Divergence 1

Introduction to the divergence of a vector field.

Let's try to get our heads
around the idea of divergence.

So first, like I did with
gradients, I'll show you the

mechanics, which are actually
pretty straightforward.

And then I'll try to
give you the intuition.

And once you have the
intuition, at first it

will seemed very, I don't
know, unintuitive, maybe.

But it once you get it,
you're like oh, that's it.

So let's see what
divergence is.

Let's say I have
a vector field.

And let's say this vector
field, just for the purposes

of visualization it could be
anything, but let's say it

represents the velocity of
particles of fluid of any

point in two dimensions.

So it's going to be a
two-dimensional vector field.

It's going to be a function of
x and y, so the velocity at any

point-- it's a vector field --
let's say it is, and I'm just

going to make up something.

Let's say it's x squared, yi.

So at any point in the
x-direction, at any point x

comma y, its velocity in the
x-direction will

be x squared, y.

And then its velocity in the
y-direction, I don't know

maybe it's just 3y, j.

That's its velocity
in the x-direction.

So its velocity in the
x-direction is actually

a function of x and y.

its velocity in the y-direction
is just a function of y.

So what is the divergence?

So a couple of ways
we can write it.

The correct way to write
it is the divergence of

our vector field, v.

But a common mnemonic to
remember the operation of

diverge and is to write the
upside down triangle, which was

the same notation we used for
gradient, but take the dot

product of that and the vector.

And if you remember from the
gradient discussion, we said

that you can view, although
it's kind of an abuse of

notation, but you could view
this upside down triangle as

being equal to the partial
derivative with respect to x in

the x-direction plus the
partial derivative with respect

to y in the y-direction,
which is the j-unit vector.

And then if we went to three
dimensions, the partial

derivative with respect to
z and the k-direction,

et cetera, et cetera.

But we're dealing with a
two-dimensional vector here,

so let's just stick with
two dimensions, x and y.

So what would this
turn out to be?

If you took the dot product of
this, which is this upside down

triangle, with this vector
field, what would you get?

Well, you would just get the
partial derivative of the x

dimension with respect to x, so
you would get-- it's actually

pretty straight forward to
memorize; you might not even

need this mnemonic right here,
this abuse of notation; you

might just know it off hand
--the x component, you take the

partial derivative with respect
to x, and the y component, you

take the partial derivative
with respect to y.

But I'll show you why it
looks like the dot product.

So if you took the dot product
of that and that, it would be

the partial derivative with
respect to x of that

expression, of x squared, y and
then plus the partial

derivative with respect to y of
that second expression, the y

component of 3y, and then
you would evaluate it.

What's the partial derivative
of this with respect to x?

We just pretended y is a
constant, just a number, so

the derivative of this with
respect to x, would be

2x times the constant.

So it'll be 2xy plus-- what's
the partial derivative

of 3y with respect to y?

Well, there's nothing else to
hold constant, so it's just

like taking the derivative
with respect to y

--so it's 2y plus 3.

So this is the divergence
at a point x, y.

You could almost view it
as a function of x and y.

So you could almost say you
know, that the divergence of

v-- I'm going to make up some
notation here --as long as you

get the point across, you can
say that the divergence

of v, that this is a
function of x and y.

That we just have an expression
that if you give me a point

anywhere in this vector
field, I can tell you the

divergence at that point.

So I think you'll find that
the computation of divergence

isn't too difficult.

You just take the partial
derivative of the x component

with respect to x, and you add
that to the partial derivative

to the y component
with respect to y.

And if you had the z, you
would do the same thing,

so on and so forth.

Actually, let me do just do one
more just hit the point home,

and then we'll work
on intuition.

So if I said that I had, I
don't know, let's say, my

vector field is cosine of yi
plus-- so it's interesting; my

x-direction is dependent on
my y-coordinate --plus, I

don't know, e to the xyj.

So then oh, that's difficult
because I have these

e's and these cosines.

But we'll see; if you just
keep your head straight on

what's constant and what's
not, it's not too bad.

So the divergence of v is equal
to the partial derivative

of this expression
with respect to x.

Well, what's the derivative
of this with respect to x?

If y is just a constant,
cosine of y is just a number.

So the derivative of this with
respect to x is just 0 plus--

what's the derivative of
this with respect to y?

Well, you could just do x,
since it's a constant,

as the coefficient on y.

So the derivative of x, y
with respect to y is just x.

And then the derivative of e
to anything is e to anything.

I just did the chain rule.

e to the x, y.

And so that is the divergence.

So you could just ignore this.

It's x, e to the x, y.

One thing to immediately
realize, even before we work on

the intuition, is when we did
gradient I gave you a surface

and it gave us a vector field.

Or I gave you a scalar field
and you got a vector field.

When you take the divergence of
something, you're going in the

opposite direction,
in some ways.

You start with the
vector field, right?

And what's a factor field?

It's something that if you
give me any point x and y,

I'll give you a vector.

So if you wanted to graph it,
in the x, y plane you'd have a

bunch of vectors, and I'll show
you how that looks in a second

when we go over to intuition.

Well, when you take the
divergence of it, you get a

value for any point x, y.

So even though a vector field
has all these vectors on it,

the divergence tells you an
actual scalar number at

any point in the field.

So let's get a little bit
of intuition of what a

divergence actually is.

Let me do it in one dimension.

Or we can even, let's do it
in two dimensions, but I'll

make it constant in the y.

So let's say that my-- let me
erase this; I'll probably

need some space.

OK, oh, I didn't want
to do that dot.

OK let's say the velocity of
fluid, or the particles in

fluid, at any point in the x, y
plane, let's say it is equal to

5xi plus, I don't know, 0y--
there's never any, sorry --0j,

right? j is the unit vector
in the y-direction.

So there's never a y component
to the velocity vector.

So what would that look like?

I don't need a computer
to draw this.

I can handle this
one myself I think.

So if that's the y-axis,
that's my x-axis.

So when x is equal-- I'll just
sample some points and draw

some vectors --when x is equal
to 1-- let's say x is 1 there

--what's the magnitude
of this vector?

It'll be 5, right?

Actually, let me make this a
different number, because

it'll make it hard to do.

Let's make this 1/2 x.

So when x is 1, the magnitude
of my vector is 1/2.

Only in the x-direction.

It has no y component;
ignore this right here.

It's 1/2xi plus 0j.

Or you could just say 1/2xi.

And when x is equal to 2-- I
could have picked any points,

but I'm just picking the
numbers that's easy to

calculate --when x is equal to
2, what is the magnitude

of the vector?

It's 1/2 times 2, which is 1.

So it's going to
be twice as big.

And remember, if I have a
particle right here in my

fluid, if this is a particle,
its velocity in the x-direction

is going to be 1 meter
per second to the right.

If I have a particle here, it's
velocity in the x-direction

is going to be 1/2 a meter
per second to the right.

Let's just do one more point.

So let's say that
x is equal to 3.

What's my velocity
to the right?

I'll do it in a different
color just so that you

don't get confused.

There's going to be 3/2; it's
going to be even longer.

But the general idea here,
and as we move up in x it

doesn't change much, right?

It doesn't change at all.

Our x value doesn't--

[COUGHS].

So for any y, the magnitude
of the vector doesn't

change, right?

It's only dependent on x.

And then for example, here,
it'll be even longer.

If we draw the vector here
it'll be even longer, right?

If you do it here.

I think you get the point.

The further you go to the
right, the faster the particles

are moving towards the right.

So now let's try to get a
little bit of intuition.

Oh, I just realized that I ran
out of time, so I will continue

this in the next video.