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GILBERT STRANG: This
is heat equation video.

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So this is the second
of the three basic

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partial differential equations.

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We had Laplace's equation,
that was-- time was not there.

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Now time comes into
the heat equation.

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We have a time derivative,
and two-- matching

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with two space derivatives.

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So I have my function.

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My solution depends
on t and on x,

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and I hope I can
separate those two parts.

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This is exactly like the way
we solved the ordinary systems

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of differential equations.

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We pulled out an
e to the lambda t,

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where lambda was the
eigenvalue, and then we

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had the eigenvector.

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Here, it's an eigenfunction
because it depends on x.

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We didn't have x before, but
now we have partial differential

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equations.

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X is also a coordinate here.

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So I look for
solutions like that.

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And just as always, I substitute
that into the differential

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equation to discover what--
what determines S, S of x.

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The time derivative
brings down a lambda.

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The space derivative
brings down--

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has two space derivatives.

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So that's what I get when I
substitute e to the lambda t S

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eigenvalue, times eigenfunction,
into the equation.

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And always, I cancel e
to the lambda t, that's

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the beauty of it, and I
have an eigenvalue equation,

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or I'm looking again
for a function.

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So the second derivative
of my function

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is lambda, some number,
times my function.

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OK.

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What I'm looking
for for functions S,

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they'll be sine functions.

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S will be, S of x, will
be the sine of K pi

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x, the sine of x, the
sign of something times x.

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What's the eigenvalue?

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Take two derivatives
of this, I get back

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sine k pi x, which is, that's
great, it's an eigenfunction.

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And out comes the
eigenvalue, lambda

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is, well when I take
two derivatives,

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k pi comes out twice
with a minus sign.

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So it's minus k
squared, pi squared.

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So I have found a bunch of
eigenvectors, eigenfunctions,

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and the eigenvalues.

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So this was a simple pair,
but now the general solution,

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the general solution,
u of t x, will be?

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If I know several solutions,
and I have a linear equation,

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I just take combinations.

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That's always our way.

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Take combinations of
those basic solutions,

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so I'll have a sum, k
going from 1 to infinity.

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In differential equation,
partial differential equations,

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I need a big sum here,
of some coefficients,

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let me call them B k,
times this solution,

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e to the lambda-- what was
lambda-- minus k squared, pi

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squared t, times S, S number k.

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I should have given this
eigenfunction it's number, k.

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So there is the, this is
my family of eigenfunctions

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and eigenvalues.

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And then, this is the
combination of these solutions.

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There is the general solution.

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And for S k, I should have
written in sine k pi x.

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So that's the dependence on t.

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Let's have a look at this.

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So the dependence
on t is fast decay.

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And if K, as K gets larger,
later terms in this sum,

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the decay is really fast.

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So the term that decays
most slowly, k equal 1,

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there'll be a B1, an e to
the minus, pi squared t.

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That's decaying
already pretty fast.

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When I'm talking about
decay, what's happening here?

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I have a bar, a material bar.

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The ends of the bar are
kept at temperature zero,

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they're frozen, and
heat is in the bar.

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Heat is flowing
around in the bar.

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And where is it going?

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It's flowing out the ends.

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The bar is approaching freezing.

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The ends are kept at freezing,
and the inside of the bar,

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whatever heat is there
at the beginning,

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is going to flow out the ends.

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So, you see, that I have
these sine functions.

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When x is 0 the sine is 0
so that's frozen at one end.

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When x is 1, I have the sine
of k pi, which is, again, 0,

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so it's frozen at the other end.

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So I'm freezing it at both ends.

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The temperature is
escaping out of the center,

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so let me graph the solution.

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So maybe I start with--
here is my bar from 0 to 1,

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and I'm keeping it frozen.

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F for frozen, f for
frozen at that end.

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Maybe I, maybe I start
off with a warm bar.

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So u at 0 and x,
I'll say it'd be 1.

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This way is x.

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So I have an
ordinary heated bar,

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and I put it in the freezer.

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So I insulate the sides so
the heat is escaping out

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the two ends, out the end
x equals 0 and out the end

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x equal 1.

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And the solution will
be, let me remember

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what the general
solution looked like,

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and I have to find
these numbers.

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OK.

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And those numbers, of
course, the numbers

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are always found by matching
the initial conditions.

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This is the initial, this
is an initial picture.

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OK.

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So I have to match that
by-- so this is t equals 0,

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I have to match the sum of B k.

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This is k going
from 1 to infinity.

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When t is, when t is zero, this
is 1, and I just have the Sk.

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The sine of k pi x,
that has to match the 1.

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And from that, I find the Bk's,
and then the final solution.

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T greater than 0 uses those Bks.

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And we're again faced with
a Fourier series problem.

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Anytime I have to find
these coefficients,

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this is a Fourier sine
series, I have only sines,

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not cosines here.

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And I'm finding
the coefficients,

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so that this will match
1, the initial condition.

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And then, for t greater than 0,
solution u will be, as we said,

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the sum of these
Bk's, which come

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from the initial conditions,
come from this-- Fourier

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coefficients, we still have to
do that video on Fourier series

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to know what these numbers will
come out to be-- times the e,

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to the minus k
squared, pi squared

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t times the sine of k pi x.

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You may think, well it's
a pretty messy solution,

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because it's an infinite sum.

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But it's not bad for a
partial differential equation.

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We have numbers, we have
something depending on time

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and decaying rapidly, and
something depending on x.

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So at time 1, if
I drew a picture,

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suppose the heat is, the
temperature starts out

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through the whole bar at 1.

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But with this kind of time
decay, a little later in time,

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the temperature's going
to be something like that.

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It'll be way down at the ends,
pretty low in the middle.

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And so at some time
t, the temperature

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will look like that, and
then soon after that,

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the temperature
will go down here,

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and the steady state, of
course, is the whole thing

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is at temperature 0.

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So that's what solutions to
the heat equation look like.

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And this is the step of finding
the-- which I didn't take,

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it's the Fourier series
step-- of finding

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the coefficients in our
infinite series of solutions.

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Once again, we have
infinitely many solutions.

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We're talking about a partial
differential equation.

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We have a whole function to
match, so we need all of those.

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And Fourier series tells
us how to do that matching,

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how to find these Bk's.

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So that's a separate
and important question,

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Fourier series.

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Thank you.