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 :-/
tiny-tim
Aug20-08, 06:41 PM
u,w are scalar functions of 3 variables
…
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 :-/
Hi Somefantastik! :smile:
(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 :wink:)
You can't just move u inside a ∂ like that (unless u is constant)!
=\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)
Now subtract the two expressions, stuff will cancel and you can factor out u(x,y,z)
Does that help?
NoMoreExams
Aug21-08, 08:20 PM
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.
Somefantastik
Aug21-08, 08:51 PM
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.
Thanks a bunch for your help, I feel better already.
NoMoreExams
Aug21-08, 11:30 PM
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.
Thanks a bunch for your help, I feel better already.
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:
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:
(btw, it's only tiny-tim, not Tiny Tim … i'm only a tiny little goldfish! :biggrin:)
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}\ +\ \cdots
I'm a lil thrown off b/c Tiny Tim said that you can't move u into a ∂ like that.
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 ∂). :smile:
Somefantastik
Aug22-08, 08:42 AM
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 :)