WEBVTT
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GILBERT STRANG: OK.
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So this is using Fourier series.
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So I had to pick
an equation where
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we were given a
function, and not just
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a couple of initial values.
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So I made the equation a
partial differential equation.
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The most famous one,
Laplace's equation.
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So this is the setup.
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And you'll see how
Fourier series comes in.
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We're in a circle.
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I'm going to make this
a nice model problem.
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So inside this circle we're
solving Laplace's equation.
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Laplace's equation was
the second derivative
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of u in the x direction,
plus the second derivative
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of u in the y direction, is 0.
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That's the way heat,
temperature, distributes itself
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when you leave it alone.
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In this problem I'm going to put
a source of heat at that point.
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So it'll be a point source.
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A delta function.
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And on the rest of the
boundary, temperature 0.
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So the boundary function
is a delta function
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with a spike at that one
point, and 0 elsewhere.
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And our problem is to solve
the Laplace's equation
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inside the circle.
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And we use polar coordinates
because we've got a circle.
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So there is the
equation with x and y,
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but we really are
thinking r and theta.
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And the reason is, you
get beautiful solutions
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to this equation
using r and theta.
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And that was a
family of solutions.
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r to the n-th cos
n theta just works.
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And so does r to the
n-th sine n theta.
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And that's for every n.
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So we have-- we can combine.
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We have a linear equation.
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We can take combinations of
solutions with coefficients
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a n in the cosines,
and bn in the sines.
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And now here's the key step.
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Put in r equal 1.
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Put r equal 1.
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And then this solution, u
at 1 and theta-- r equal 1--
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is the boundary.
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It's the circle.
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And that's where we're given u
of 1 to be the delta function.
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The point source.
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The delta function.
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Delta of theta.
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The point source
at theta equals 0.
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So you see our job.
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That function, that
boundary condition,
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is supposed to tell us
the a's and the b's.
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And then we have our solution.
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So by putting r equal
to 1 in this formula,
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we're supposed to get
the delta function.
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So let me put r equal 1.
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Easy to do.
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It's the sum of a n, 1
to the n-th, cos n theta,
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plus the sum of bn, 1 to
the n-th, sine n theta,
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is supposed to match
the delta function.
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So that's the Fourier series
for the delta function.
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That's the whole point.
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That we use a Fourier
series expression
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for the boundary function,
whatever that boundary function
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is.
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Here it's a particularly
nice neat one.
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And actually, the delta
function is an even function.
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It's 0 at theta and
it's 0 at minus theta.
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So changing theta to minus theta
still leaves me the spike at 0.
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So because it's
an even function,
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I won't see any signs.
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I won't see any odd functions.
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The sine theta.
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And I have an easy time
to find the coefficients
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a n of the cosines.
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Actually, we did that directly
from the formula for the a n's.
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Let me just remember
that formula.
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The formula was a0 was 1
over 2 pi times the average.
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a0 is the average value
of the temperature.
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And the temperature on the
boundary is delta theta.
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And that integrates to 1, and
we get the answer 1 over 2 pi.
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That's the average temperature.
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Isn't that a little weird?
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The temperature 0
except at one point.
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At that point it's a delta
function with the coefficient 1
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outside it.
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And then we get 1 over
2 pi as the average.
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The other a n's, the
coefficients of the cosines,
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are 1 over pi, times the
integral of our delta function,
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times cos n theta d theta.
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And the delta function,
that point source,
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picks out that number
at theta equals 0.
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And that number is 1.
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So I'm getting 1 over pi.
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So finally I now
know the a's and b's.
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When I put those in, that
tells me the solution.
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The solution-- now I can
put r back in the picture--
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it's a sum.
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Well, let me take the a0 term.
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The a0 is 1 over 2 pi--
that's the constant, that's
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the average-- plus the sum of
1 over pi cos n theta, from n
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equals 1 to infinity.
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And r to the n-th.
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Sorry. r to the n-th.
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So you see what happens.
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When r is 1, we have the Fourier
series for the delta function.
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That's the very
exceptional function
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that's given on the boundary.
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As soon as r is less than
1, these r to the n-th's
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get small, and we
have a series that
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adds up to a reasonable sum.
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And we can actually-- it's
possible to add up that series.
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It's possible to
add up that series.
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It's a geometric series
if you switch from cosines
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to exponentials.
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That's usually the good
way to get good formulas.
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And here, so you can add it up.
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And I think there's a 1
over 2 pi still there.
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And I think it's 1 minus r
squared, over 1 plus r squared,
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minus 2r cos theta.
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Let me just be sure
I got that right.
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Yep, looks good.
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Looks good.
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And we could check if it's good.
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Let's take theta equal 0.
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So if we take theta equal 0.
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Let me draw that circle again.
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Theta equal 0.
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We're coming out on that ray.
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And we're expecting to see
infinity when we get there,
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at r equal 1.
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So theta equal 0.
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So let me just put that.
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On the ray, theta equal 0.
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This is what you should do.
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We have a formula
for all r and theta,
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but let's look at
some particular points
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to see what's happening.
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So along that ray,
where theta is 0,
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I have 1 over 2 pi, 1
minus r squared, over 1
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plus r squared, minus 2r.
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Because cos theta is 1.
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And 1 plus r squared, minus
2r, is 1 minus r squared.
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Right?
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Because cos theta
is 1 on this ray.
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Theta is 0.
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Cos theta is 1.
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And now 1 minus r will
factor out of this.
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And I think we get 1 plus r.
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And we still have a
1 minus r down below.
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I like that.
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You don't often, for partial
differential equations,
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get some nice expression
for the solution.
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So that's the solution.
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And as r goes to 1,
this solution blows up.
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Right.
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The temperature is
infinite on the boundary,
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but the temperature is
something reasonable inside.
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And at r equals 0,
I have 1 over 2 pi.
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Well, of course.
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It's the average value.
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Right at the center
that temperature
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is going to be the
average on the boundary.
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That's a natural key property
of Laplace's equation.
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It averages everything.
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Actually, if I take a
little circle in anywhere,
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those temperature in the
center of that circle
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would be the average
of the temperatures
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around the little circle.
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For all the circles.
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It's just the
Laplace's equation.
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Solving Laplace's equation
averages everything.
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And the result is that the
temperature function sort of
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smoothes out as I come in.
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Around the boundary
it's far from smooth.
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There's a big jolt
at theta equals 0.
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But if I look on that circle,
or that circle, or this circle,
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the temperature is a
nice smooth function.
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And it's never going to be above
the maximum on the boundary.
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And it's never going to be below
the minimum on the boundary.
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Everything's being averaged.
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So that's, you see, a use
of the Fourier series.
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For one particular function.
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I could do another function,
but I don't think I will.
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I could take the function that's
1 on the top of the circle
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and minus 1 on the bottom
half of the circle.
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OK, that's a function with a
jump, but not a delta function.
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So we would see a Fourier
series that would give us
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the a's and the b's.
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There would probably
only be b's in that case.
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Sine.
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Sine terms.
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And we'd get an answer.
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May I just, while I'm
talking about averages,
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add one final comment.
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Usually, for a
complicated region,
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we can't solve Laplace's
equation with formulas.
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It's not possible.
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We can't find sines and cosines
that match some crazy boundary.
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So we have to replace
Laplace's equation.
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So I'll write Laplace's
equation again.
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That goes into u--
we have a region.
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And we carve it out with a grid.
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And then at each
point on the grid,
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we have an equation connecting
the value of u at that point.
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Say u0 at the center.
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With u east, maybe, u
west, u north, and u south.
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So we have an
equation, and I want
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to write that equation down.
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u center is just going
to be the average.
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It's just going to be 1/4 of
u east, u west, u north, and u
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south.
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So that'll be true at--
that equation will hold.
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The unknowns are all these u's.
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The u's of all the mesh points.
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And I have an equation
at every mesh point.
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So I have the same number of
equations from the mesh points
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as unknowns at the mesh points.
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I solve that big system, and
that gives me a solution u.
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An approximate solution
u to Laplace's equation.
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So this would be called
Laplace's difference equation,
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or Laplace's five-point
scheme, because it uses
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five points in that average.
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OK.
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That's an important problem
in numerical analysis.
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Thank you.