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PROFESSOR: Ready to go on
lecture five, I guess this is.

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So this -- we've done the
one-way wave equation,

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first-order equation and now
it's just natural to see what

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about the ordinary, second-order
wave equation and on this

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board, I've written it two ways.

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I've written it as a
second-order equation,

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so second time
derivative appears.

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That's going to make a change.

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It will have second
differences in time,

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so we're going to
have more steps,

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and of the matching second
difference in space.

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That's one way.

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Then this is another
important way

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to write the same equation
as a first-order system.

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So I can get back to first-order
systems and I can use all those

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methods -- Friedrichs,
Lax-Friedrichs for example,

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Lax-Wendroff, whatever
improvements we think of --

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on this first order system, d
by dt of a vector is a matrix

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times d by dx of a vector.

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So that's -- you really
have a choice here.

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Written this way, well,
notice a little bit

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of a good thing,
one good thing here.

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The matrix that shows
up is symmetric.

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That was by the careful choice
of putting a c in there.

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Then you see that the first
equation in this system is u_tt

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is c time c*u_xx.

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So the first equation
is the wave equation.

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And what about the
second equation?

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This is typical, when
you reduce things

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to a lower-order system,
that you get an equation here

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which is just an identity.

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The derivative of c -- the
time -- this is c*u_xt --

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it's the derivative in
x and t from there --

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and we have c times the
x derivative of u_t,

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the cross derivative.

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So that is an identity because
u_tx is the same as u_xt.

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We can take time and x and t
derivatives in either order,

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for a function of x and t.

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So we'll see that.

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I guess what I want to
do in the lecture is not

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to repeat Lax-Friedrichs or
Lax-Wendroff for this system.

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That would be unexciting.

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Let me just mention, what are
the eigenvalues of that matrix?

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So the eigenvalues
-- for a system,

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it's the eigenvalues of the
matrix that tell us what waves

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we've got, what speeds they go,
where it -- when it was 1 by 1,

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of course, it was just
that single number c,

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but what are the
eigenvalues of that matrix?

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Now, having made it
a symmetric matrix,

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we know right away it
has real eigenvalues,

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and so it's a wave problem.

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For a heat equation, we would
see something different,

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but this'll be a wave equation
and this has two eigenvalues

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and they are c and minus c.

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You could quickly check
that that 2 by 2 matrix

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has eigenvalues c and minus c.

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They add up to 0,
which is the trace.

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They multiply -- c times
minus c is minus c squared,

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which is the determinant.

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So that's -- they're
the right eigenvalues.

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So that's going to tell us
what we'll find every way,

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that the speeds of the waves
are c and minus c, which

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means a wave going
one way with speed c

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and a wave going the
other way with speed c.

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So it's a two-way wave equation.

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And this is the equation of
real -- that really happens,

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and I wanted to say just
a little bit about it --

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just two words
about real problems,

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because we can't,
in these first days,

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tackle the difficult
problems of electromagnetism,

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antenna design, all
kinds of wave problems.

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I guess I'm hoping some of
you may be meeting real wave

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problems in other courses and
would make a project that tries

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our methods on real problems.

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I just wanted to
write down here one

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sort of typical real problem.

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So in a real problem, we have
a variable coefficient x,

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because maybe our material
is not homogeneous,

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it's not uniform, and we
have a forcing term --

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I just put F for the amplitude
to emphasize that's forcing

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term, so F standing for forcing,
but it's an oscillating forcing

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term, it's a -- this is
what you're going to have.

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In electromagnetism,
the frequency

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is going to be very, very high.

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So this is a problem that's
not easy numerically.

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Very, very high
frequencies would normally

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require very, very small
wavelengths, very, very

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small delta x,
because you remember,

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there'll be a k
delta x in the error.

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So we took a big delta x or
even an ordinary size delta

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x and let the frequency go way,
way up, high, high frequency

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oscillation, our k
delta x would be big

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and the accuracy would be
poor and the results useless.

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So a lot of -- I'm really --
guess I'm thinking here about

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a course on wave propagation
would study this as a first

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model problem, do some analysis
and then go into 2-D and 3-D.

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That's the reality
of wave theory,

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is you can get asymptotic
results for k going

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to infinity.

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You get things -- you're
looking in matching powers of k,

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so to speak.

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You're getting plane waves
and interesting stuff

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that we can't do
already, right away here.

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So I'm going back to the
ordinary wave equation,

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either second or --
start with second order,

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u_tt is c squared u_xx.

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That's our goal.

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As I said, we don't want to see
Lax-Friedrichs and Lax-Wendroff

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again.

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We know those.

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It's rather -- so one new method
is going to show up and it's --

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leapfrog is the right name,
the natural name to give it.

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Leapfrog because in
the time direction,

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we have time n minus
1, n and n plus 1.

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There's a leap there, but let
me start with semidiscrete.

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We haven't done
this until today.

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Semidiscrete means, as you
see, that the time variable

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stays continuous.

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We have ordinary
differential equations in t,

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but the space variable is
what's getting discrete.

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so the problem is like half
discrete -- discrete in x,

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there's a delta x, but
there's no delta t yet.

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So it's natural to
analyze this problem.

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So what do we have, really?

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In the x direction, we have
a mesh of width delta x.

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So we have unknowns.

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If that's 0, j is measuring
-- so j delta x is the typical

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point, point number
j along there.

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And then in the time direction,
we're going continuously.

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So I don't have discrete
steps, so it's called --

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because those are lines in
the time direction, this is --

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semidiscrete is also referred
to as the method of lines.

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So the method of lines is what
we're doing here and it's --

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all I want to do is
what I always do.

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Take exponentials, look
at their growth factors,

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understand the equation.

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Actually, I better do it first
with a wave equation itself.

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So I guess my
organization for today is,

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first step is follow e to the
i*k*x for the wave equation.

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So I'll do that here.

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u_tt equals c squared u_xx,
and I'm going to look at u

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proportional to -- and I'll
call this factor G again --

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G of t and times
the frequency i*k*x.

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So I'm going to look
for pure exponentials

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and we'll quickly find them.

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And then I'm going to do
the same for semidiscrete,

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where that x becomes j
delta x and I'm at point j.

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And then I'm going to eventually
look at the discrete, now fully

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discrete, discrete in time
and discrete in space,

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and the question of
stability arises.

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Accuracy we can --
you pretty much --

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we pretty much know the
accuracy for second differences,

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that's a second-order accurate
thing because it's centered,

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but stability is
going to come in.

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So that's -- this
is our first step.

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Then I'll just mention
what other point

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remains in this wave
equation, is to look

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at the first-order system.

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How does that look
in space and time?

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As I said, the
obvious thing to do

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would be Lax-Friedrichs
or Lax-Wendroff,

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but there's another idea and
it brings in a staggered grid.

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You'll see it.

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Somehow that staggered grid
is really an important idea

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that we haven't met yet.

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That some unknowns are on
the standard grid, U_(j, n),

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but the other unknowns are on
a grid that's half a delta x

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and half a delta t different.

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It's just right, actually.

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So that's a premier method
in electromagnetism.

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So that's what's
coming, but let's start

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with the second-order
method, because we haven't

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seen the second-order method.

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Second-order equation --
plug in G. So I have --

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the second time derivative
gives me G double prime times e

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to the i*k*x -- I'm separating
variables, of course;

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00:13:09,750 --> 00:13:12,460
t here and x there.

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00:13:12,460 --> 00:13:15,030
On the other side,
I have c squared.

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Now I have the x variable,
so that brings down an i*k --

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00:13:19,600 --> 00:13:27,620
so that's i*k squared
times G e to the i*k*x.

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Good.

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So simple, but substituting this
is just a natural first step

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to see what's happening.

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Then, of course, canceling e
to the i*k*x leaves us with

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00:13:44,860 --> 00:13:50,790
the equation for the
G. G double prime is --

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00:13:50,790 --> 00:13:57,940
I'll make that i squared minus
-- minus c squared k squared G.

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00:13:57,940 --> 00:14:03,100
Of course, the new aspect
is it's second order.

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00:14:03,100 --> 00:14:06,890
So it's got two solutions
instead of just one.

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00:14:06,890 --> 00:14:09,920
It's constant
coefficients, of course,

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00:14:09,920 --> 00:14:13,840
and so we look for exponentials.

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00:14:13,840 --> 00:14:22,800
And the solution then would be
-- two solutions are some --

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00:14:22,800 --> 00:14:28,410
there's an e to the i -- I
guess I'm bringing back the i --

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00:14:28,410 --> 00:14:44,100
e to the i*c*k*t e to the i*k*x
and e to the minus i*c*k*t e

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00:14:44,100 --> 00:14:46,540
to the i*k*x.

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00:14:46,540 --> 00:14:49,550
Those are the two possible G's.

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00:14:49,550 --> 00:14:57,280
Those are the -- if we take
their second time derivative,

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00:14:57,280 --> 00:15:02,690
we get the same as taking
the second x derivative.

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00:15:02,690 --> 00:15:07,010
We get the factor c
is in the right place.

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00:15:07,010 --> 00:15:08,200
You'll see it.

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00:15:08,200 --> 00:15:11,300
The second time derivative
brings this factor down,

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00:15:11,300 --> 00:15:16,910
just as we wanted, but the point
is that there are two of them.

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00:15:16,910 --> 00:15:25,270
So this I can write as
e to the i*k x plus ct.

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00:15:25,270 --> 00:15:27,650
That's our old friend.

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00:15:27,650 --> 00:15:34,650
This one I can write as
e to the i*k x minus ct,

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00:15:34,650 --> 00:15:37,920
which is our new
friend, the new wave.

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00:15:37,920 --> 00:15:39,660
The wave that's
going to the right.

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00:15:39,660 --> 00:15:45,600
So this x plus c*t follows
the characteristic lines

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00:15:45,600 --> 00:15:52,000
to the left, just the way we had
in the one-way wave equation.

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00:15:52,000 --> 00:15:59,910
This x minus c*t stays constant
on characteristic lines going

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00:15:59,910 --> 00:16:10,580
up to the right; it travels with
speed c and the signal travels

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00:16:10,580 --> 00:16:14,270
along those characteristics
with speed c, so that a --

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00:16:14,270 --> 00:16:23,910
an initial value, now -- maybe
I draw a picture over here --

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00:16:23,910 --> 00:16:27,620
so here is t equals --
this is x as always.

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00:16:27,620 --> 00:16:29,320
This is t equals 0.

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00:16:29,320 --> 00:16:33,630
I have some initial
value u at (0, 0),

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00:16:33,630 --> 00:16:39,400
and I also have an initial
velocity, because I'm in --

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00:16:39,400 --> 00:16:44,010
got a second-order equation,
but information somehow,

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00:16:44,010 --> 00:16:49,270
pure exponential information
travels along characteristics

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00:16:49,270 --> 00:16:51,880
this way -- that's
the new one --

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00:16:51,880 --> 00:16:55,370
and characteristics this
way -- that's the old one.

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00:16:55,370 --> 00:16:58,790
This is time, space.

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00:16:58,790 --> 00:17:02,490
I'm graphing the two
characteristic lines now.

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00:17:02,490 --> 00:17:08,520
x minus c*t equals --
if it starts from 0,

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00:17:08,520 --> 00:17:10,520
then it would be 0.

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00:17:10,520 --> 00:17:15,670
This is the x plus c*t equals
0, the one we had before.

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00:17:15,670 --> 00:17:17,090
The two characteristic lines.

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00:17:19,790 --> 00:17:24,540
Waves are going
both ways, right?

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00:17:24,540 --> 00:17:29,205
Now, what we were able to do
with the first-order equation

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00:17:29,205 --> 00:17:37,400
-- we even got a completely
explicit solution

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00:17:37,400 --> 00:17:40,120
to the one-way wave equation
and we can do the same

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00:17:40,120 --> 00:17:44,480
for the two-way wave equation,
so I better write it down.

238
00:17:44,480 --> 00:17:47,730
I better write
down the solution.

239
00:17:47,730 --> 00:17:52,130
So let me just continue
this same idea.

240
00:17:52,130 --> 00:18:01,730
The general, the complete
solution will be u of x and t

241
00:18:01,730 --> 00:18:10,820
is some function of x minus
c*t and some function of --

242
00:18:10,820 --> 00:18:16,860
call that f_2 -- of x plus c*t.

243
00:18:16,860 --> 00:18:22,500
Now I'm taking the jump, the
jump from one exponential,

244
00:18:22,500 --> 00:18:30,430
a particular k that gave
me these particular guys,

245
00:18:30,430 --> 00:18:35,360
to all exponentials and
assembling all exponentials

246
00:18:35,360 --> 00:18:36,350
into functions.

247
00:18:36,350 --> 00:18:37,830
So why is that OK?

248
00:18:37,830 --> 00:18:42,070
It's only OK in this because
of this very special situation

249
00:18:42,070 --> 00:18:45,620
that all frequencies are
behaving the same here.

250
00:18:45,620 --> 00:18:50,830
All frequencies -- I
have no dispersion.

251
00:18:50,830 --> 00:18:54,490
Dispersion is going to
be different behavior

252
00:18:54,490 --> 00:18:58,580
of different frequencies,
so that's an important fact,

253
00:18:58,580 --> 00:19:01,500
which we don't face here.

254
00:19:01,500 --> 00:19:06,330
There will be dispersion
in the discrete methods,

255
00:19:06,330 --> 00:19:09,890
the semidiscrete and
the discrete method.

256
00:19:09,890 --> 00:19:13,210
Different frequencies will
travel at different speeds

257
00:19:13,210 --> 00:19:17,400
and that's what produces
the oscillating error.

258
00:19:17,400 --> 00:19:22,830
Whatever shape the error has is
because different k's are doing

259
00:19:22,830 --> 00:19:25,300
slightly different
things, as we'll see

260
00:19:25,300 --> 00:19:28,680
when we do the discrete ones.

261
00:19:28,680 --> 00:19:33,560
But in the continuous case,
every k, we have the same --

262
00:19:33,560 --> 00:19:38,200
traveling either with speed
c or minus c and therefore,

263
00:19:38,200 --> 00:19:41,570
we put them all together and we
have a whole function traveling

264
00:19:41,570 --> 00:19:44,450
with speed c or minus c.

265
00:19:44,450 --> 00:19:48,280
Now we've got two
functions and we

266
00:19:48,280 --> 00:19:51,440
have to do we have to match
two initial conditions.

267
00:19:51,440 --> 00:19:59,540
We have to match a u of x
and 0 and the time derivative

268
00:19:59,540 --> 00:20:03,080
has to match a u_t
of x and 0, so these

269
00:20:03,080 --> 00:20:05,850
are the initial conditions.

270
00:20:10,010 --> 00:20:12,690
You just match them up
and you find the answer.

271
00:20:12,690 --> 00:20:15,990
So you discover what these
functions really are.

272
00:20:15,990 --> 00:20:19,190
It turns out that, I think,
we had -- you remember before,

273
00:20:19,190 --> 00:20:26,740
we had a u of x plus c*t and 0?

274
00:20:26,740 --> 00:20:30,380
That was everything in the
solution to the one-way wave

275
00:20:30,380 --> 00:20:31,400
equation.

276
00:20:31,400 --> 00:20:44,990
Now we will have u of x minus
c*t, 0 and this divided by 2 --

277
00:20:44,990 --> 00:20:51,270
so there, you see, a wave's
going each way and the division

278
00:20:51,270 --> 00:20:58,660
by 2 makes us -- now we're
matching the initial condition,

279
00:20:58,660 --> 00:21:04,740
so that would be the total
answer if the equation,

280
00:21:04,740 --> 00:21:08,940
if the body starts from
rest with zero velocity,

281
00:21:08,940 --> 00:21:14,760
but I haven't matched -- so I've
only matched du/dt equals 0 --

282
00:21:14,760 --> 00:21:18,040
think the time derivative
of that would be 0.

283
00:21:18,040 --> 00:21:20,570
Yeah, because the time
derivative of this would bring

284
00:21:20,570 --> 00:21:25,740
out a c, this would bring out
a minus c and a t equals 0,

285
00:21:25,740 --> 00:21:26,940
they'd cancelled.

286
00:21:26,940 --> 00:21:34,200
So I need another term
and it turns out to be --

287
00:21:34,200 --> 00:21:35,680
I'll just write it down.

288
00:21:35,680 --> 00:21:41,360
It's got to depend on x
minus c*t and x plus c*t

289
00:21:41,360 --> 00:21:50,500
and it's this -- it's the
integral of this velocity dx.

290
00:21:50,500 --> 00:21:55,840
I won't -- there's no reason
to take time to derive that.

291
00:21:55,840 --> 00:21:59,190
I think it's right
and could check.

292
00:21:59,190 --> 00:22:00,290
Right.

293
00:22:00,290 --> 00:22:09,510
So what this means is that
the initial velocity is --

294
00:22:09,510 --> 00:22:12,560
it's entering,
somehow, all along --

295
00:22:12,560 --> 00:22:23,600
somehow the initial velocity
is affecting things within this

296
00:22:23,600 --> 00:22:31,520
cone, this characteristic
cone that's defined

297
00:22:31,520 --> 00:22:34,480
by the characteristic lines.

298
00:22:34,480 --> 00:22:37,880
In two dimensions, that cone
will really look like a cone.

299
00:22:37,880 --> 00:22:40,300
It'll look like an
ice cream cone --

300
00:22:40,300 --> 00:22:41,950
or maybe that's in
three dimensions.

301
00:22:44,670 --> 00:22:45,170
Whatever.

302
00:22:48,560 --> 00:22:51,600
In other words, if
the initial condition

303
00:22:51,600 --> 00:22:54,760
was a delta function
at the origin,

304
00:22:54,760 --> 00:23:01,130
you could quickly see what
the solution would be,

305
00:23:01,130 --> 00:23:03,330
and if the delta, if
the initial condition

306
00:23:03,330 --> 00:23:06,830
was a step function at the
origin, what would happen then?

307
00:23:06,830 --> 00:23:10,810
Suppose I started with this
step function at the origin.

308
00:23:10,810 --> 00:23:14,910
So, like, a wall of water
and suppose it's at rest.

309
00:23:14,910 --> 00:23:19,060
So nothing -- at
moment t equals 0,

310
00:23:19,060 --> 00:23:26,200
I remove the dam and the water
starts -- the water flows.

311
00:23:26,200 --> 00:23:32,110
Then this term is 0 because
there was no initial velocity

312
00:23:32,110 --> 00:23:35,780
and this term is just half
of that wall goes one way

313
00:23:35,780 --> 00:23:38,560
and half goes the other way.

314
00:23:38,560 --> 00:23:46,670
The notes draw a picture of
that, so that -- so that's --

315
00:23:46,670 --> 00:23:52,712
I've kind of quickly disposed
of the solution of the wave

316
00:23:52,712 --> 00:23:58,270
equation and of course,
that's too easy.

317
00:23:58,270 --> 00:24:07,010
This is -- I mean, that's good
to know and good to compare

318
00:24:07,010 --> 00:24:12,340
when we meet real problems
where c depends on x,

319
00:24:12,340 --> 00:24:15,520
where there's a forcing
term, where there's a --

320
00:24:15,520 --> 00:24:18,840
real complications.

321
00:24:18,840 --> 00:24:22,580
But it just pays to stay
with the model problems

322
00:24:22,580 --> 00:24:23,950
at the beginning.

323
00:24:23,950 --> 00:24:28,850
So, I'm going to stay with
the model problem and move

324
00:24:28,850 --> 00:24:36,470
on to discrete, so
try to understand --

325
00:24:36,470 --> 00:24:42,930
let me try to understand what
happens if I make the space

326
00:24:42,930 --> 00:24:45,670
variable discrete, right?

327
00:24:49,260 --> 00:24:54,270
I guess I should say,
another step toward reality

328
00:24:54,270 --> 00:24:58,540
would be, have some boundaries.

329
00:24:58,540 --> 00:25:01,990
Here, this is on the whole
line, x from minus infinity

330
00:25:01,990 --> 00:25:06,640
to infinity, so reality would
be that there would be --

331
00:25:06,640 --> 00:25:14,290
this picture would have some
boundary, maybe there at, say,

332
00:25:14,290 --> 00:25:22,430
minus L and L, and some
boundary conditions there,

333
00:25:22,430 --> 00:25:25,180
on that boundary.

334
00:25:25,180 --> 00:25:32,000
And that -- so now we
have a finite problem.

335
00:25:32,000 --> 00:25:34,750
That means the
calculations are fine,

336
00:25:34,750 --> 00:25:37,470
but it means some new
effects are coming in.

337
00:25:37,470 --> 00:25:43,230
Waves can bounce back
off the boundary, partly.

338
00:25:43,230 --> 00:25:50,120
They can go out,
be lost, go beyond

339
00:25:50,120 --> 00:25:54,840
or disappear, depending what
the boundary conditions are.

340
00:25:54,840 --> 00:25:57,130
So that's another
part of reality,

341
00:25:57,130 --> 00:26:00,290
is boundary conditions.

342
00:26:00,290 --> 00:26:06,720
I'm probably suggesting,
then, some more realistic

343
00:26:06,720 --> 00:26:08,330
numerical experiments.

344
00:26:08,330 --> 00:26:10,060
So a realistic
numerical experiment

345
00:26:10,060 --> 00:26:14,140
would be, put in
some boundaries,

346
00:26:14,140 --> 00:26:17,040
make this a finite system.

347
00:26:17,040 --> 00:26:24,910
So if that's 0, let's just have
a finite number, stop there.

348
00:26:24,910 --> 00:26:27,260
This would be one boundary,
this would be the other.

349
00:26:27,260 --> 00:26:33,120
Then I have just seven
unknowns in that problem.

350
00:26:33,120 --> 00:26:36,840
That'd be a seven by
seven matrix over there

351
00:26:36,840 --> 00:26:38,580
and it would be the
second difference

352
00:26:38,580 --> 00:26:49,810
matrix that is so fundamental
in all of scientific computing.

353
00:26:49,810 --> 00:26:52,180
So we have here a second
difference matrix,

354
00:26:52,180 --> 00:26:54,420
but it's infinite
because I haven't

355
00:26:54,420 --> 00:26:56,410
got a boundary right now.

356
00:26:56,410 --> 00:26:59,130
So the fact of not
having a boundary

357
00:26:59,130 --> 00:27:02,540
means Fourier is ready to go.

358
00:27:02,540 --> 00:27:08,870
So I'm going to plug in --
again, I'm going to try U.

359
00:27:08,870 --> 00:27:12,840
This is capital U now, because
I've discretized something.

360
00:27:17,010 --> 00:27:21,470
I better write this in a
form where we see the j,

361
00:27:21,470 --> 00:27:28,400
so can I write this equation,
just again for U number j --

362
00:27:28,400 --> 00:27:32,353
so tracking up this line,
for example, or this one.

363
00:27:32,353 --> 00:27:38,550
So I'm tracking up this line
-- do you see that U is now

364
00:27:38,550 --> 00:27:39,370
a system?

365
00:27:39,370 --> 00:27:43,310
In this case -- let's see.

366
00:27:43,310 --> 00:27:46,240
I shouldn't have put this one in
so that I'd have 1, 2, 3, 4, 5,

367
00:27:46,240 --> 00:27:47,560
6, 7 -- that's good.

368
00:27:51,650 --> 00:27:58,020
So my unknowns are U_j
tracking up that line.

369
00:27:58,020 --> 00:28:03,930
Each U_j is going in
the time direction,

370
00:28:03,930 --> 00:28:09,960
so this equation is really
the time derivative.

371
00:28:09,960 --> 00:28:12,640
It's an ordinary derivative.

372
00:28:12,640 --> 00:28:15,710
I can write d and not
partial derivative,

373
00:28:15,710 --> 00:28:21,450
because the only derivatives
here are in time of U_j --

374
00:28:21,450 --> 00:28:31,700
is c squared over delta x
squared times U_(j+1) minus

375
00:28:31,700 --> 00:28:37,700
2*U_j plus U_(j-1).

376
00:28:37,700 --> 00:28:40,410
That's my equation.

377
00:28:40,410 --> 00:28:42,340
My system of equations, sorry.

378
00:28:42,340 --> 00:28:45,140
My system of equations.

379
00:28:45,140 --> 00:28:48,530
For the moment, I'm thinking
j goes from minus infinity

380
00:28:48,530 --> 00:28:49,280
to infinity.

381
00:28:49,280 --> 00:28:53,860
I'm not going to put in a
boundary, but what I will do,

382
00:28:53,860 --> 00:29:05,150
of course, is try U_j is
some growth factor that --

383
00:29:05,150 --> 00:29:09,160
j is a space, is
counting space steps,

384
00:29:09,160 --> 00:29:14,400
so the growth factor is in time
and it's continuous in time,

385
00:29:14,400 --> 00:29:22,800
because time's continuous,
but it's always e to the i*k*j

386
00:29:22,800 --> 00:29:28,250
delta x.

387
00:29:28,250 --> 00:29:30,070
That's the right
thing to plug in.

388
00:29:30,070 --> 00:29:36,870
I mean, I guess what we're
-- what I'm really --

389
00:29:36,870 --> 00:29:42,490
the reason for really doing this
is just to reinforce the idea

390
00:29:42,490 --> 00:29:47,410
that trying a pure exponential
is a good thing to do.

391
00:29:47,410 --> 00:29:51,270
It's even a good thing to do
with variable coefficients

392
00:29:51,270 --> 00:29:54,450
and real problems.

393
00:29:54,450 --> 00:29:59,100
It's absolutely a good thing to
do with these model problems.

394
00:29:59,100 --> 00:30:02,970
Plug it in and what happens?

395
00:30:02,970 --> 00:30:07,830
So I get G double prime
times the exponential,

396
00:30:07,830 --> 00:30:10,190
but of course you
know in advance,

397
00:30:10,190 --> 00:30:12,390
I'm going to cancel
that exponential --

398
00:30:12,390 --> 00:30:18,230
is c squared over delta
x squared times --

399
00:30:18,230 --> 00:30:19,450
what goes there?

400
00:30:19,450 --> 00:30:23,320
What will cancel
the exponential --

401
00:30:23,320 --> 00:30:27,180
so I'm imagining that
exponential as far out,

402
00:30:27,180 --> 00:30:29,340
but here it is at j plus 1.

403
00:30:29,340 --> 00:30:35,845
So I've incremented j by 1, so
that is going to give me an e

404
00:30:35,845 --> 00:30:39,670
to the i*k delta x, right?

405
00:30:39,670 --> 00:30:44,080
Here I haven't incremented it
and here I've incremented it

406
00:30:44,080 --> 00:30:51,180
by minus 1, so that will give me
an e to the minus i*k delta x,

407
00:30:51,180 --> 00:30:55,780
all times the same exponential
which appeared there

408
00:30:55,780 --> 00:30:58,820
and appeared there
and got cancelled.

409
00:30:58,820 --> 00:31:00,870
So that's it there.

410
00:31:00,870 --> 00:31:07,890
G double prime -- oh there's
got to be a G -- right,

411
00:31:07,890 --> 00:31:15,924
there's a G. Right.

412
00:31:15,924 --> 00:31:17,590
When I plugged in on
the right-hand side

413
00:31:17,590 --> 00:31:21,230
I got a G that was so
uninteresting I lost it.

414
00:31:23,890 --> 00:31:28,330
What does -- what's G there?

415
00:31:31,200 --> 00:31:41,660
I guess the main point to see
is, this quantity, which is --

416
00:31:41,660 --> 00:31:48,930
divided by delta x squared,
which is coming because we have

417
00:31:48,930 --> 00:31:50,940
differences instead
of derivatives.

418
00:31:50,940 --> 00:31:55,790
When we had derivatives, that
quantity was c squared i*k

419
00:31:55,790 --> 00:31:56,290
squared.

420
00:31:59,650 --> 00:32:02,560
Or, in other words, minus
c squared k squared.

421
00:32:05,230 --> 00:32:18,300
This was the correct factor and
this one is the discrete factor

422
00:32:18,300 --> 00:32:21,380
and it's worth just paying
a moment of attention

423
00:32:21,380 --> 00:32:23,860
to compare the two.

424
00:32:27,500 --> 00:32:30,320
Where am I going to
write that comparison?

425
00:32:30,320 --> 00:32:33,560
I guess that's going to be
on the board that's hidden,

426
00:32:33,560 --> 00:32:37,990
so I'm going to -- can
you try to remember that?

427
00:32:37,990 --> 00:32:41,830
Let me focus just on this
quantity for a minute, or even

428
00:32:41,830 --> 00:32:47,490
just on this one, just because
we see it so many times,

429
00:32:47,490 --> 00:32:50,860
because it's coming directly
from second differences.

430
00:32:50,860 --> 00:32:53,350
For the moment, let
me take k delta x.

431
00:32:53,350 --> 00:32:54,600
I'll just call it theta.

432
00:32:54,600 --> 00:32:56,420
It's some angle.

433
00:32:56,420 --> 00:33:03,180
So can I just do a little
bit of trig with this factor?

434
00:33:03,180 --> 00:33:07,890
2 minus 2 cos theta.

435
00:33:07,890 --> 00:33:09,490
Let's see.

436
00:33:09,490 --> 00:33:12,760
I'm going to, I'm going
to bring out a minus sign.

437
00:33:12,760 --> 00:33:15,600
So I'm going to put
plus signs on those

438
00:33:15,600 --> 00:33:18,500
and minus signs on those.

439
00:33:18,500 --> 00:33:22,100
You see the main
point of course,

440
00:33:22,100 --> 00:33:26,990
that e to the plus
something, e to the i theta

441
00:33:26,990 --> 00:33:33,370
and e to the minus i theta
combine to give 2 cos theta.

442
00:33:33,370 --> 00:33:35,550
Right, so that's the key
simplification that's

443
00:33:35,550 --> 00:33:38,260
going to make this much neater.

444
00:33:38,260 --> 00:33:41,000
So I have 2 cos theta minus 2.

445
00:33:41,000 --> 00:33:47,230
I have 2 cos theta minus 2, and
if I bring out a minus sign,

446
00:33:47,230 --> 00:33:49,540
I'll have 2 minus 2
cos theta and that's

447
00:33:49,540 --> 00:33:50,670
what I want to write.

448
00:33:50,670 --> 00:33:52,560
I want to write 2
minus 2 cos theta.

449
00:33:59,450 --> 00:34:02,690
I want to see what
that expression is.

450
00:34:02,690 --> 00:34:05,325
Right, so that brought
out a minus sign just

451
00:34:05,325 --> 00:34:13,430
to match that minus sign
and the c squareds match,

452
00:34:13,430 --> 00:34:17,570
but what's different is
this k squared is now

453
00:34:17,570 --> 00:34:25,570
gone into this 2 minus 2
cos k delta x I'll write it,

454
00:34:25,570 --> 00:34:29,120
divided by delta x squared.

455
00:34:29,120 --> 00:34:33,710
You see that this is
-- that's the change.

456
00:34:33,710 --> 00:34:36,790
k squared, which is
a positive quantity

457
00:34:36,790 --> 00:34:41,720
became this, which is also a
positive quantity, 2 minus 2

458
00:34:41,720 --> 00:34:46,370
cos theta, but it's
not a pure k squared.

459
00:34:46,370 --> 00:34:49,670
We are getting dispersion.

460
00:34:49,670 --> 00:34:54,070
Different frequencies are
showing up inside the cosine

461
00:34:54,070 --> 00:34:57,210
and that's not a
linear function.

462
00:34:57,210 --> 00:35:01,150
So we're seeing
different frequencies

463
00:35:01,150 --> 00:35:10,160
going their own way in a more
interesting way then just what

464
00:35:10,160 --> 00:35:12,970
we saw here with k squared.

465
00:35:12,970 --> 00:35:17,580
Of course, when we take the
square root, as we did here,

466
00:35:17,580 --> 00:35:24,280
that produced linear in k and
no dispersion and something

467
00:35:24,280 --> 00:35:28,200
simple, but here, when we
take the square root --

468
00:35:28,200 --> 00:35:29,954
let's get ready to
take the square root.

469
00:35:29,954 --> 00:35:31,620
How can we take the
square root of that?

470
00:35:36,100 --> 00:35:37,520
By writing it as a square.

471
00:35:40,200 --> 00:35:44,170
It's never negative, and if I
use a little trigonometry --

472
00:35:44,170 --> 00:35:45,840
so I'm just going
to write it down.

473
00:35:45,840 --> 00:35:52,960
This is 4 -- I think it's
sine squared of theta over 2.

474
00:35:55,600 --> 00:35:59,170
That's just a handy fact.

475
00:35:59,170 --> 00:36:04,500
It's handy because I can take
its square root and really see

476
00:36:04,500 --> 00:36:10,060
what -- so theta -- remember
this theta is standing for k

477
00:36:10,060 --> 00:36:12,420
delta x.

478
00:36:12,420 --> 00:36:15,580
So then, I want to divided
it by delta x squared.

479
00:36:15,580 --> 00:36:19,290
Can I bring down that quantity
just so you see it again?

480
00:36:19,290 --> 00:36:25,130
So this was the 2 minus 2 cos
theta with the minus sign,

481
00:36:25,130 --> 00:36:29,470
and dividing by delta
x squared and that's --

482
00:36:29,470 --> 00:36:32,680
so it was G double prime
-- let's just write it.

483
00:36:32,680 --> 00:36:37,850
So I have G -- I'm now
going to copy what --

484
00:36:37,850 --> 00:36:42,670
G double prime was a minus sign
times the c squared times this

485
00:36:42,670 --> 00:36:52,330
-- 4 sine squared of k delta
x on 2 divided by delta x

486
00:36:52,330 --> 00:36:53,490
squared.

487
00:36:53,490 --> 00:36:56,789
Maybe I'll put the delta
x squared over there,

488
00:36:56,789 --> 00:36:57,830
where it kind of belongs.

489
00:37:03,970 --> 00:37:16,130
Times G, always
forgetting G. So we're

490
00:37:16,130 --> 00:37:19,840
just following the
golden way here,

491
00:37:19,840 --> 00:37:23,130
plugging in an exponential
and seeing what happens,

492
00:37:23,130 --> 00:37:27,240
and what happens is pretty nice.

493
00:37:27,240 --> 00:37:29,300
We now know what
the solutions are.

494
00:37:29,300 --> 00:37:31,000
There'll be two solutions.

495
00:37:31,000 --> 00:37:33,670
It's a linear problem,
so the solutions

496
00:37:33,670 --> 00:37:35,730
will be exponential in t.

497
00:37:38,840 --> 00:37:42,080
The solutions will
have a plus or minus i,

498
00:37:42,080 --> 00:37:44,280
because it's a minus sign.

499
00:37:44,280 --> 00:37:46,960
So G -- can I write it this way?

500
00:37:46,960 --> 00:37:52,940
G of -- the solution to this,
or the two exponential solutions

501
00:37:52,940 --> 00:37:57,930
will be e to the
plus or minus --

502
00:37:57,930 --> 00:38:00,520
now I'm going to take
this square root --

503
00:38:00,520 --> 00:38:06,810
i times c times the
square root of this.

504
00:38:06,810 --> 00:38:08,480
Actually, that's not so hard.

505
00:38:08,480 --> 00:38:11,440
2 -- that was our
point, wasn't it?

506
00:38:11,440 --> 00:38:18,960
Sorry, I forgot about that --
sine k delta x over 2 divided

507
00:38:18,960 --> 00:38:23,550
by delta x -- and look,
here's something nice.

508
00:38:23,550 --> 00:38:25,780
Bring the 2 down there.

509
00:38:28,570 --> 00:38:31,060
Make it 1/2 of the denominator.

510
00:38:31,060 --> 00:38:35,100
Do you see what
this expression is?

511
00:38:35,100 --> 00:38:38,920
Actually, do you know its name?

512
00:38:38,920 --> 00:38:42,310
It's the sinc function.

513
00:38:42,310 --> 00:38:45,040
It's the sinc function, right?

514
00:38:45,040 --> 00:38:52,060
And for very small delta
x -- oh, times t, right?

515
00:38:52,060 --> 00:38:55,000
It just -- we just
have an exponential.

516
00:38:55,000 --> 00:38:56,500
Of course we do.

517
00:38:56,500 --> 00:39:00,410
This was a constant
coefficient problem in t,

518
00:39:00,410 --> 00:39:04,940
and a second order, so
I had two exponentials,

519
00:39:04,940 --> 00:39:08,290
and I just took the square
root of that coefficient,

520
00:39:08,290 --> 00:39:10,410
put it there, times t.

521
00:39:10,410 --> 00:39:11,950
Absolutely straightforward.

522
00:39:16,750 --> 00:39:20,050
This quantity is
the key one and this

523
00:39:20,050 --> 00:39:28,970
is the quantity which, for
small k or small delta x,

524
00:39:28,970 --> 00:39:32,060
is approximately what?

525
00:39:32,060 --> 00:39:37,165
So if k and delta x are
small, or if even just delta x

526
00:39:37,165 --> 00:39:40,560
is small, what does this
sinc function look like?

527
00:39:40,560 --> 00:39:48,030
What is that ratio
approximately, when k is small?

528
00:39:48,030 --> 00:39:51,400
What does the sine of
theta look like when

529
00:39:51,400 --> 00:39:54,280
theta is a small angle?

530
00:39:54,280 --> 00:39:57,460
Looks like theta, right?

531
00:39:57,460 --> 00:40:00,200
Sine theta is about theta.

532
00:40:00,200 --> 00:40:04,070
So this thing is about k delta
x over 2 divided by delta --

533
00:40:04,070 --> 00:40:15,650
it's k, so it's approximately
k for small delta x, let's say.

534
00:40:15,650 --> 00:40:18,550
Delta x is small.

535
00:40:18,550 --> 00:40:23,370
That's the key thing to know
about the sinc function,

536
00:40:23,370 --> 00:40:27,730
that it begins by approximating
the sine function,

537
00:40:27,730 --> 00:40:32,390
but then of course, when k delta
x gets large, it wanders away.

538
00:40:32,390 --> 00:40:33,850
It's not linear in k.

539
00:40:33,850 --> 00:40:37,620
See, it starts out
practically linear in k,

540
00:40:37,620 --> 00:40:40,480
and that k is
exactly the right k.

541
00:40:40,480 --> 00:40:42,220
That's the right factor.

542
00:40:42,220 --> 00:40:43,400
That's the k here.

543
00:40:47,680 --> 00:40:49,280
But in the
differential equation,

544
00:40:49,280 --> 00:40:53,930
it stayed k and in the
difference equation,

545
00:40:53,930 --> 00:41:02,790
for k delta x moving away,
it changes and that's why --

546
00:41:02,790 --> 00:41:06,380
of course, that's the
source of the error.

547
00:41:06,380 --> 00:41:09,790
We're seeing now why
the method is probably

548
00:41:09,790 --> 00:41:11,720
second-order accuracy.

549
00:41:11,720 --> 00:41:16,140
We know that these second
differences are second --

550
00:41:16,140 --> 00:41:18,130
give second-order accuracy.

551
00:41:18,130 --> 00:41:21,050
This would be -- if I
looked at the next term,

552
00:41:21,050 --> 00:41:23,650
what's the next
term in sine theta?

553
00:41:23,650 --> 00:41:26,750
Sine theta starts
out theta and then,

554
00:41:26,750 --> 00:41:28,440
what's the term after that?

555
00:41:31,810 --> 00:41:38,480
Everybody's Taylor series or --
[UNINTELLIGIBLE] theta cubed,

556
00:41:38,480 --> 00:41:40,880
it's a theta cubed term.

557
00:41:40,880 --> 00:41:45,490
Exactly, it's a minus theta
cubed over 6, three factorial,

558
00:41:45,490 --> 00:41:47,700
but it's a theta cubed term.

559
00:41:47,700 --> 00:41:53,110
And so the theta gave
us the right thing

560
00:41:53,110 --> 00:41:55,580
and then two more
powers of theta

561
00:41:55,580 --> 00:41:58,070
will give us a
second-order error, right?

562
00:42:00,880 --> 00:42:08,600
I better move since -- so that's
the story for semidiscrete.

563
00:42:08,600 --> 00:42:12,620
Semidiscrete produced
a pure exponential,

564
00:42:12,620 --> 00:42:21,360
but with a phase
factor, you could say,

565
00:42:21,360 --> 00:42:23,990
that wasn't linear in k
where the real one was.

566
00:42:26,590 --> 00:42:33,350
Now, what about -- let me
move up to that board above,

567
00:42:33,350 --> 00:42:38,970
where now we're going to have
-- we have it's discrete in time

568
00:42:38,970 --> 00:42:43,394
as well, so we're going
to have single steps.

569
00:42:43,394 --> 00:42:44,810
What's going to
be the difference?

570
00:42:47,580 --> 00:42:50,530
Again, I'm going
to try -- you know,

571
00:42:50,530 --> 00:42:56,920
I only have one idea here
-- U_(j, n) will be some G,

572
00:42:56,920 --> 00:43:04,750
some growth factor, times
e to the i*j*k delta x.

573
00:43:04,750 --> 00:43:09,740
The same -- I'm going to
try an e to the i*k*x,

574
00:43:09,740 --> 00:43:14,830
and in the space variable, it
will be G. And after n steps,

575
00:43:14,830 --> 00:43:17,730
it'll be G n times.

576
00:43:17,730 --> 00:43:19,910
So this is the discrete, right?

577
00:43:19,910 --> 00:43:25,400
This is the ansatz,
to use a fancy word.

578
00:43:25,400 --> 00:43:26,590
This is what you plug in.

579
00:43:29,550 --> 00:43:31,800
Little crazy to use
the word ansatz, which

580
00:43:31,800 --> 00:43:36,110
sounds fancy and the word
plug, which is far from fancy.

581
00:43:36,110 --> 00:43:40,460
I should say substitute.

582
00:43:40,460 --> 00:43:41,560
Substitute in it.

583
00:43:41,560 --> 00:43:43,710
All right.

584
00:43:43,710 --> 00:43:46,990
Do we know what the
process is going to do?

585
00:43:46,990 --> 00:43:51,900
When I plug it in -- sorry,
substitute it in on the left,

586
00:43:51,900 --> 00:43:59,710
we get -- and I'm going to
cancel this from every term --

587
00:43:59,710 --> 00:44:03,090
so all I'm interested
in is the --

588
00:44:03,090 --> 00:44:05,780
maybe I'll cancel
G to the n minus 1.

589
00:44:05,780 --> 00:44:08,090
So let me write down what I get.

590
00:44:08,090 --> 00:44:10,580
This is going to give me
-- in the time direction,

591
00:44:10,580 --> 00:44:19,080
I'll have a G squared minus a 2G
plus a 1 over delta t squared,

592
00:44:19,080 --> 00:44:23,650
and in the space direction,
which is at the centered time,

593
00:44:23,650 --> 00:44:28,330
so it's going to multiply a
G, it's going to be my same,

594
00:44:28,330 --> 00:44:29,620
my very same thing.

595
00:44:29,620 --> 00:44:33,260
I could maybe give
it a name here.

596
00:44:33,260 --> 00:44:35,960
I don't know whether --
should I give it a name?

597
00:44:35,960 --> 00:44:37,770
I know what it is
anyway, because it's

598
00:44:37,770 --> 00:44:42,510
the same thing that I had
in the semidiscrete method.

599
00:44:42,510 --> 00:44:49,130
It's this -- let's see
what it looked like.

600
00:44:49,130 --> 00:44:57,980
At this point, it was --
look, let's do one thing.

601
00:44:57,980 --> 00:45:02,010
Let me get this delta t squared
up there times the c squared

602
00:45:02,010 --> 00:45:03,670
divided by the delta x squared.

603
00:45:03,670 --> 00:45:08,810
Let me just clean
up those constants,

604
00:45:08,810 --> 00:45:13,250
so that when I multiply up
by the delta t squared times

605
00:45:13,250 --> 00:45:18,640
the c squared divided by the
delta x squared, what's that?

606
00:45:18,640 --> 00:45:24,030
c delta t over delta x squared
is my old ratio, the Courant

607
00:45:24,030 --> 00:45:31,020
number r, squared, and
then it's this times G.

608
00:45:31,020 --> 00:45:33,980
So this will all be
just exactly the same

609
00:45:33,980 --> 00:45:38,380
and that's what I
simplified here.

610
00:45:38,380 --> 00:45:43,030
Let me just leave it
as -- with a minus.

611
00:45:43,030 --> 00:45:48,370
Let me remember that I did take
a minus so that I could write

612
00:45:48,370 --> 00:45:58,400
it as 2 minus 2 cosine of k
delta x times G. Is that --

613
00:45:58,400 --> 00:46:01,950
I really think I have kept
track of everything here.

614
00:46:01,950 --> 00:46:07,640
The r squared coming from
all this stuff, the minus 2

615
00:46:07,640 --> 00:46:11,560
coming here and the plus 2
cosine coming from there.

616
00:46:11,560 --> 00:46:12,370
You OK with that?

617
00:46:16,670 --> 00:46:20,510
But there's one new
ingredient here.

618
00:46:20,510 --> 00:46:27,240
Again, G to the n -- I cancelled
G to the n minus 1 times this

619
00:46:27,240 --> 00:46:30,340
from every term.

620
00:46:30,340 --> 00:46:33,850
So the new ingredient, that we
just have five minutes to deal

621
00:46:33,850 --> 00:46:39,530
with, is the fact
that we have second --

622
00:46:39,530 --> 00:46:42,760
G squared just showed up.

623
00:46:42,760 --> 00:46:44,330
So why did G squared show up?

624
00:46:44,330 --> 00:46:52,000
Because it's a multistep
method and therefore,

625
00:46:52,000 --> 00:46:53,240
I've got two G's.

626
00:46:53,240 --> 00:46:54,320
Of course.

627
00:46:54,320 --> 00:46:57,690
I had two G's in all
these other cases

628
00:46:57,690 --> 00:46:59,440
because I had second
derivatives and now I

629
00:46:59,440 --> 00:47:01,160
have second differences.

630
00:47:01,160 --> 00:47:04,180
So I have two G's, so I
have to solve for G --

631
00:47:04,180 --> 00:47:07,220
and what is stability
going to depend on?

632
00:47:07,220 --> 00:47:11,760
Stability is going
to be whether G --

633
00:47:11,760 --> 00:47:14,090
both G's, because there
are two of them now --

634
00:47:14,090 --> 00:47:15,750
are less than or equal to 1.

635
00:47:15,750 --> 00:47:18,510
That's what I have to check
and that's the equation.

636
00:47:22,490 --> 00:47:24,610
What do we figure?

637
00:47:24,610 --> 00:47:30,160
We figure that if r is pretty
big -- if r is big, no way.

638
00:47:30,160 --> 00:47:34,770
If r is big, the
Courant test will fail

639
00:47:34,770 --> 00:47:37,040
and stability will fail.

640
00:47:37,040 --> 00:47:40,050
But if r is small,
then I expect --

641
00:47:40,050 --> 00:47:42,590
I have a right to
hope for stability.

642
00:47:42,590 --> 00:47:45,500
So it's going to
be the size of r.

643
00:47:45,500 --> 00:47:48,370
So I just have to -- I've
got this quadratic equation.

644
00:47:48,370 --> 00:47:50,530
I'm going to bring
this thing over here.

645
00:47:50,530 --> 00:47:56,520
I'm going to write this as
G squared minus 2a*G plus 1

646
00:47:56,520 --> 00:48:01,860
equals 0, where a is this
quantity -- So I had minus 2a,

647
00:48:01,860 --> 00:48:08,240
so a is a 1 from there, and when
I bring this over to the other

648
00:48:08,240 --> 00:48:13,810
side, it's going to be a plus
r squared -- factoring out --

649
00:48:13,810 --> 00:48:16,000
notice the 2 is everywhere.

650
00:48:16,000 --> 00:48:19,190
So I put the 2 here.

651
00:48:19,190 --> 00:48:23,460
The a comes from the 1
there and from this times

652
00:48:23,460 --> 00:48:30,670
the r squared 1 plus r squared
minus r squared cos k delta x.

653
00:48:39,380 --> 00:48:45,560
I'm just doing algebra and
it's coming out nicely.

654
00:48:45,560 --> 00:48:47,510
It's coming out
to a nice equation

655
00:48:47,510 --> 00:48:51,080
here with a slightly
messy expression for a,

656
00:48:51,080 --> 00:48:57,730
but now that I've named this
quantity a, I can solve this.

657
00:48:57,730 --> 00:48:59,120
So what's the solution?

658
00:48:59,120 --> 00:49:07,430
I remember the quadratic formula
and I get G as 1, I think,

659
00:49:07,430 --> 00:49:13,450
plus or minus the square
root of 1 minus a squared.

660
00:49:16,010 --> 00:49:24,280
That's G. Oh, maybe this
is an a; probably is.

661
00:49:24,280 --> 00:49:31,450
I guess the quadratic equation
takes a little memorizing,

662
00:49:31,450 --> 00:49:34,780
but I guess the first point
is that that coefficient shows

663
00:49:34,780 --> 00:49:38,940
up there and then the
square root that we know.

664
00:49:41,550 --> 00:49:42,380
So what's up?

665
00:49:47,440 --> 00:49:50,030
Again, none of this is deep.

666
00:49:50,030 --> 00:49:56,330
I'm just following
my one-way path.

667
00:49:56,330 --> 00:49:58,290
Check stability.

668
00:49:58,290 --> 00:50:02,140
Check the size of G.
Get an equation for G.

669
00:50:02,140 --> 00:50:06,970
Get a formula for
G and look at it.

670
00:50:06,970 --> 00:50:12,010
So a is some positive
number and the key

671
00:50:12,010 --> 00:50:15,790
is going to be whether
a is bigger than 1

672
00:50:15,790 --> 00:50:17,950
or smaller than 1.

673
00:50:17,950 --> 00:50:25,160
If a is -- is that right, 1
minus a squared or should that

674
00:50:25,160 --> 00:50:28,760
be a squared minus 1?

675
00:50:28,760 --> 00:50:30,150
Is it a squared minus 1?

676
00:50:32,780 --> 00:50:36,080
Professors are not responsible
for the quadratic formula

677
00:50:36,080 --> 00:50:41,450
because we proved that we
could do it in third grade

678
00:50:41,450 --> 00:50:44,110
and lost it since.

679
00:50:44,110 --> 00:50:46,370
I guess it's right.

680
00:50:46,370 --> 00:50:49,050
Is that right?

681
00:50:49,050 --> 00:50:52,150
Because look, here's the story.

682
00:50:52,150 --> 00:50:55,350
Suppose a is bigger than 1.

683
00:50:55,350 --> 00:50:59,440
Suppose -- I'm really looking
at the roots of this equation.

684
00:50:59,440 --> 00:51:01,810
Notice, by the way, the
roots of that equation

685
00:51:01,810 --> 00:51:04,020
multiply to give 1.

686
00:51:04,020 --> 00:51:07,240
So I'm on the edge here.

687
00:51:07,240 --> 00:51:14,220
Either both roots have size 1 or
one of those roots is too big.

688
00:51:14,220 --> 00:51:17,650
The test is going to be, is
a greater than 1? a greater

689
00:51:17,650 --> 00:51:28,020
than 1 will be unstable. a
less than 1 will be stable

690
00:51:28,020 --> 00:51:35,420
and I'll connect a less
than 1 to r less than 1.

691
00:51:35,420 --> 00:51:39,330
That will give me r less than 1.

692
00:51:39,330 --> 00:51:39,830
All right.

693
00:51:39,830 --> 00:51:44,990
I've given away the result
because time is pressing.

694
00:51:44,990 --> 00:51:49,760
Let me just take the
remaining time to see this.

695
00:51:49,760 --> 00:51:53,300
Suppose a is bigger than 1.

696
00:51:53,300 --> 00:51:55,520
I'm just looking at the
roots of this equation

697
00:51:55,520 --> 00:51:56,820
and here they are.

698
00:51:56,820 --> 00:52:03,190
If a is bigger than 1, like 2,
I have 2 plus square root of 3,

699
00:52:03,190 --> 00:52:05,470
I'm way up, right?

700
00:52:05,470 --> 00:52:07,990
I'm bigger than 1, no question.

701
00:52:07,990 --> 00:52:11,550
Suppose a is less than 1.

702
00:52:11,550 --> 00:52:14,420
Suppose a is 1/2.

703
00:52:14,420 --> 00:52:19,880
Then I have 1/2 plus or minus
the square root of -- what?

704
00:52:19,880 --> 00:52:21,280
What's up here?

705
00:52:21,280 --> 00:52:28,440
If a is smaller than 1, this is
negative and if it's negative,

706
00:52:28,440 --> 00:52:31,800
its square root is imaginary.

707
00:52:31,800 --> 00:52:33,510
So this is good.

708
00:52:33,510 --> 00:52:39,200
I have a real part and
then an imaginary part

709
00:52:39,200 --> 00:52:43,070
and then on some magic
little bit of board,

710
00:52:43,070 --> 00:52:44,650
I'm going to add their squares.

711
00:52:44,650 --> 00:52:50,240
So I just want to take the real
parts squared and the imaginary

712
00:52:50,240 --> 00:52:54,230
part squared -- now the
imaginary part is going to be 1

713
00:52:54,230 --> 00:53:00,090
minus a squared because
when I bring out an i,

714
00:53:00,090 --> 00:53:04,310
then I have a 1 minus a squared
there and when I add them,

715
00:53:04,310 --> 00:53:06,850
I get 1.

716
00:53:06,850 --> 00:53:09,810
So let me just say
again what it is.

717
00:53:09,810 --> 00:53:11,880
I just figured out
what is absolute value

718
00:53:11,880 --> 00:53:16,010
of G squared as
real part squared

719
00:53:16,010 --> 00:53:19,420
and imaginary part squared
and I got the answer 1.

720
00:53:19,420 --> 00:53:25,620
So the conclusion is, the roots
are stable when a is below 1

721
00:53:25,620 --> 00:53:30,120
and I have instability
when a is above 1.

722
00:53:30,120 --> 00:53:34,620
Then if I track that down,
that reduces to the tests --

723
00:53:34,620 --> 00:53:37,700
it just happens to be
the same test on r,

724
00:53:37,700 --> 00:53:41,810
whether r is below 1, which
is the Courant-Friedrichs-Lewy

725
00:53:41,810 --> 00:53:46,120
condition, that I'm staying
inside the characteristics,

726
00:53:46,120 --> 00:53:50,920
or r is bigger than 1 and
I'm taking a delta t that

727
00:53:50,920 --> 00:53:53,540
isn't stable.

728
00:53:53,540 --> 00:53:56,530
I'm sorry to go
slightly over the time.

729
00:53:56,530 --> 00:53:58,240
Thanks for staying with me.

730
00:53:58,240 --> 00:54:03,130
I'll begin next time
with this staggered grid.

731
00:54:03,130 --> 00:54:05,470
That won't take
the whole lecture,

732
00:54:05,470 --> 00:54:10,680
but it's worth knowing because
we see staggered grids in many

733
00:54:10,680 --> 00:54:13,210
of the best methods.

734
00:54:13,210 --> 00:54:13,760
Thanks.

735
00:54:13,760 --> 00:54:17,510
Have a good weekend and I guess
it's Tuesday rather than Monday

736
00:54:17,510 --> 00:54:19,940
that I see you next.