V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

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V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that: (V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)

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You need to prove this?
yes sir
and i dont think it's hard...just long...and i think i'm missing steps

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Other answers:

Yes. Calculate both sides separately and show they are equal. Just work through it patiently.
when do i use the (1/2)?
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.
what do you mean when? It's attached to the ∇(V∙V) term.
right...do i calculate the ∇(V∙V) first then take 1/2...or distribute the 1/2 first
whatever you like. I would calculate ∇(V∙V) first. You'll see every term has a 2.
right..i'll get partial squared..which should cancel with with the 1/2 correct to just make it the original again?
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. ;-)
alright you gonna be on for a while?
i'll post if i get stuck here
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.
hint: there will be a product rule somewhere in there!
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 \[u=u(x,y,z), v=v(x,y,z), w=w(x,y,z)\]
and by the way, if you have been torturing yourself over this, there's a very good reason.
I will have to put you out of your misery...
It's not \[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V) - (\nabla \times V)\times V.\]
It is this: \[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V) + (\nabla \times V)\times V\] !!!!!!!!!!!!!!!!!!!!!!!!!
i'm still working on breaking down the left side of the equation as much as i can
yes, but the last equation you've written down is exactly equal to the question is written at the top as \[ (\nabla \times V) \times V = - V \times (\nabla \times V) \]
ohh yeah! lol thanks. Yeah that's a cross product rule isn't it.
okay...so just for the section of 1/2∇(V dot V)
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)
what happens with the 1/2 now?
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.
agreed?
i'm trying to figure out if we have the same thing here
same as what i wrote or not
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.
use any of the above notations?
I don't understand what you're asking.
okay...so the 1/2 cancels with the 2's leaving me with (uux + uuy + uuz + vvx + vvy + vvz + wwx + wwy + wwz)
\[ V \cdot V = u^2 + v^2 + w^2 \] hence \[ \nabla(V \cdot V) = \nabla(u^2 + v^2 + w^2) \] \[ = \left( \hat{i}\frac{\partial \ }{\partial x} + \hat{j}\frac{\partial \ }{\partial y} + \hat{k}\frac{\partial \ }{\partial z} \right) (u^2 + v^2 + w^2) \] \[ = \hat{i} 2(uu_x + vv_x + ww_x) + \hat{j} ... + \hat{k} ... \]
I should have included the components.
Therefore \[ \frac{1}{2} \nabla(V \cdot V) = (uu_x + vv_x + ww_x) \hat{i} + (uu_y + vv_y + ww_y) \hat{j} + ... \hat{k} \]
yes?
okay...uux can be rewritten as just du/dx correct where d is the partial
is that correct?
yes, it's notation that's faster the write and in many ways easier to read.
i'm just trying to make sure i understand the notation your using
\[ u_x = \frac{\partial u}{\partial x} \] hence \[ uu_x = u\frac{\partial u}{\partial x} \]
alright
so that finishes up that part of the equation...now i gotta work on breaking up the -V X( ∇ X V) portion
just a quick question....is that the entire equation - v cross del cross v
nevermind
\[ \nabla \times V = (w_y - v_z) \hat{i} + ... \]
then i cross product that answer with V or negative V?
exactly as you have written it at the top of the question.
minus sign is killing me
i think i've shortcircuited
i'm getting nowhere
(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.
I shall now shoot myself because entering all that has been a very annoying experience...
so we all seem to be getting to the point of vxdelxv and going mad lol
The challenge wasn't to dream up the math, but to enter the damn stuff! :D

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