WEBVTT
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Today I'm speaking about
the first of the three great
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partial differential equations.
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So this one is called Laplace's
equation, named after Laplace.
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And you see partial derivatives.
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So we have-- I don't have time.
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This equation is
in steady state.
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I have x and y, I'm
in the xy plane.
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And I have second
derivatives in x and then y.
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So I'm looking for
solutions to that equation.
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And of course I'm given
some boundary values.
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So time is not here.
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The boundary
values, the boundary
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is in the xy plane,
maybe a circle.
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Think about a circle
in the xy plane.
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And on the circle, I
know the solution u.
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So the boundary values
around the circle are given.
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And I have to find
the temperature
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u inside the circle.
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So I know the temperature
on the boundary.
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I let it settle down and I want
to know the temperature inside.
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And the beauty is, it solves
that basic partial differential
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equation.
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So let's find some solutions.
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They might not match
the boundary values,
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but we can use them.
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So u equal constant certainly
solves the equation.
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U equal x, the second
derivatives will be 0.
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U equal y.
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Here is a better one, x
squared minus y squared.
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So the second derivative
in the x direction is 2.
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The second derivative in
the y direction is minus 2.
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So I have 2, minus 2,
it solves the equation.
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Or this one, the second
derivative in x is 0.
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Second derivative in y is 0,
those are simple solutions.
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But those are only
a few solutions
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and we need an infinite
sequence because we're going
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to match boundary conditions.
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So is there a pattern here?
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So this is degree 0, constant.
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These are degree 1, linear.
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These are degree 2, quadratic.
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So I hope for two cubic ones.
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And then I hope for
two fourth degree ones.
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And that's the pattern,
that's the pattern.
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Let me find-- let me
spot the cubic ones.
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X cubed, if I start with x
cubed, of course the second x
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derivative is probably 6x.
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So I need the second y
derivative to be minus 6x.
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And I think minus
3xy squared does it.
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The second derivative
in y is 2 times
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the minus 3x is
minus 6x, cancels
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the 6x from the second
derivative there, and it works.
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So that fits the pattern,
but what is the pattern?
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Here it is.
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It's fantastic.
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I get these crazy polynomials
from taking x plus iy
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to the different powers.
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Here to the first power, if n
is 1, and I just have x plus iy
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and I take the real
part, that's x.
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So I'll take the
real part of this.
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The real part of this when
n is 1, the real part is x.
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What about when n is 2?
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Can you square
that in your head?
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So we have x squared and we
have i squared y squared,
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i squared being minus 1.
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So I have x squared and
I have minus y squared.
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Look, the real
part of this when n
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is 2, the real part
of x plus iy squared,
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the real part is x
squared minus y squared.
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And the imaginary
part was the 2ixy.
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So the imaginary part that
multiplies i is the 2xy.
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This is our pattern when n is 2.
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And when n is 3, I take
x plus iy cubed, and that
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begins with x cubed like that.
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And then I think that
the other real part
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would be a minus 3xy squared.
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I think you should check that.
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And then there will
be an imaginary part.
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Well, I think I could figure out
the imaginary part as I think.
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Maybe something like
minus-- maybe it's
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3yx squared minus y cubed,
something like that.
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That would be the real
part and that would be
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the imaginary part when n is 3.
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And wonderfully,
wonderfully, it works
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for all powers, exponents n.
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So I have now sort of a pretty
big family of solutions.
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A list, a double list,
really, the real parts
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and the imaginary
parts for every n.
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So I can use those
to find the solution
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u, which I'm looking for, the
temperature inside the circle.
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Now of course, I have
a linear equation.
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So if I have several
solutions, I can combine them
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and I still have a solution.
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X plus 7y will be a solution.
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Plus 11x squared minus
y squared, no problem.
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Plus 56 times 2xy.
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Those are all solutions.
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So I'm going to find a
solution, my final solution
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u will be a combination of this,
this, this, this, this, this,
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this, and all the
others for higher n.
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That's going to be my solution.
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And I will need that
infinite family.
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See, partial
differential equations,
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we move up to infinite family
of solutions instead of just
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a couple of null solutions.
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So let me take an example.
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Let me take an example.
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We're taking the
region to be a circle.
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So in that circle, I'm looking
for the solution u of x and y.
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And actually in a circle,
it's pretty natural
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to use polar coordinates.
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Instead of x and y
inside a circle that's
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inconvenient in the
xy plane, its equation
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involves x equals square root of
1 minus y squared or something,
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I'll switch to polar
coordinates r and theta.
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Well, you might say
you remember we had
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these nice family of solutions.
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Is it still good in
polar coordinates?
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Well the fact is,
it's even better.
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So the solution of u
will be the real part
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and the imaginary part.
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Now what is x plus
iy in r and theta?
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Well, we all know x is r Cos
theta plus ir sine theta.
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And that's r times Cos
theta plus i sine theta,
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the one unforgettable
complex Euler's formula, e
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to the I theta.
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Now, I need its nth power.
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The nth power of
this is wonderful.
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The real part and imaginary
part of the nth power
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is r to the nth e
to the in theta.
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That's my x plus iy to the nth.
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Much nicer in polar
coordinates, because I
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can take the real part and
the imaginary part right away.
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It's r to the nth Cos n theta
and r to the nth sine n theta.
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These are my solutions,
my long list of solutions,
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to Laplace's equation.
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And it's some
combination of those,
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my final thing is going to
be some combination of those,
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some combination.
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Maybe coefficients a sub n.
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I can use these and
I can use these.
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So maybe b sub n r to
the nth sine n theta.
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You may wonder what I'm
doing, but what I'm achieved,
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it's done now, is to
find the general solution
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of Laplace's equation.
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Instead of two
constants that we had
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for an ordinary differential
equation, a C1 and a C2,
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here I have these guys
go from up to infinity.
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N goes up to infinity.
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So I have many solutions.
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And any combination working,
so that's the general solution.
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That's the general solution.
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And I would have to
match that-- now here's
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the final step and not
simple, not always simple--
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I have to match this to
the boundary conditions.
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That's what will tell me
the constants, of course.
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As usual, c1 and c2 came from
the matching the conditions.
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Now I don't have just c1 and
c2, I have this infinite family
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of a's, infinite family of b's.
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And I have a lot more to
match because on the boundary,
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here I have to match
u0, which is given.
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So I might be
given, suppose I was
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given u0 equal to the
temperature was equal 1
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on the top half.
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And on the bottom half, say
the temperature is minus 1.
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That's a typical problem.
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I have a circular region.
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The top half is held
at one temperature,
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the lower half is held at
a different temperature.
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I reach equilibrium.
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Everybody knows that
along that line,
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probably the temperature
would be 0 by symmetry.
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But once the temperature
there halfway up, not so easy,
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or anywhere in there.
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Well, the answer is u in
the middle, u of r and theta
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inside is given by that formula.
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And again, the ANs
and the BNs come
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by matching the-- getting the
right answer on the boundary.
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Well, there's a big theory
there how do I match these?
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That's called a Fourier series.
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That's called a Fourier series.
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So I'm finding the coefficients
for a Fourier series, the A's
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and B's, that match a
function around the boundary.
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And I could match any
function, and Fourier series
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is another entirely
separate video.
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We've done the job with
Laplace's equation in a circle.
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We've reduced the problem
to a Fourier series problem.
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We have found the
general solution.
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And then to match it to
a specific given boundary
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value, that's a
Fourier series problem.
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So I'll have to put that off
to the Fourier series video.
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Thank you.