Question

V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that:

(V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

Answer 1

You need to prove this?

Answer 2

yes sir

Answer 3

and i dont think it's hard...just long...and i think i'm missing steps

Answer 4

Yes. Calculate both sides separately and show they are equal. Just work through it patiently.

Answer 5

when do i use the (1/2)?

Answer 6

I remember I did lots of these in fluid dynamics, we used to take the messy side (the right hand side) and play with it until the simple side comes out.

Answer 7

what do you mean when? It's attached to the ∇(V∙V) term.

Answer 8

right...do i calculate the ∇(V∙V) first then take 1/2...or distribute the 1/2 first

Answer 9

whatever you like. I would calculate ∇(V∙V) first. You'll see every term has a 2.

Answer 10

right..i'll get partial squared..which should cancel with with the 1/2 correct to just make it the original again?

Answer 11

Not quite. You'll have partial derivatives there as well.

Just muster up your courage and do it. We can talk you through only so much. You've got to bungee jump yourself. ;-)

Answer 12

alright you gonna be on for a while?

Answer 13

i'll post if i get stuck here

Answer 14

If you do, do it again. When I first learnt these sorts of identities and proofs, I had to do them all at least twice because it's easy to make an error.

Answer 15

hint: there will be a product rule somewhere in there!

Answer 16

i.e.
\[\frac{1}{2}\nabla(v\cdot v) = \frac{1}{2}\left[\frac{\partial}{\partial x}\left(u^2 + v^2 + w^2\right)i + \frac{\partial}{\partial y}\left(u^2 + v^2 + w^2\right)j + \frac{\partial}{\partial z}\left(u^2 + v^2 + w^2\right)k\right] \]
on which using the product rule will get rid of the 1/2 and also give you products of functions and their partial derivatives, which will help when you later...also you should specify that
\[u=u(x,y,z), v=v(x,y,z), w=w(x,y,z)\]

Answer 17

and by the way, if you have been torturing yourself over this, there's a very good reason.

Answer 18

I will have to put you out of your misery...

Answer 19

It's not
\[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V) - (\nabla \times V)\times V.\]

Answer 20

It is this:
\[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V) + (\nabla \times V)\times V\]
!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 21

i'm still working on breaking down the left side of the equation as much as i can

Answer 22

yes, but the last equation you've written down is exactly equal to the question is written at the top as
\[ (\nabla \times V) \times V = - V \times (\nabla \times V) \]

Answer 23

ohh yeah! lol thanks. Yeah that's a cross product rule isn't it.

Answer 24

okay...so just for the section of 1/2∇(V dot V)

Answer 25

i get 1/2( du^2/dx + du^2/dy+du^2/dz + dv^2/dx+dv^2/dy+dv^2/dz+dw^2/dx+dw^2/dy+dw^2/dz)

Answer 26

what happens with the 1/2 now?

Answer 27

You get 9 terms, yes. Easier to write this with subscripts. x,y,z for the partials. Using the chain rule you should find that this term in total is
u.ux + u.uy + u.uz
+ v.vx + v.vy + v.vz
+ w.wx + w.wy + w.wz

where . here just means multiplication and I've added them just to make it a little more readable.

Answer 28

agreed?

Answer 29

i'm trying to figure out if we have the same thing here

Answer 30

same as what i wrote or not

Answer 31

yes, it is, because for instance,
\[ \frac{\partial \ }{\partial x}u^2 = 2u \frac{\partial u}{\partial x} = 2uu_x \]

Now repeat that another 8 times. Then multiply everything by 1/2.

Answer 32

use any of the above notations?

Answer 33

I don't understand what you're asking.

Answer 34

okay...so the 1/2 cancels with the 2's leaving me with (uux + uuy + uuz + vvx + vvy + vvz + wwx + wwy + wwz)

Answer 35

\[ V \cdot V = u^2 + v^2 + w^2 \]
hence
\[ \nabla(V \cdot V) = \nabla(u^2 + v^2 + w^2) \]
\[ = \left( \hat{i}\frac{\partial \ }{\partial x} + \hat{j}\frac{\partial \ }{\partial y} + \hat{k}\frac{\partial \ }{\partial z} \right) (u^2 + v^2 + w^2) \]
\[ = \hat{i} 2(uu_x + vv_x + ww_x) + \hat{j} ... + \hat{k} ... \]

Answer 36

I should have included the components.

Answer 37

Therefore
\[ \frac{1}{2} \nabla(V \cdot V) = (uu_x + vv_x + ww_x) \hat{i} + (uu_y + vv_y + ww_y) \hat{j} + ... \hat{k} \]

Answer 38

yes?

Answer 39

okay...uux can be rewritten as just du/dx correct where d is the partial

Answer 40

is that correct?

Answer 41

yes, it's notation that's faster the write and in many ways easier to read.

Answer 42

i'm just trying to make sure i understand the notation your using

Answer 43

\[ u_x = \frac{\partial u}{\partial x} \]
hence
\[ uu_x = u\frac{\partial u}{\partial x} \]

Answer 44

alright

Answer 45

so that finishes up that part of the equation...now i gotta work on breaking up the -V X( ∇ X V) portion

Answer 46

just a quick question....is that the entire equation - v cross del cross v

Answer 47

nevermind

Answer 48

\[ \nabla \times V = (w_y - v_z) \hat{i} + ... \]

Answer 49

then i cross product that answer with V or negative V?

Answer 50

exactly as you have written it at the top of the question.

Answer 51

minus sign is killing me

Answer 52

i think i've shortcircuited

Answer 53

i'm getting nowhere

Answer 54

(V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV) let's do it term by term....... Letting
\[V=(u(x,y,z), v(x,y,z), w(x,y,z))\] and start with the term
\[\frac{1}{2}\nabla(V\cdot V).\]
\[V\cdot V = u^2 + v^2 + w^2\]
and this is a scalar quantity. Then \[\nabla (u^2 + v^2 + w^2)\] is a vector quantity, and as above, is equal to:
\[\left[\frac{\partial}{\partial x}\left(u^2 + v^2 + w^2\right)i + \frac{\partial}{\partial y}\left(u^2 + v^2 + w^2\right)j + \frac{\partial}{\partial z}\left(u^2 + v^2 + w^2\right)k\right]\]
in calculating each component of this vector, we will use subscripts to indicate partial differentiation wrt the subscripted variable. We have
\[\left[(2uu_x + 2vv_x + 2ww_x)i + (2uu_y + 2vv_y + 2ww_y)j + (2uu_z + 2vv_z + 2ww_z)k\right]\]
So multiplying by the 1/2 will remove all these 2s meaning:
\[\frac{1}{2}\nabla(V\cdot V) = \left[(uu_x + vv_x + ww_x)i + (uu_y + vv_y + ww_y)j + (uu_z + vv_z + ww_z)k\right]\]

Next we want to deal with the term:
\[V\times (\nabla \times V)\]
First we calculate
\[\nabla \times V = (w_y-v_z)i + (u_z-w_x)j + (v_x-u_y)k\]
Next we calculate
\[V\times [(w_y-v_z)i + (u_z-w_x)j + (v_x-u_y)k]\]
\[=[v(v_x-u_y)-w(u_z-w_x)]i + [w(w_y-v_z)-u(v_x-u_y)]j\]
\[ + [u(u_z-w_x) -v(w_y-v_z)]k\]
and this equals
\[V\times (\nabla \times V).\]
Now we do the subtraction. In the i component we get:
\[[uu_x + vv_x + ww_x - v(v_x-u_y)+w(u_z-w_x)]i\]
which is equal to
\[[uu_x +vu_y + wu_z]i\]
Looking good? Let's do the j and k component now:
\[[uu_y + vv_y + ww_y - w(w_y-v_z)+u(v_x-u_y)]j\]
which is equal to
\[[uv_x +vv_y +wv_z]j\]
and at long last, the k component...
\[[uu_z + vv_z + ww_z - u(u_z-w_x) + v(w_y-v_z)]k\]
which is equal to
\[[uw_x+vw_y + ww_z]k\]
THEREFORE we have shown that the right hand side reduces to the vector:
\[[uu_x + vu_y + wu_z]i + [uv_x + vv_y + wv_z]j + [uw_x+vw_y + ww_z]k\]
Now let's stop here and think for a bit.

\[V\cdot \nabla = u\frac{\partial }{\partial x} + v \frac{\partial }{\partial y} + w\frac{\partial }{\partial z}\]

and this is an operator which acts on vectors to the right of it, if you like, so we get:
\[(V\cdot \nabla )V = \left[u\frac{\partial }{\partial x}u + v \frac{\partial }{\partial y}u + w\frac{\partial }{\partial z}u\right]i + \left[u\frac{\partial }{\partial x}v + v \frac{\partial }{\partial y}v + w\frac{\partial }{\partial z}v\right]j\] \[+ \left[u\frac{\partial }{\partial x}w + v \frac{\partial }{\partial y}w + w\frac{\partial }{\partial z}w\right]k\]

Notice that this is exactly what we just proved, therefore the identity has been verified.

Answer 55

I shall now shoot myself because entering all that has been a very annoying experience...

Answer 56

so we all seem to be getting to the point of vxdelxv and going mad lol

Answer 57

The challenge wasn't to dream up the math, but to enter the damn stuff! :D