
at that point or at that point
00:09:46.090 --> 00:09:50.430
or at any of these other
points, P of u plus v
00:09:50.430 --> 00:09:54.400
is bigger then what
it is at the minimum.
00:09:54.400 --> 00:09:58.110
That tells us that
u is the winner.
00:09:58.110 --> 00:10:04.040
Now how do we get an
equation out of that?
00:10:04.040 --> 00:10:05.860
Calculus comes in now.
00:10:05.860 --> 00:10:10.800
We expand this thing -- this is
some small movement away from
00:10:10.800 --> 00:10:11.890
u.
00:10:11.890 --> 00:10:15.220
So we expand it, we look at
the leading term -- well,
00:10:15.220 --> 00:10:18.420
of course, the leading
term is P of u.
00:10:18.420 --> 00:10:22.730
Then what is the next term?
00:10:22.730 --> 00:10:27.190
What's the first-order,
first variation
00:10:27.190 --> 00:10:32.910
in P when I vary u to u plus v?
00:10:32.910 --> 00:10:37.550
Well, it's the whole point
of calculus, actually.
00:10:37.550 --> 00:10:42.290
The central point of calculus
is that this is some function
00:10:42.290 --> 00:10:49.940
that we call P prime of u times
v, plus order of v squared.
00:10:52.960 --> 00:10:57.420
When v is small, v
squared is very small.
00:10:57.420 --> 00:10:59.640
So what's our equation?
00:10:59.640 --> 00:11:03.490
Well, if this has to be
bigger than P of u --
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I could just cancel P of u --
so this thing has to be bigger
00:11:07.050 --> 00:11:09.600
equal zero now.
00:11:09.600 --> 00:11:12.090
I've squeezed it in a corner,
but since it's calculus
00:11:12.090 --> 00:11:14.580
we kind of remember it.
00:11:14.580 --> 00:11:19.205
Also, it's easy to learn
calculus and forget
00:11:19.205 --> 00:11:20.460
the main point.
00:11:20.460 --> 00:11:23.250
So this has to be
greater equal zero.
00:11:23.250 --> 00:11:26.530
Now this is going to
be -- if v is small,
00:11:26.530 --> 00:11:29.630
this is going to be very
small, so it won't help.
00:11:29.630 --> 00:11:35.910
So this thing had
better be zero, right?
00:11:35.910 --> 00:11:38.750
That had better be zero.
00:11:38.750 --> 00:11:42.700
Because if it isn't zero, I
could take v of the right sign
00:11:42.700 --> 00:11:47.040
to make it positive,
and take v small, so
00:11:47.040 --> 00:11:48.610
that this would dominate this.
00:11:51.440 --> 00:11:53.730
Maybe I wanted to take
v of the right sign
00:11:53.730 --> 00:11:57.190
to make that negative anyway.
00:11:57.190 --> 00:12:00.930
I need P prime of u to be zero.
00:12:00.930 --> 00:12:04.300
So that's what I end up with,
of course, as everybody knew.
00:12:04.300 --> 00:12:08.810
That P prime of
u had to be zero.
00:12:08.810 --> 00:12:13.130
So in that tiny picture,
I've remembered what we know.
00:12:13.130 --> 00:12:19.920
Now let me come
back to what we have
00:12:19.920 --> 00:12:21.310
to do when we have functions.
00:12:23.900 --> 00:12:26.690
So what happened?
00:12:26.690 --> 00:12:31.060
I compare P of u -- u is
now a function, u of x --
00:12:31.060 --> 00:12:39.360
with P of u plus v.
So I plug in u plus v,
00:12:39.360 --> 00:12:42.840
I compare with what
I get with only u,
00:12:42.840 --> 00:12:44.020
and what's the difference?
00:12:44.020 --> 00:12:47.670
I look at the difference
and the difference
00:12:47.670 --> 00:12:56.724
will have a linear term from
u prime plus v prime squared
00:12:56.724 --> 00:12:59.265
and then I'm going to take away
the u prime squared because I
00:12:59.265 --> 00:13:01.070
gotta compare the two.
00:13:01.070 --> 00:13:02.070
So what's left?
00:13:02.070 --> 00:13:04.600
There will be a 2
u prime, v prime.
00:13:04.600 --> 00:13:07.500
I'm maybe just being lazy here.
00:13:07.500 --> 00:13:09.740
I'm asking you to do
it mentally and then
00:13:09.740 --> 00:13:11.720
I'll do it a little better.
00:13:11.720 --> 00:13:15.170
So the difference in
this comparison is a 2 u
00:13:15.170 --> 00:13:18.670
prime v prime times
c, and the 2's cancel.
00:13:18.670 --> 00:13:22.170
There's the difference
right there.
00:13:22.170 --> 00:13:24.540
What's the difference
over in this term?
00:13:24.540 --> 00:13:26.150
Well, I have that
term and then I
00:13:26.150 --> 00:13:29.370
have the same term with u plus
v, and then when I subtract,
00:13:29.370 --> 00:13:34.020
I just have the term with v.
00:13:34.020 --> 00:13:46.610
So this is the dP/du that
has to be zero for every v.
00:13:46.610 --> 00:13:49.440
Now I'm really saying
the important thing.
00:13:49.440 --> 00:13:56.470
This weak form is
like saying this
00:13:56.470 --> 00:14:02.990
has to be zero for every
v. Then, of course,
00:14:02.990 --> 00:14:06.800
in this scalar case, it
was like a very small step
00:14:06.800 --> 00:14:09.580
to decide, well, if this
is zero for every v,
00:14:09.580 --> 00:14:11.720
then that's zero, which
is the strong form.
00:14:15.320 --> 00:14:16.540
Are you with me?
00:14:16.540 --> 00:14:20.190
So we have a minimum
problem, minimize P,
00:14:20.190 --> 00:14:25.600
we have a weak form,
the first variation,
00:14:25.600 --> 00:14:27.760
the first derivative,
the first-order term
00:14:27.760 --> 00:14:33.730
has to be zero for every v. Then
if it's zero for every v, that
00:14:33.730 --> 00:14:39.630
forces the derivative to be
zero and that's the strong form.
00:14:39.630 --> 00:14:47.570
Now over here it
took more space.
00:14:47.570 --> 00:14:50.600
But the ultimate
idea is the same.
00:14:50.600 --> 00:14:53.550
We looked at P of
u plus v compared
00:14:53.550 --> 00:14:56.480
with P of u, subtracted,
looked at the linear term
00:14:56.480 --> 00:14:59.550
and there it is,
the first variation.
00:14:59.550 --> 00:15:05.990
That has to be zero
for every v. I'll just
00:15:05.990 --> 00:15:07.590
mentioned boundary
conditions now,
00:15:07.590 --> 00:15:10.440
as long as we're
at this weak form.
00:15:10.440 --> 00:15:15.570
Don't think of this weak form as
just some mathematical nonsense
00:15:15.570 --> 00:15:17.680
to get to the
differential equation,
00:15:17.680 --> 00:15:20.860
because the weak form is the
foundation for the finite
00:15:20.860 --> 00:15:25.730
element method -- all
sorts of discrete methods,
00:15:25.730 --> 00:15:30.510
discretization methods will
begin with the weak form,
00:15:30.510 --> 00:15:36.980
the weighted integral form,
rather than the strong form.
00:15:36.980 --> 00:15:39.720
I was just going to say a word
about boundary conditions.
00:15:39.720 --> 00:15:41.810
What are the boundary
conditions on v?
00:15:46.460 --> 00:15:55.730
So you could say well, v stands
for a virtual displacement.
00:15:55.730 --> 00:15:59.510
Virtual meaning kind of
we just imagining it,
00:15:59.510 --> 00:16:05.700
it's a displacement that we can
imagine moving by that amount,
00:16:05.700 --> 00:16:11.690
but the whole point is the
nature fix the minimum.
00:16:11.690 --> 00:16:12.940
What's the boundary condition?
00:16:12.940 --> 00:16:17.270
Well, all the candidates have
to satisfy these boundary
00:16:17.270 --> 00:16:17.900
conditions.
00:16:17.900 --> 00:16:22.870
So I have to have u plus v
at zero also has to equal a,
00:16:22.870 --> 00:16:27.050
and u plus v at 1
also has to equal b,
00:16:27.050 --> 00:16:31.370
if I took these simple
boundary conditions.
00:16:31.370 --> 00:16:36.190
So, by subtraction, I learn
the boundary conditions
00:16:36.190 --> 00:16:44.880
on v. When I say all v, I
mean v of zero has to be what?
00:16:44.880 --> 00:16:46.740
Zero.
00:16:46.740 --> 00:16:49.800
And v of 1 has to be zero --
when we had those boundary
00:16:49.800 --> 00:16:50.809
conditions.
00:16:50.809 --> 00:16:53.350
Different problems could bring
different boundary conditions,
00:16:53.350 --> 00:16:58.560
of course, but this is
easier than -- simplest.
00:16:58.560 --> 00:17:01.350
So when I say all v,
I mean every function
00:17:01.350 --> 00:17:05.930
that starts at zero
and ends at zero
00:17:05.930 --> 00:17:07.870
is a candidate in
this weak form.
00:17:10.450 --> 00:17:14.900
I have to get the answer
zero for all those functions.
00:17:14.900 --> 00:17:21.320
Now somehow, I want
to get to this point,
00:17:21.320 --> 00:17:27.710
the differential equation, and
why don't we give it the name
00:17:27.710 --> 00:17:34.100
that everybody -- the two guys'
names, Euler-Lagrange -- well,
00:17:34.100 --> 00:17:35.320
pretty famous names.
00:17:37.850 --> 00:17:39.350
This is the
Euler-Lagrange equation.
00:17:45.580 --> 00:17:49.330
So maybe before where I said
Kuhn-Tucker or something,
00:17:49.330 --> 00:17:52.060
if I'm talking about
differential equations,
00:17:52.060 --> 00:17:55.160
I go back to these guys.
00:17:55.160 --> 00:18:00.720
Now, how did I get from
weak form to strong form?
00:18:00.720 --> 00:18:03.130
That's a key.
00:18:03.130 --> 00:18:07.540
If you see these two steps
from the minimum principle
00:18:07.540 --> 00:18:13.820
to the weak form, that's
just, again, plug in u plus v,
00:18:13.820 --> 00:18:18.430
subtract and take
the linear part.
00:18:18.430 --> 00:18:22.100
Then it's true for all v's.
00:18:22.100 --> 00:18:23.780
Now, how do I get
from here to here?
00:18:26.710 --> 00:18:32.780
Notice that this form
is an integrated form.
00:18:32.780 --> 00:18:35.370
This form is at every point.
00:18:35.370 --> 00:18:39.620
So it's much stronger
and much more demanding.
00:18:42.910 --> 00:18:46.060
You could say OK, that
gives us the equation --
00:18:46.060 --> 00:18:49.450
that's the equation as
we usually see them.
00:18:49.450 --> 00:18:53.060
I'll do 2D and that'll be
Laplace's equation or somebody
00:18:53.060 --> 00:18:59.100
else's equation, but
minimal surface equation,
00:18:59.100 --> 00:19:00.610
all sorts of equations.
00:19:00.610 --> 00:19:01.590
Everything.
00:19:01.590 --> 00:19:04.460
All sorts of
applications including
00:19:04.460 --> 00:19:07.080
this afternoon's lecture.
00:19:07.080 --> 00:19:08.820
How to get from here to here?
00:19:08.820 --> 00:19:14.440
Well, there's one trick in
advanced calculus, actually.
00:19:14.440 --> 00:19:17.050
The most important trick
in advanced calculus
00:19:17.050 --> 00:19:20.550
is integration by parts.
00:19:20.550 --> 00:19:23.070
Well, we use those words,
integration by parts,
00:19:23.070 --> 00:19:24.120
in 1D here.
00:19:26.720 --> 00:19:28.330
So I'll do that.
00:19:28.330 --> 00:19:30.870
We use maybe somebody
else's name --
00:19:30.870 --> 00:19:33.000
Green's formula or
Green's theorem,
00:19:33.000 --> 00:19:39.060
or Green-Gauss or the
divergence theorem or whatever,
00:19:39.060 --> 00:19:40.930
in more dimensions.
00:19:40.930 --> 00:19:43.740
But 1D, I just
integrate by parts.
00:19:43.740 --> 00:19:45.760
Do you remember how
integration by parts goes?
00:19:51.860 --> 00:19:58.120
I want to get v by
itself, but I got v prime,
00:19:58.120 --> 00:20:02.260
because when I plugged in
u plus v here, out came --
00:20:02.260 --> 00:20:04.920
it's the derivative so
I've got a derivative.
00:20:04.920 --> 00:20:07.250
So how do I get rid
of a derivative?
00:20:07.250 --> 00:20:08.770
Integrate by parts.
00:20:08.770 --> 00:20:12.570
Take the derivative off of v --
can I just do that this quick
00:20:12.570 --> 00:20:14.170
way?
00:20:14.170 --> 00:20:19.090
Put the derivative onto
-- I almost said onto u.
00:20:22.070 --> 00:20:25.670
I'm doing integration by parts
and saying how important it is
00:20:25.670 --> 00:20:29.140
but not writing out every step.
00:20:29.140 --> 00:20:33.260
So I take the derivative off
of this and I put it onto this,
00:20:33.260 --> 00:20:42.530
and it's gotta be -- and a minus
sign appears when I do that.
00:20:42.530 --> 00:20:46.130
Where the heck am I gonna
put that minus sign?
00:20:46.130 --> 00:20:48.210
Right in there.
00:20:48.210 --> 00:20:51.950
Minus.
00:20:51.950 --> 00:20:55.060
Then everybody knows that
there's also a boundary term,
00:20:55.060 --> 00:20:56.770
right?
00:20:56.770 --> 00:20:59.660
So I have to squeeze somewhere
in this boundary term.
00:20:59.660 --> 00:21:02.380
So now that I've done
an integration by parts,
00:21:02.380 --> 00:21:09.930
the boundary term will be
the v times the c u prime
00:21:09.930 --> 00:21:16.165
at the ends of the interval.
00:21:16.165 --> 00:21:17.040
I think that's right.
00:21:24.440 --> 00:21:28.670
What we hope is
that that goes away.
00:21:28.670 --> 00:21:32.230
Well, of course, if it
doesn't, then we have to --
00:21:32.230 --> 00:21:35.520
that there's a good reason that
we don't want it to, but here
00:21:35.520 --> 00:21:37.490
it's nice if it goes away.
00:21:37.490 --> 00:21:40.630
You see that it
does, because we just
00:21:40.630 --> 00:21:43.110
decided that the
boundary conditions on v
00:21:43.110 --> 00:21:48.070
were zero at both ends.
00:21:48.070 --> 00:21:52.900
So, v being zero at both
ends kills that term.
00:21:52.900 --> 00:21:55.410
So now, do you see
what I have here?
00:21:55.410 --> 00:21:59.250
I could write it a
little more cleanly.
00:21:59.250 --> 00:22:02.910
The whole point is that the v --
I now have v there and I have v
00:22:02.910 --> 00:22:05.180
there so I can
factor v out of this.
00:22:09.080 --> 00:22:09.900
Just put it there.
00:22:14.100 --> 00:22:15.120
That was a good move.
00:22:18.020 --> 00:22:20.500
This minus sign
is still in here.
00:22:20.500 --> 00:22:28.170
So I now have the integral of
some function times v is zero.
00:22:28.170 --> 00:22:30.240
That's what I'm looking for.
00:22:30.240 --> 00:22:36.710
The integral of some
function, some stuff, times v,
00:22:36.710 --> 00:22:40.230
and v can be
anything -- is zero.
00:22:40.230 --> 00:22:44.200
What happens now?
00:22:44.200 --> 00:22:47.360
This integral has to
be zero for every v.
00:22:47.360 --> 00:22:50.930
So if this stuff had
a little bump up,
00:22:50.930 --> 00:22:54.030
I could take a v to
have the same bump
00:22:54.030 --> 00:22:57.640
and the integral
wouldn't be zero.
00:22:57.640 --> 00:23:01.740
So this stuff can't bump
up, it can't bump down,
00:23:01.740 --> 00:23:02.840
it can't do anything.
00:23:02.840 --> 00:23:06.560
It has to be zero and
that's the strong form.
00:23:06.560 --> 00:23:10.030
So the strong form is with
this minus sign in there,
00:23:10.030 --> 00:23:16.120
minus the derivative -- see,
an extra derivative came onto
00:23:16.120 --> 00:23:20.230
the c u prime because
it came off the v,
00:23:20.230 --> 00:23:25.980
and the f was just sitting
there in the linear,
00:23:25.980 --> 00:23:28.310
in the no derivative.
00:23:33.640 --> 00:23:35.630
So, do you see that pattern?
00:23:35.630 --> 00:23:39.060
You may have seen it before,
but calculus variations
00:23:39.060 --> 00:23:41.560
have sort of
disappeared as a subject
00:23:41.560 --> 00:23:50.590
to teach in advanced calculus.
00:23:50.590 --> 00:23:54.110
It used to be here in courses
that Professor Hildebrand
00:23:54.110 --> 00:23:55.420
taught.
00:23:55.420 --> 00:23:59.560
But actually it comes back
because we so much need
00:23:59.560 --> 00:24:05.600
the weak form in finite
elements and other methods.
00:24:05.600 --> 00:24:14.250
What I wrote over here is
the discrete equivalent.
00:24:14.250 --> 00:24:24.580
I can't resist looking at the
matrix form, for two reasons.
00:24:24.580 --> 00:24:27.300
First, it's simpler
and it copies this.
00:24:27.300 --> 00:24:29.960
Do you see how this
is a copy of that?
00:24:29.960 --> 00:24:32.760
That matrix form
is supposed to be
00:24:32.760 --> 00:24:36.560
exactly analogous to
this continuous form.
00:24:36.560 --> 00:24:37.880
Why is that?
00:24:37.880 --> 00:24:42.540
Because u transpose f, that's
an inner product of u with f --
00:24:42.540 --> 00:24:44.040
that's what this is.
00:24:44.040 --> 00:24:45.320
That integral.
00:24:45.320 --> 00:24:52.610
A*u is u prime in this analogy.
00:24:52.610 --> 00:24:57.950
So this is u prime times c
times u prime with the 1/2,
00:24:57.950 --> 00:25:02.711
and that, again, that transpose
is telling us inner product
00:25:02.711 --> 00:25:03.210
integral.
00:25:06.910 --> 00:25:11.210
So if I forget u prime and
think of it as a matrix problem,
00:25:11.210 --> 00:25:13.860
that's my minimum
problem for matrices.
00:25:20.730 --> 00:25:25.560
I want to find an equation
for the winning u.
00:25:25.560 --> 00:25:29.710
In the end, this is
going to be the equation.
00:25:29.710 --> 00:25:31.850
That's the equation
that minimizes that.
00:25:34.750 --> 00:25:39.390
Half of 18.085 was
about this problem.
00:25:39.390 --> 00:25:41.979
Well, I concentrated
in 18.085 on this one,
00:25:41.979 --> 00:25:44.520
because minimum principles are
just that little bit trickier,
00:25:44.520 --> 00:25:47.160
so that's 18.086.
00:25:47.160 --> 00:25:52.650
And then, in between,
something people seldom
00:25:52.650 --> 00:25:55.470
write about but, of
course, it's going to work,
00:25:55.470 --> 00:26:07.070
is that I change u to u plus v,
I multiply it out, I subtract,
00:26:07.070 --> 00:26:09.960
I look at the term linear
in v, and that's it.
00:26:14.280 --> 00:26:18.390
That would be if I make
that just a minus sign
00:26:18.390 --> 00:26:20.650
and put it all together
the way I did here,
00:26:20.650 --> 00:26:22.900
that's the same thing.
00:26:22.900 --> 00:26:24.270
So this is the weak form.
00:26:24.270 --> 00:26:27.824
This is the minimum form,
this is the weak form,
00:26:27.824 --> 00:26:28.990
and this is the strong form.
00:26:36.900 --> 00:26:40.020
You see that weak form?
00:26:40.020 --> 00:26:48.560
Somehow in the discrete case,
it's pretty clear that --
00:26:48.560 --> 00:26:49.290
let's see.
00:26:49.290 --> 00:26:55.440
I could write this as --
you see, it's u transpose,
00:26:55.440 --> 00:27:00.770
A transpose C*A*v
equal f transpose v.
00:27:00.770 --> 00:27:04.280
The conclusion is if
this holds for every v,
00:27:04.280 --> 00:27:07.820
then this is the same as this.
00:27:07.820 --> 00:27:11.560
If two things have the same
inner product with every vector
00:27:11.560 --> 00:27:14.630
v, they're the same, and
that's the strong form.
00:27:17.410 --> 00:27:20.730
You'd have to transpose the
whole thing, but no problem.
00:27:23.530 --> 00:27:32.490
So now I guess I've tried to
give the main sequence of logic
00:27:32.490 --> 00:27:37.260
in the continuous
case, and it's parallel
00:27:37.260 --> 00:27:42.020
in the discrete case
for this example.
00:27:42.020 --> 00:27:45.320
For this specific example
because it's the easiest.
00:27:45.320 --> 00:27:54.460
Let me do what Euler and
Lagrange did by extending
00:27:54.460 --> 00:27:57.170
to a larger class of examples.
00:27:57.170 --> 00:28:01.680
So now our minimization
is still an integral --
00:28:01.680 --> 00:28:04.880
I'll still stay in 1D, I'll
still keep these boundary
00:28:04.880 --> 00:28:10.620
conditions, but I'm going
to allow some more general
00:28:10.620 --> 00:28:13.390
expression here.
00:28:13.390 --> 00:28:17.040
Instead of that pure quadratic,
this could be whatever.
00:28:19.630 --> 00:28:21.650
Now I'm going to do
calculus of variations.
00:28:26.340 --> 00:28:27.950
Still in 1D.
00:28:27.950 --> 00:28:33.700
Calculus of variations,
minimize the integral
00:28:33.700 --> 00:28:45.270
of some function of u and
u prime with the boundary
00:28:45.270 --> 00:28:47.170
conditions, and I'll
keep those nice.
00:28:47.170 --> 00:28:50.150
So that integral's
still 0 to 1 and I'll
00:28:50.150 --> 00:28:54.050
keep these nice
boundary conditions just
00:28:54.050 --> 00:28:56.570
to make my life easy.
00:28:56.570 --> 00:29:01.080
So that will lead to v of
zero being zero, and v of 1
00:29:01.080 --> 00:29:02.770
being zero.
00:29:10.930 --> 00:29:12.690
What do I have to do?
00:29:12.690 --> 00:29:16.460
Again, I have to
plug in u plus v
00:29:16.460 --> 00:29:20.350
and compare this result with
the same thing having u plus v.
00:29:20.350 --> 00:29:24.840
So essentially I've
got to compare F
00:29:24.840 --> 00:29:31.900
at u plus v, u plus u prime
plus v prime with F of u and u
00:29:31.900 --> 00:29:34.350
prime.
00:29:34.350 --> 00:29:40.550
I have to find the leading
term in the difference.
00:29:44.720 --> 00:29:46.450
So I'll just find
out leading term,
00:29:46.450 --> 00:29:50.170
and then will come the integral.
00:29:50.170 --> 00:29:53.040
But the first job is
really the leading term,
00:29:53.040 --> 00:29:56.790
and it's calculus, of course.
00:29:56.790 --> 00:29:59.730
Now can we do that one?
00:29:59.730 --> 00:30:00.970
It's pure calculus.
00:30:06.090 --> 00:30:11.130
I have a function
at two variables,
00:30:11.130 --> 00:30:13.610
the function of u and u prime.
00:30:13.610 --> 00:30:15.920
Actually, I did here.
00:30:15.920 --> 00:30:19.050
I had a u prime
there and a u there.
00:30:26.650 --> 00:30:30.000
Once I write it down you're
going to say sure, of course,
00:30:30.000 --> 00:30:32.220
I knew that.
00:30:32.220 --> 00:30:34.050
So I have a function
of two variables
00:30:34.050 --> 00:30:39.070
and I'm looking for
a little change.
00:30:39.070 --> 00:30:48.700
So a little change in the first
argument produces the dF --
00:30:48.700 --> 00:30:55.040
the derivative of F with respect
to that first argument times
00:30:55.040 --> 00:31:01.160
the delta u, which is
what I'm calling v.
00:31:01.160 --> 00:31:05.431
That's the part that the
dependence on u is responsible
00:31:05.431 --> 00:31:05.930
for.
00:31:05.930 --> 00:31:09.400
Now there's also a
dependence on u prime.
00:31:09.400 --> 00:31:13.340
So I have the derivative
of F with respect
00:31:13.340 --> 00:31:19.400
to u prime times the little
movement in u prime, which
00:31:19.400 --> 00:31:20.200
is v prime.
00:31:26.540 --> 00:31:29.800
I can't leave it
with an equal there,
00:31:29.800 --> 00:31:34.080
because that's only
the linearized part,
00:31:34.080 --> 00:31:36.630
but that's all I
really care about.
00:31:36.630 --> 00:31:43.700
This is order of v squared
and v prime squared.
00:31:43.700 --> 00:31:48.920
Higher order, which
is not going to --
00:31:48.920 --> 00:31:52.750
when I think of v as
small, v prime as small,
00:31:52.750 --> 00:31:58.200
then the linear part dominates.
00:31:58.200 --> 00:32:04.247
So can you see
what dP/du is now?
00:32:04.247 --> 00:32:05.330
Now I've got to integrate.
00:32:09.170 --> 00:32:12.510
I integrate this,
that very same thing.
00:32:12.510 --> 00:32:18.460
This -- dx.
00:32:18.460 --> 00:32:28.320
That has to be zero for all
v. You don't mind if I lazily
00:32:28.320 --> 00:32:35.110
don't copy that into the
-- that's the weak form.
00:32:38.750 --> 00:32:43.930
This was the minimum form,
now I've got to the weak form.
00:32:43.930 --> 00:32:46.310
This is the first variation.
00:32:46.310 --> 00:32:56.470
The integral of the change in F,
which has two components, when
00:32:56.470 --> 00:32:59.510
there's a little change in u.
00:33:04.960 --> 00:33:08.170
I should be doing an
example, but allow
00:33:08.170 --> 00:33:14.280
me to just keep going here
until we get to the strong form.
00:33:14.280 --> 00:33:16.690
So this is the weak
form for every v.
00:33:16.690 --> 00:33:19.600
Let me just repeat that the
weak form is quite important
00:33:19.600 --> 00:33:22.880
because, in the
finite element method,
00:33:22.880 --> 00:33:27.280
we have the v's are
the test functions,
00:33:27.280 --> 00:33:31.770
and we discretize
the v's -- you know,
00:33:31.770 --> 00:33:33.730
we have a finite number
of test functions.
00:33:33.730 --> 00:33:41.050
Well, I can't go entirely
-- I'll come back to finite
00:33:41.050 --> 00:33:43.250
elements.
00:33:43.250 --> 00:33:45.815
Let me stay with this
continuous problem, calculus
00:33:45.815 --> 00:33:47.550
of variations problem.
00:33:47.550 --> 00:33:55.490
v is v of x here, and it
satisfies these boundary
00:33:55.490 --> 00:33:55.990
conditions.
00:33:55.990 --> 00:34:01.880
That's the only requirement that
we need to think about here.
00:34:01.880 --> 00:34:02.950
What's the strong form?
00:34:05.780 --> 00:34:11.510
Also called the
Euler-Lagrange equation.
00:34:14.100 --> 00:34:15.530
How do I get to
that strong form?
00:34:15.530 --> 00:34:17.350
How did I get to it before?
00:34:17.350 --> 00:34:23.230
I would like to get this
into something times v. Here
00:34:23.230 --> 00:34:25.570
I've got something
times v but I've also
00:34:25.570 --> 00:34:27.380
got something times v prime.
00:34:31.560 --> 00:34:34.910
I wanted to get it
like up here where
00:34:34.910 --> 00:34:37.560
it was times v. What do I do?
00:34:37.560 --> 00:34:40.630
You know what I do.
00:34:40.630 --> 00:34:43.420
There's only the one idea here.
00:34:43.420 --> 00:34:46.020
Integrate by parts.
00:34:46.020 --> 00:34:54.450
So I have the integral --
well, dF/du*v was no problem --
00:34:54.450 --> 00:34:56.190
that had the v that I like.
00:34:59.130 --> 00:35:06.000
But it's this other guy that
has a v prime and I want v. So,
00:35:06.000 --> 00:35:11.410
it doesn't take
too much thinking.
00:35:11.410 --> 00:35:15.740
Integrate by parts, take
the derivative off of v,
00:35:15.740 --> 00:35:19.240
get a minus sign, and
put a derivative --
00:35:19.240 --> 00:35:22.650
can I do it with a prime,
but I'll do better below --
00:35:22.650 --> 00:35:26.260
onto this, and then
there's a boundary term,
00:35:26.260 --> 00:35:29.110
but the boundary term goes
away because of the boundary
00:35:29.110 --> 00:35:32.620
condition.
00:35:32.620 --> 00:35:45.450
So now I have the v. Can
I make it on this board?
00:35:45.450 --> 00:35:49.910
There's the dF*du multiplying
v, and then there's the minus --
00:35:49.910 --> 00:35:56.810
this is the derivative d
by dx of dF / d u prime.
00:35:56.810 --> 00:36:02.680
Now all that is
multiplying v and giving me
00:36:02.680 --> 00:36:04.500
the integral of zero.
00:36:04.500 --> 00:36:10.450
I promise to write
that bigger now.
00:36:10.450 --> 00:36:13.180
But, again, the
central point was
00:36:13.180 --> 00:36:18.740
to get the linear term times v.
That's always the main point.
00:36:18.740 --> 00:36:20.940
Then what's the conclusion?
00:36:20.940 --> 00:36:24.780
What's the
Euler-Lagrange equation?
00:36:24.780 --> 00:36:32.150
This integral is this quantity
times v and v can be anything,
00:36:32.150 --> 00:36:34.680
and I've got to get zero.
00:36:34.680 --> 00:36:38.120
So, what's the equation?
00:36:38.120 --> 00:36:41.430
That stuff in brackets is zero.
00:36:41.430 --> 00:36:43.440
That's the
Euler-Lagrange equation.
00:36:43.440 --> 00:36:48.050
Finally, let me just
write it down here.
00:36:48.050 --> 00:36:52.020
Euler-Lagrange strong form.
00:36:55.740 --> 00:37:07.440
In this general problem it would
be dF/du minus d by dx of dF /
00:37:07.440 --> 00:37:11.840
d u prime equals zero.
00:37:11.840 --> 00:37:14.250
Would you like me
to put on what a --
00:37:14.250 --> 00:37:17.940
if F happened to depend
on u double prime,
00:37:17.940 --> 00:37:22.390
it would be a plus -- this
would be what would happen --
00:37:22.390 --> 00:37:29.640
just so you get the
pattern -- equaling zero.
00:37:29.640 --> 00:37:33.430
So I've gone one step
further by allowing
00:37:33.430 --> 00:37:38.730
f to depend on curvature,
and writing down
00:37:38.730 --> 00:37:45.320
what the resulting term would be
in the Euler-Lagrange equation.
00:37:45.320 --> 00:37:49.650
Could you figure out
why it would be that?
00:37:49.650 --> 00:37:52.420
Where would this
thing have come from?
00:37:52.420 --> 00:37:55.705
It would have come from --
there would have been a dF / d u
00:37:55.705 --> 00:38:01.690
double prime times v double
prime in the weak form.
00:38:01.690 --> 00:38:06.020
Then I would have done two
integrations by parts --
00:38:06.020 --> 00:38:12.530
two minus signs making a plus,
two derivatives moving off of v
00:38:12.530 --> 00:38:19.310
and onto the other thing
and it would give me that.
00:38:19.310 --> 00:38:23.910
All times v, but now I've
got everything times v,
00:38:23.910 --> 00:38:29.640
it's true for every v, so
the quantity has to be zero.
00:38:29.640 --> 00:38:32.440
That's the
Euler-Lagrange equation,
00:38:32.440 --> 00:38:34.690
strong form for 1D problem.
00:38:41.850 --> 00:38:42.620
Yes?
00:38:42.620 --> 00:38:45.810
AUDIENCE: [INAUDIBLE PHRASE]?
00:38:45.810 --> 00:38:47.040
PROFESSOR: Oh, you're right.
00:38:47.040 --> 00:38:48.750
You're right.
00:38:48.750 --> 00:38:53.900
If we had this situation
then we would be up to --
00:38:53.900 --> 00:38:57.880
this would typically be a
fourth-order equation and we
00:38:57.880 --> 00:39:00.390
would have two boundary
conditions at each end.
00:39:00.390 --> 00:39:02.450
Absolutely.
00:39:02.450 --> 00:39:06.310
When I slipped in this just
to sort of show the pattern,
00:39:06.310 --> 00:39:09.180
I didn't account for
the boundary conditions.
00:39:09.180 --> 00:39:14.490
That would be one
level up as well.
00:39:14.490 --> 00:39:15.620
Exactly.
00:39:15.620 --> 00:39:19.560
And nature -- well, fortunately
we don't get many equations
00:39:19.560 --> 00:39:23.220
of higher degree than four.
00:39:23.220 --> 00:39:28.050
This would be like the beam
equation, the parallel to that
00:39:28.050 --> 00:39:32.790
would be a beam
equation or a plate
00:39:32.790 --> 00:39:35.810
equation or a shell
equation, God forbid.
00:39:35.810 --> 00:39:38.630
In shell theory
they're incredibly
00:39:38.630 --> 00:39:41.500
complicated because
they're on surfaces.
00:39:41.500 --> 00:39:44.870
But the pattern is always this.
00:39:47.790 --> 00:39:49.250
Well, and, of
course, they're also
00:39:49.250 --> 00:39:52.060
complicated because
they're in 2D.
00:39:52.060 --> 00:39:54.340
So maybe I should
say a little bit.
00:39:54.340 --> 00:40:00.830
Could I maybe write -- I'm
trying to do a lot today.
00:40:00.830 --> 00:40:05.260
Trying to do kind of
the formal stuff today.
00:40:05.260 --> 00:40:07.420
So another step in
the formal stuff
00:40:07.420 --> 00:40:12.520
would be to get into 2D,
which I haven't done.
00:40:15.400 --> 00:40:18.260
What would be the
famous 2D problem that
00:40:18.260 --> 00:40:20.430
leads to Laplace's equation?
00:40:20.430 --> 00:40:25.470
So 2D, what would I
minimize, just to --
00:40:25.470 --> 00:40:35.270
so P of u would be a double
integral of du/dx squared --
00:40:35.270 --> 00:40:41.830
maybe times a c
-- du/dy squared.
00:40:41.830 --> 00:40:47.120
I would really like 1/2 on
that just to make life good.
00:40:47.120 --> 00:40:51.580
Then minus a double
integral of f times u.
00:40:51.580 --> 00:41:02.860
dx*dy, to emphasize these
are double integrals.
00:41:02.860 --> 00:41:06.870
That's a very, very
important problem.
00:41:06.870 --> 00:41:10.410
Many people have
tried to study that.
00:41:10.410 --> 00:41:14.980
Euler and Lagrange would
produce an equation for it.
00:41:14.980 --> 00:41:17.100
You might say OK, solve
the equation, that
00:41:17.100 --> 00:41:18.350
finishes the problem.
00:41:18.350 --> 00:41:23.890
But mathematicians are always
worried, is there a solution?
00:41:23.890 --> 00:41:25.840
Is there a minimum?
00:41:25.840 --> 00:41:30.230
So I'm dodging the
bullet on that one.
00:41:30.230 --> 00:41:33.490
When I say minimize
over all functions u,
00:41:33.490 --> 00:41:40.530
I could create problems where
worse and worse and worse
00:41:40.530 --> 00:41:43.710
functions got closer
and closer to a minimum
00:41:43.710 --> 00:41:47.150
and there was no
limiting minimizer.
00:41:47.150 --> 00:41:49.610
I won't do that.
00:41:49.610 --> 00:41:52.860
This is a problem
that works fine.
00:41:52.860 --> 00:41:54.280
What happens to it?
00:41:54.280 --> 00:41:59.340
Well, should we try to
do the same weak form?
00:42:02.410 --> 00:42:07.610
We'd have a dP/du, and what do
you think it would look like?
00:42:07.610 --> 00:42:12.260
It would have the
integral, double integral.
00:42:12.260 --> 00:42:13.650
What would it have?
00:42:13.650 --> 00:42:21.100
It will have a c*du/dx
and there will be a dv/dx.
00:42:21.100 --> 00:42:23.040
But then integration
by parts will
00:42:23.040 --> 00:42:28.190
take that x derivative off of
v and onto this, with a minus.
00:42:35.690 --> 00:42:39.430
I did it fast, did it way fast.
00:42:39.430 --> 00:42:43.980
Then out of this thing will
come, if I look at the v term
00:42:43.980 --> 00:42:48.760
they'll be a dv/dy and I
take that y derivative off
00:42:48.760 --> 00:42:51.290
of that and onto this.
00:42:51.290 --> 00:42:55.200
Oh, I can't use d anymore,
I have to use partials.
00:42:55.200 --> 00:42:57.910
d by du of c*du/dy.
00:43:00.870 --> 00:43:05.060
Then the f is just
sitting there.
00:43:08.050 --> 00:43:17.030
Oh, it's all multiplied by
v. dx/dy I didn't put yet.
00:43:17.030 --> 00:43:32.770
Equals zero for all v.
Again, this is the same thing
00:43:32.770 --> 00:43:38.330
that we had in ordinary calculus
where it was just P prime of u,
00:43:38.330 --> 00:43:39.740
v equals zero.
00:43:43.940 --> 00:43:49.980
This is a level of
sophistication up.
00:43:49.980 --> 00:43:53.820
It's producing
differential equations, not
00:43:53.820 --> 00:43:56.480
scalar equations.
00:43:56.480 --> 00:43:59.180
So what's the deal?
00:43:59.180 --> 00:44:01.060
I've done the
integration by parts,
00:44:01.060 --> 00:44:04.000
so I've got everything
multiplying a v.
00:44:04.000 --> 00:44:15.020
So what's the strong
form of this problem?
00:44:17.970 --> 00:44:25.550
Well, if this integral has
to be zero for every v,
00:44:25.550 --> 00:44:30.910
then the conclusion is that this
stuff in the brackets is zero.
00:44:30.910 --> 00:44:33.100
That's always the same.
00:44:33.100 --> 00:44:36.040
That's the strong form.
00:44:36.040 --> 00:44:38.590
So that's Laplace's
equation or actually
00:44:38.590 --> 00:44:41.970
Poisson's equation because I
have a right-hand side, f of x.
00:44:45.470 --> 00:44:47.360
Well, those are the mechanics.
00:44:47.360 --> 00:44:55.210
Now, what else comes into
the calculus of variations?
00:44:55.210 --> 00:44:57.630
You've seen the pattern here.
00:44:57.630 --> 00:45:01.560
There's one important
further possibility
00:45:01.560 --> 00:45:07.940
that we met last time,
which was constraints.
00:45:07.940 --> 00:45:10.320
We're dealing here with
a pure minimization.
00:45:13.370 --> 00:45:17.060
I didn't impose
any side conditions
00:45:17.060 --> 00:45:21.740
on u except maybe the
boundary condition.
00:45:21.740 --> 00:45:27.910
So let me give you
an example which --
00:45:27.910 --> 00:45:35.460
I'll just close with an example
that I'm going to follow up,
00:45:35.460 --> 00:45:39.340
and it's going to
have a constraint.
00:45:39.340 --> 00:45:44.960
So it'll look like the
original problem, but, well,
00:45:44.960 --> 00:45:46.930
there will be two u's.
00:45:46.930 --> 00:45:49.510
So can I try to
get it right here?
00:45:53.740 --> 00:46:04.580
My minimization, my unknown
will have two components, u_1
00:46:04.580 --> 00:46:05.080
and u_2.
00:46:09.770 --> 00:46:13.980
And it'll be in 2D actually.
00:46:13.980 --> 00:46:21.860
What I'm going to produce here
is called the Stokes problem.
00:46:21.860 --> 00:46:25.270
I'll study it next time, so if
I run out of time, as I probably
00:46:25.270 --> 00:46:28.020
will, that's part of the plan.
00:46:31.010 --> 00:46:34.710
So it's Stokes and
not Navier-Stokes.
00:46:34.710 --> 00:46:38.050
All I want to do is to write
down a problem in which there
00:46:38.050 --> 00:46:40.120
is a constraint.
00:46:40.120 --> 00:46:48.730
So I write it as a minimization,
say, dv_1/ d -- oh,
00:46:48.730 --> 00:46:50.830
probably all these
guys are in here.
00:46:50.830 --> 00:47:00.990
dv_1/dx and dv_1/dy and
dv_2/dx and dv_2/dy --
00:47:00.990 --> 00:47:02.470
sorry about all this stuff.
00:47:05.360 --> 00:47:13.290
Probably an f_1*v --
oh, I've written v,
00:47:13.290 --> 00:47:16.150
because my mind is saying
that the usual notation,
00:47:16.150 --> 00:47:20.050
I should be writing u
because that would fit with
00:47:20.050 --> 00:47:22.120
today's lecture.
00:47:22.120 --> 00:47:25.400
It's a velocity and that's
why many people call it v,
00:47:25.400 --> 00:47:30.450
and then they have to call
the perturbation some w.
00:47:30.450 --> 00:47:35.240
f_1*u_1, f_2*u_2,
all that stuff.
00:47:35.240 --> 00:47:36.960
No problem.
00:47:36.960 --> 00:47:39.820
That would lead to Laplace's
equation, just the same,
00:47:39.820 --> 00:47:42.440
Poisson's equation.
00:47:42.440 --> 00:47:47.140
But I'm going to
add a constraint.
00:47:47.140 --> 00:47:54.980
This [u 1, u 2] is a
velocity, despite the letter.
00:47:54.980 --> 00:48:00.530
I want to make the
material incompressible.
00:48:00.530 --> 00:48:06.740
I have flow here and
it's like flow of water,
00:48:06.740 --> 00:48:09.580
probably incompressible.
00:48:09.580 --> 00:48:24.042
So incompressible means that
dv_1/dx plus dv_2/dx is zero.
00:48:24.042 --> 00:48:25.000
So that's a constraint.
00:48:31.160 --> 00:48:32.550
How the heck do we deal with it?
00:48:35.790 --> 00:48:38.870
So I could do this minimization
but with the constraint.
00:48:45.380 --> 00:48:50.260
So all this stuff I'm
totally cool with now.
00:48:50.260 --> 00:48:53.310
That would just be
calculus of variations,
00:48:53.310 --> 00:48:56.130
that would get me
the dP/du, but I have
00:48:56.130 --> 00:48:57.490
to account for the constraint.
00:48:57.490 --> 00:49:01.230
So how do you account
for a constraint?
00:49:01.230 --> 00:49:06.790
You build it into the problem
with a Lagrange multiplier.
00:49:06.790 --> 00:49:11.140
So I multiply this thing by
some Lagrange multiplier,
00:49:11.140 --> 00:49:14.360
and as I emphasized last time,
Lagrange multipliers always
00:49:14.360 --> 00:49:16.940
turn out to mean
something physically,
00:49:16.940 --> 00:49:18.980
and here it's the pressure.
00:49:18.980 --> 00:49:26.650
So it's natural to call the
Lagrange multiplier p of x, y.
00:49:26.650 --> 00:49:34.320
I build that in, so I subtract
the Lagrange multiplier
00:49:34.320 --> 00:49:38.820
times this thing
that has to be zero.
00:49:38.820 --> 00:49:42.050
That gets in the problem.
00:49:42.050 --> 00:49:50.380
Now my function now depends
on u and the pressure.
00:49:54.690 --> 00:50:00.420
I'm not going to push
this to the very limit
00:50:00.420 --> 00:50:02.520
to find the strong form.
00:50:02.520 --> 00:50:04.940
But the strong form is
the Stokes equations
00:50:04.940 --> 00:50:06.960
that we'll study.
00:50:06.960 --> 00:50:12.960
So we have a lot to
do here to make this
00:50:12.960 --> 00:50:18.060
into practical calculations
where we can compute something.
00:50:18.060 --> 00:50:22.560
And finite elements is
a powerful way to do it.
00:50:22.560 --> 00:50:25.570
So we have to turn these
continuous problems
00:50:25.570 --> 00:50:27.460
into discrete problems.
00:50:27.460 --> 00:50:32.490
And then, later, we have to
turn this type of problem, which
00:50:32.490 --> 00:50:35.910
will be a saddle point problem
because it's got this Lagrange
00:50:35.910 --> 00:50:40.010
multiplier in there,
into a discrete problem.
00:50:40.010 --> 00:50:45.910
Let me just stop by putting
the words saddle point there,
00:50:45.910 --> 00:50:52.690
and just as in the lecture
this afternoon, saddle points
00:50:52.690 --> 00:50:57.190
appear as soon as you have
constraints and Lagrange
00:50:57.190 --> 00:50:59.320
multipliers.
00:50:59.320 --> 00:51:00.750
Well, thanks for your patience.
00:51:00.750 --> 00:51:05.100
That's a lot of material
that will quickly -- Now,
00:51:05.100 --> 00:51:11.260
those basic steps will be
section 7.2 and will go up
00:51:11.260 --> 00:51:18.160
on the web quickly, just as
soon as we get them revised.
00:51:18.160 --> 00:51:22.060
And I'm writing notes
on your projects
00:51:22.060 --> 00:51:25.430
and I hope I'll have
them ready for Friday.
00:51:25.430 --> 00:51:29.220
I'll aim for Friday because
Monday is Patriot's Day
00:51:29.220 --> 00:51:32.160
and you have to
run the marathon.
00:51:32.160 --> 00:51:33.440
So I'll see you Friday.
00:51:33.440 --> 00:51:33.940
Good.
00:51:33.940 --> 00:51:35.190
Thanks.