# Vector calculus

1. Aug 20, 2008

### Somefantastik

u,w are scalar functions of 3 variables

the book says you can verify by direct computation:

$$u\nabla^{2}w = \nabla \bullet\left(u\nabla w\right) - \left(\nabla u\right)\bullet\left(\nabla w \right)$$

and

$$\nabla^{2}w$$ is $$\Delta w$$, the Laplacian operator.

Every time I've worked this, I just can't seem to come close. Can someone get me started on this? It's out of a PDE book so of course they don't bother expounding on this and I haven't looked at vector calculus in years...

My first thoughts were:

$$u \nabla ^{2}w$$
= $$u \nabla \nabla w$$
= $$u \ div \ grad \ w$$
=$$u \left(\frac{\delta^{2}w}{\delta x^{2}}$$ $$+ \frac{ \delta^{2}w}{ \delta y^{2}} + \frac{ \delta^{2}w}{ \delta z^{2}}} \right)$$

I can't get that RHS parens in there, sorry :-/

2. Aug 20, 2008

### tiny-tim

Hi Somefantastik!

(you put a /tex and tex in the middle of your brackets, and then the second pair refused to recognise a \right without a preceding \left )

$$\frac{\partial}{\partial x_i}\left(u\,\frac{\partial w}{\partial x_i}\right)\ =\ u\,\frac{\partial^2 w}{\partial x_i^2}\ +\ \frac{\partial u}{\partial x_i}\,\frac{\partial w}{\partial x_i}$$

3. Aug 20, 2008

### Somefantastik

Ah ok thanks so much.

4. Aug 21, 2008

### Somefantastik

I get this:

$$u \nabla ^{2}w$$

$$= u \ div \ grad \ w$$

$$= u \left(\frac{\delta^{2}w}{\delta x^{2}} + \frac{ \delta^{2}w}{ \delta y^{2}} + \frac{ \delta^{2}w}{ \delta z^{2}}} \right)$$
$$= \frac{\delta^{2}uw}{\delta x^{2}}+ \frac{\delta^{2}uw}{\delta y^{2}} + \frac{\delta^{2}uw }{\delta z^{2}}$$

But I also get this:

$$\left( \nabla u \right)\bullet\left(\nabla w \right)$$

$$=\left( \frac{\delta u}{\delta x}, \frac{\delta u}{\delta y},\frac{\delta u}{\delta z} \right) \bullet \left(\frac{\delta w}{\delta x},\frac{\delta w}{\delta y},\frac{\delta w}{\delta z} \right)$$

$$= \frac{\delta^{2}uw}{\delta x^{2}}+ \frac{\delta^{2}uw}{\delta y^{2}} + \frac{\delta^{2}uw }{\delta z^{2}}$$

They can't be equal; can someone show me where I messed up my algebra?

5. Aug 21, 2008

### tiny-tim

Hi Somefantastik

(use \partial for ∂, not \delta, and \cdot, not \bullet )
Nooo

You can't just move u inside a ∂ like that (unless u is constant)!
Again, you can't say (∂u/∂x)(∂w/∂x) = ∂2uw/∂x2.

6. Aug 21, 2008

### NoMoreExams

Basically if "d" is your differential operator there's a difference between, for example,

$$\frac{d}{dx} \left(\frac{dy}{dx} \right) = \frac{d^{2}y}{dx^{2}}$$

and

$$\left(\frac{dy}{dx} \right) \cdot \left(\frac{dy}{dx} \right) = \left(\frac{dy}{dx} \right)^{2}$$

For example let

$$y = x^{2}$$

The first method will yield 2 whereas the 2nd one will yield $$(2x)^{2} = 4x^{2}$$

7. Aug 21, 2008

### Somefantastik

Ok, thank you. Can you give me a small nudge in the right direction? What can I do?

8. Aug 21, 2008

### Somefantastik

This is clear to me now after looking at it

$$\left(\frac{dy}{dx} \right) \cdot \left(\frac{dy}{dx} \right) = \left(\frac{dy}{dx} \right)^{2}$$

And I can get somewhere with that I think.

Can someone nudge me on the other algebraic error?

"You can't just move u inside a ∂ like that (unless u is constant)!"

I can tell that it's the same error, but I just cannot see how to manipulate it.

Thanks for all input so far, I really appreciate it.

9. Aug 21, 2008

### NoMoreExams

I used the \cdot to denote multiplication, if you are doing a dot product wouldn't you just use the definition i.e. you have a sumproduct so

$$(a, b, c) \cdot (d, e, f) = ad + be + cf$$

10. Aug 21, 2008

### Somefantastik

I understand how the dot product works, but what I don't understand is what happens here. I know u = u(x,yz), but I don't know how to compute

$$= u \left(\frac{\delta^{2}w}{\delta x^{2}} + \frac{ \delta^{2}w}{ \delta y^{2}} + \frac{ \delta^{2}w}{ \delta z^{2}}} \right)$$

11. Aug 21, 2008

### NoMoreExams

Leave that side as it is, so on the LHS you have

$$u \left(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}} \right)$$

Now on RHS you have

$$\nabla \cdot \left(u \nabla w \right) - \left(\nabla u \right) \cdot \left(\nabla w \right)$$

So let's example this part by part, starting with:

$$\nabla \cdot \left(u \nabla w \right) = \nabla \cdot \left(u \frac{\partial w}{\partial x} + u \frac{\partial w}{\partial y} + u \frac{\partial w}{\partial z} \right) = u \frac{\partial^{2} w}{\partial x^{2}} + \frac{\partial u}{\partial x} \frac{\partial w}{\partial x} +u \frac{\partial^{2} w}{\partial y^{2}} + \frac{\partial u}{\partial y} \frac{\partial w}{\partial y} + u \frac{\partial^{2} w}{\partial z^{2}} + \frac{\partial u}{\partial z} \frac{\partial w}{\partial z}$$

Next let's look at

$$\left(\nabla u \right) \cdot \left(\nabla w \right) = \left(\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z}\right) \cdot \left(\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}\right) = \frac{\partial u}{\partial x} \frac{\partial w}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial w}{\partial y} + \frac{\partial u}{\partial z} \frac{\partial w}{\partial z}$$

Now subtract the two expressions, stuff will cancel and you can factor out u(x,y,z)

Does that help?

12. Aug 21, 2008

### NoMoreExams

I think your biggest confusion here, as it often is, is confusing when to multiply and when to apply the differential operators (d or \nabla or whatever else). Hopefully the above helps.

13. Aug 21, 2008

### Somefantastik

I'm a lil thrown off b/c Tiny Tim said that you can't move u into a ∂ like that. I see that your way works out, though, so you must be right. I'm so confused! Can you point me to a website that describes this sort of algebra? All my books seem to gloss right over the differential operators and how to manipulate them.

14. Aug 21, 2008

### NoMoreExams

I think tiny-tim was saying, and please correct me if I'm wrong, was that if you have (where \cdot is multiplication here)

$$u \cdot \left( \frac{\partial^2 w}{\partial x^2} + ... \right)$$

Then that becomes

$$u \frac{\partial^2 w}{\partial x^2} + ...$$

So you just multiply your function u by your partial derivative, that's different from what you wrote:

$$\frac{\partial^2 uw}{\partial x^2}$$

Similarly (and let's let \cdot denote dot product now) when you had:

$$\left(\frac{\partial u}{\partial x}, ... \right) \cdot \left(\frac{\partial w}{\partial x}, ... \right)$$

Then you just apply the dot product rule and multiply the appropriate components out and then sum them to get

$$\frac{\partial u}{\partial x} \frac{\partial w}{\partial x} + ...$$

Which is different from what you had:

$$\frac{\partial uw}{\partial x^2} + ...$$

Does that make sense?

It might be that this notation isn't too clear for you so let's work through an example

Let's examine the difference between

$$\frac{\partial^2 f}{\partial u w}$$ and $$\frac{\partial f}{\partial u} \frac{\partial f}{\partial w}$$

Define $$f(u, w) = u^{2}w^{2}$$

The first expression means you differentiate with respect to u and then differentiate with respect to w (or vice versa mixed partials commute under some nice conditions which our function satisfies) so let's see what we get

$$\frac{\partial f}{\partial u} = 2uw^{2}$$

and now applying the derivative wrt to get (note that here we are applying the differential operator to our derivative:

$$\frac{\partial}{\partial w} \left(\frac{\partial f}{\partial u} \right) = \frac{\partial^2 f}{\partial u w} = 4uw$$

As opposed to taking our f, differentiating wrt u, then taking our f and differentiating wrt w and multiplying the result together, i.e.

$$\frac{\partial f}{\partial u} = 2uw^{2}, \, \frac{\partial f}{\partial w} = 2u^{2}w$$

and therefore

$$\frac{\partial f}{\partial u} \frac{\partial f}{\partial w} = 4u^{3}w^{3}$$

Does that make more sense?

Last edited: Aug 21, 2008
15. Aug 22, 2008

### tiny-tim

product rule

Hi Somefantastik!

(btw, it's only tiny-tim, not Tiny Tim … i'm only a tiny little goldfish! )
ah, but NoMoreExams didn't just move a u inside a ∂ …

he split it into two parts …

he did move a u inside a ∂ in the first part,

but for the second part he multiplied by ∂u/∂x outside the ∂ …

this is the product rule: ∂(uA)/∂x = u∂(A)/∂x + A∂(u)/∂x (and in this case A itself is a ∂).

16. Aug 22, 2008

### Somefantastik

Ok, yeah that makes a lot of sense. I was really frustrated for awhile. Sorry you all had to tell me the same thing over and over again. I really appreciate it :)