- [Voiceover] In the last video, I started introducing
the intuition for the Laplacian operator
in the context of the function with this graph
and with the gradient field pictured below it.
And here, I'd like to go through the computation
involved in that.
So the function that I had there was defined,
it's a two-variable function.
And it's defined as f(x,y)
is equal to three
multiplied by sin(y/2).
And then the Laplacian which we define
with this right side up triangle
is an operator of f.
And it's defined to be the divergence,
so kind of this nabla dot
times the gradient which is just nabla of f.
So two different things going on.
It's kind of like a second derivative.
And the first thing we need to do
is take the gradient of f.
And the way we do that,
we kind of imagine expanding
this upside down triangle
as a vector full of partial differential operators:
partial partial x
and partial partial y.
And with a gradient, you just kind of imagine
multiplying that by the function.
So if you imagine multiplying that
by the function, what it looks like
is just a vector full of partial derivatives.
So you're taking the partial of f
with respect to x and the partial of f
with respect to y.
Those are the two different components
of this vector-valued function that is the gradient.
And in our specific example,
when we take the partial derivative of f
with respect to x
what we get, is we look over here.
Three just looks like a constant,
so nothing happens.
The derivative of that with respect to x,
we kind of take out that 1/2.
So 1/2, and the derivative of cos is -sin.
So that's -sin(x/2).
And sin(y/2), well, y just looks like a constant.
So sin(y/2) is just some other constant.
So in our derivative, we just keep that constant in there,
And then, for the second component,
the partial derivative of f
with respect to y.
Three still looks like a constant
cause it is a constant.
Now, cos(x/2) looks like a constant
because as far as y is concerned,
x is a constant.
So cos(x) is a constant.
But then, the sin(y) has a derivative of cos.
And we also take out that 1/2.
So you take out that 1/2
when you take the derivative of the inside
and then the derivative of the outside
is cosine of whatever was in there.
So in this case, y/2.
And we're multiplying it by that original constant,
So, still we have our cos(x/2)
since it was a constant times a certain variable thing, x/2.
So that's the gradient.
And then, the next step here
is to take the divergence of that.
So with the divergence, we're going to imagine
taking that del operator and dot producting
with this guy.
So if I scroll down to give some room here,
we're going to take that,
that vector that's kind of the same vector,
the partial partial x.
And I say vector, but vector-ish thing,
partial partial y.
And now we're going to take
the dot product with this entire guy.
So I'll go ahead and just copy it over.
Just kind of copy it over here.
And let's see.
So, I need a little bit more room to evaluate this.
So here, when you imagine taking the dot product,
you kind of multiply these top components together.
So we're taking the partial derivative
with respect to x of this whole guy.
And when you do that, well you get,
you still have that 1/2
and then the derivative of -sin(x/2).
So that 1/2 gets pulled out
when you're kind of taking
the derivative of the inside.
And the derivative of -sin is -cos.
So -cos of that stuff on the inside,
And of course, we still multiply it by this.
This looks like a constant, the sin(y/2).
And we multiply by that,
And then we add that
because it's kind of like a dot product.
You add that to what it looks like
when you multiply these next two components.
So we're going to add.
And you have that 1/2.
And then cos(y/2), when we differentiate that
you also pull out the 1/2.
So again, you have that pulled-out 1/2.
And the derivative of cos is -sin.
So now we're taking -sin of,
and then that stuff on the inside, y/2.
And we continue multiplying by the constant.
As far as y is concerned, cos(x/2) is a constant.
So we multiply it by that, cos(x/2).
And then that, so that is the divergence
of that gradient field.
So the divergence of the gradient
of our original function gives us the Laplacian.
And in fact, we could simplify this further
because both of these terms
kind of look identical.
But the main point of this video
is kind of how you go through that process
where you imagine taking the gradient
of your function and then the divergence of that.
And that's what the Laplacian is.
See you next video.