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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,

55
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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.

150
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And we still have a
1 minus r down below.

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I like that.

152
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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.

155
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And as r goes to 1,
this solution blows up.

156
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Right.

157
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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.

162
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Right at the center
that temperature

163
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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

174
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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.

181
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Everything's being averaged.

182
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So that's, you see, a use
of the Fourier series.

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For one particular function.

184
00:10:59,350 --> 00:11:04,260
I could do another function,
but I don't think I will.

185
00:11:04,260 --> 00:11:08,320
I could take the function that's
1 on the top of the circle

186
00:11:08,320 --> 00:11:11,630
and minus 1 on the bottom
half of the circle.

187
00:11:11,630 --> 00:11:15,710
OK, that's a function with a
jump, but not a delta function.

188
00:11:15,710 --> 00:11:19,500
So we would see a Fourier
series that would give us

189
00:11:19,500 --> 00:11:20,655
the a's and the b's.

190
00:11:20,655 --> 00:11:24,070
There would probably
only be b's in that case.

191
00:11:24,070 --> 00:11:24,690
Sine.

192
00:11:24,690 --> 00:11:25,950
Sine terms.

193
00:11:25,950 --> 00:11:29,900
And we'd get an answer.

194
00:11:29,900 --> 00:11:32,090
May I just, while I'm
talking about averages,

195
00:11:32,090 --> 00:11:34,470
add one final comment.

196
00:11:34,470 --> 00:11:39,020
Usually, for a
complicated region,

197
00:11:39,020 --> 00:11:42,880
we can't solve Laplace's
equation with formulas.

198
00:11:42,880 --> 00:11:45,500
It's not possible.

199
00:11:45,500 --> 00:11:50,810
We can't find sines and cosines
that match some crazy boundary.

200
00:11:50,810 --> 00:11:55,600
So we have to replace
Laplace's equation.

201
00:11:55,600 --> 00:11:58,320
So I'll write Laplace's
equation again.

202
00:12:05,160 --> 00:12:13,300
That goes into u--
we have a region.

203
00:12:13,300 --> 00:12:19,050
And we carve it out with a grid.

204
00:12:19,050 --> 00:12:22,680
And then at each
point on the grid,

205
00:12:22,680 --> 00:12:27,400
we have an equation connecting
the value of u at that point.

206
00:12:27,400 --> 00:12:30,690
Say u0 at the center.

207
00:12:30,690 --> 00:12:42,040
With u east, maybe, u
west, u north, and u south.

208
00:12:45,200 --> 00:12:47,640
So we have an
equation, and I want

209
00:12:47,640 --> 00:12:49,690
to write that equation down.

210
00:12:49,690 --> 00:12:56,010
u center is just going
to be the average.

211
00:12:56,010 --> 00:13:05,270
It's just going to be 1/4 of
u east, u west, u north, and u

212
00:13:05,270 --> 00:13:05,770
south.

213
00:13:09,070 --> 00:13:13,150
So that'll be true at--
that equation will hold.

214
00:13:13,150 --> 00:13:15,730
The unknowns are all these u's.

215
00:13:15,730 --> 00:13:17,690
The u's of all the mesh points.

216
00:13:17,690 --> 00:13:20,490
And I have an equation
at every mesh point.

217
00:13:20,490 --> 00:13:26,640
So I have the same number of
equations from the mesh points

218
00:13:26,640 --> 00:13:29,890
as unknowns at the mesh points.

219
00:13:29,890 --> 00:13:36,500
I solve that big system, and
that gives me a solution u.

220
00:13:36,500 --> 00:13:40,170
An approximate solution
u to Laplace's equation.

221
00:13:40,170 --> 00:13:44,180
So this would be called
Laplace's difference equation,

222
00:13:44,180 --> 00:13:49,520
or Laplace's five-point
scheme, because it uses

223
00:13:49,520 --> 00:13:53,040
five points in that average.

224
00:13:53,040 --> 00:13:53,810
OK.

225
00:13:53,810 --> 00:13:57,650
That's an important problem
in numerical analysis.

226
00:13:57,650 --> 00:13:59,360
Thank you.