Question

If f, g are vectors in R^3, then
\(<\triangle f, g>=?\)
Please, help

Answer 1

I know that \(\triangle f = \dfrac{\partial ^2f}{\partial x^2}+\dfrac{\partial^2f}{\partial y^2}+\dfrac{\partial^2f}{\partial z^2}\)
but not sure how to take dot product with g

Answer 2

@IrishBoy123 Hi, friend!! I got part a) try to apply to part b but not get anything yet. Can you give me a hand?

Answer 3

hey!

the laplacian is a scalar so that expression doesn't mean anything. you can't dot a scalar with a vector

but once again, i am all at sea with the formalism :-))

enoying watching these adventures though!

Answer 4

hihihii.... have fun. thanks anyway

Answer 5

Perhaps what you want is something like this?

\[\vec g \cdot \vec \nabla f\]

Where \(\vec \nabla f\) is the gradient of the scalar field f. Then the dot product of some vector g with the gradient is the derivative of f in the direction of g.

Answer 6

I am using Laplacian here

Answer 7

Let me post my problem, please, take a look

Answer 8

Ok yeah that'd be good.

Answer 9

http://assets.openstudy.com/updates/attachments/56ba982ee4b0548ff6113db7-loser66-1455069241057-doc1.pdf

Answer 10

I got part a)

Answer 11

Need apply it to the second part, which is from "and if F and G are vectors......"

Answer 12

Ah ok I see, I thought you were doing vector calculus.

I am not sure myself yet, but it looks like you can plug in the definition of the Laplacian here and then use the definitions above to simplify it.

The Laplacian is the divergence of the gradient for scalars, but for vectors it's the divergence of the curl I believe. I am not entirely sure because I am familiar with how I would do something similar to this with tensors and that's what it would be.

Answer 13

:)
My plan to extend the LHS and simplify to get the RHS, but not know how to yet.

Answer 14

Actually no, that second definition can't be right what I've said. I think the key is to look up what the definition of the Laplacian is for vectors since it is different than it is for scalars in your notation. In tensor notation there's no difference, but it's hard to translate it over because you end up with a bunch of nasty dot product garbage.

Answer 15

How about this. \(\bigtriangledown (\bigtriangledown F) = \partial_1(\bigtriangledown f), \ \partial_2(\bigtriangledown f)\partial_3(\bigtriangledown f),\)

Answer 16

so that we can take dot product with G, which give us
\[G1\partial_1(\bigtriangledown f)+G2\partial_2(\bigtriangledown f)+G3\partial_3(\bigtriangledown f)\]