GILBERT STRANG: This
is heat equation video. So this is the second
of the three basic partial differential equations. We had Laplace's equation,
that was-- time was not there. Now time comes into
the heat equation. We have a time derivative,
and two-- matching with two space derivatives. So I have my function. My solution depends
on t and on x, and I hope I can
separate those two parts. This is exactly like the way
we solved the ordinary systems of differential equations. We pulled out an
e to the lambda t, where lambda was the
eigenvalue, and then we had the eigenvector. Here, it's an eigenfunction
because it depends on x. We didn't have x before, but
now we have partial differential equations. X is also a coordinate here. So I look for
solutions like that. And just as always, I substitute
that into the differential equation to discover what--
what determines S, S of x. The time derivative
brings down a lambda. The space derivative
brings down-- has two space derivatives. So that's what I get when I
substitute e to the lambda t S eigenvalue, times eigenfunction,
into the equation. And always, I cancel e
to the lambda t, that's the beauty of it, and I
have an eigenvalue equation, or I'm looking again
for a function. So the second derivative
of my function is lambda, some number,
times my function. OK. What I'm looking
for for functions S, they'll be sine functions. S will be, S of x, will
be the sine of K pi x, the sine of x, the
sign of something times x. What's the eigenvalue? Take two derivatives
of this, I get back sine k pi x, which is, that's
great, it's an eigenfunction. And out comes the
eigenvalue, lambda is, well when I take
two derivatives, k pi comes out twice
with a minus sign. So it's minus k
squared, pi squared. So I have found a bunch of
eigenvectors, eigenfunctions, and the eigenvalues. So this was a simple pair,
but now the general solution, the general solution,
u of t x, will be? If I know several solutions,
and I have a linear equation, I just take combinations. That's always our way. Take combinations of
those basic solutions, so I'll have a sum, k
going from 1 to infinity. In differential equation,
partial differential equations, I need a big sum here,
of some coefficients, let me call them B k,
times this solution, e to the lambda-- what was
lambda-- minus k squared, pi squared t, times S, S number k. I should have given this
eigenfunction it's number, k. So there is the, this is
my family of eigenfunctions and eigenvalues. And then, this is the
combination of these solutions. There is the general solution. And for S k, I should have
written in sine k pi x. So that's the dependence on t. Let's have a look at this. So the dependence
on t is fast decay. And if K, as K gets larger,
later terms in this sum, the decay is really fast. So the term that decays
most slowly, k equal 1, there'll be a B1, an e to
the minus, pi squared t. That's decaying
already pretty fast. When I'm talking about
decay, what's happening here? I have a bar, a material bar. The ends of the bar are
kept at temperature zero, they're frozen, and
heat is in the bar. Heat is flowing
around in the bar. And where is it going? It's flowing out the ends. The bar is approaching freezing. The ends are kept at freezing,
and the inside of the bar, whatever heat is there
at the beginning, is going to flow out the ends. So, you see, that I have
these sine functions. When x is 0 the sine is 0
so that's frozen at one end. When x is 1, I have the sine
of k pi, which is, again, 0, so it's frozen at the other end. So I'm freezing it at both ends. The temperature is
escaping out of the center, so let me graph the solution. So maybe I start with--
here is my bar from 0 to 1, and I'm keeping it frozen. F for frozen, f for
frozen at that end. Maybe I, maybe I start
off with a warm bar. So u at 0 and x,
I'll say it'd be 1. This way is x. So I have an
ordinary heated bar, and I put it in the freezer. So I insulate the sides so
the heat is escaping out the two ends, out the end
x equals 0 and out the end x equal 1. And the solution will
be, let me remember what the general
solution looked like, and I have to find
these numbers. OK. And those numbers, of
course, the numbers are always found by matching
the initial conditions. This is the initial, this
is an initial picture. OK. So I have to match that
by-- so this is t equals 0, I have to match the sum of B k. This is k going
from 1 to infinity. When t is, when t is zero, this
is 1, and I just have the Sk. The sine of k pi x,
that has to match the 1. And from that, I find the Bk's,
and then the final solution. T greater than 0 uses those Bks. And we're again faced with
a Fourier series problem. Anytime I have to find
these coefficients, this is a Fourier sine
series, I have only sines, not cosines here. And I'm finding
the coefficients, so that this will match
1, the initial condition. And then, for t greater than 0,
solution u will be, as we said, the sum of these
Bk's, which come from the initial conditions,
come from this-- Fourier coefficients, we still have to
do that video on Fourier series to know what these numbers will
come out to be-- times the e, to the minus k
squared, pi squared t times the sine of k pi x. You may think, well it's
a pretty messy solution, because it's an infinite sum. But it's not bad for a
partial differential equation. We have numbers, we have
something depending on time and decaying rapidly, and
something depending on x. So at time 1, if
I drew a picture, suppose the heat is, the
temperature starts out through the whole bar at 1. But with this kind of time
decay, a little later in time, the temperature's going
to be something like that. It'll be way down at the ends,
pretty low in the middle. And so at some time
t, the temperature will look like that, and
then soon after that, the temperature
will go down here, and the steady state, of
course, is the whole thing is at temperature 0. So that's what solutions to
the heat equation look like. And this is the step of finding
the-- which I didn't take, it's the Fourier series
step-- of finding the coefficients in our
infinite series of solutions. Once again, we have
infinitely many solutions. We're talking about a partial
differential equation. We have a whole function to
match, so we need all of those. And Fourier series tells
us how to do that matching, how to find these Bk's. So that's a separate
and important question, Fourier series. Thank you.
