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anonymous
 4 years ago
V = u i + v j + w k and ∇= ∂╱∂x i + ∂╱∂y j + ∂╱∂z k that:
(V∙∇)V = (1╱2)∇(V∙V)−V X (∇XV)
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and i dont think it's hard...just long...and i think i'm missing steps

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. Calculate both sides separately and show they are equal. Just work through it patiently.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when do i use the (1/2)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean when? It's attached to the ∇(V∙V) term.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right...do i calculate the ∇(V∙V) first then take 1/2...or distribute the 1/2 first

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0whatever you like. I would calculate ∇(V∙V) first. You'll see every term has a 2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right..i'll get partial squared..which should cancel with with the 1/2 correct to just make it the original again?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Not 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. ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright you gonna be on for a while?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'll post if i get stuck here

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hint: there will be a product rule somewhere in there!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i.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 \[u=u(x,y,z), v=v(x,y,z), w=w(x,y,z)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and by the way, if you have been torturing yourself over this, there's a very good reason.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will have to put you out of your misery...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's not \[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V)  (\nabla \times V)\times V.\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is this: \[(V\cdot \nabla )V = \frac{1}{2}\nabla(V\cdot V) + (\nabla \times V)\times V\] !!!!!!!!!!!!!!!!!!!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm still working on breaking down the left side of the equation as much as i can

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0yes, but the last equation you've written down is exactly equal to the question is written at the top as \[ (\nabla \times V) \times V =  V \times (\nabla \times V) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh yeah! lol thanks. Yeah that's a cross product rule isn't it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...so just for the section of 1/2∇(V dot V)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what happens with the 1/2 now?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm trying to figure out if we have the same thing here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same as what i wrote or not

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0yes, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use any of the above notations?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand what you're asking.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...so the 1/2 cancels with the 2's leaving me with (uux + uuy + uuz + vvx + vvy + vvz + wwx + wwy + wwz)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0\[ V \cdot V = u^2 + v^2 + w^2 \] hence \[ \nabla(V \cdot V) = \nabla(u^2 + v^2 + w^2) \] \[ = \left( \hat{i}\frac{\partial \ }{\partial x} + \hat{j}\frac{\partial \ }{\partial y} + \hat{k}\frac{\partial \ }{\partial z} \right) (u^2 + v^2 + w^2) \] \[ = \hat{i} 2(uu_x + vv_x + ww_x) + \hat{j} ... + \hat{k} ... \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I should have included the components.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Therefore \[ \frac{1}{2} \nabla(V \cdot V) = (uu_x + vv_x + ww_x) \hat{i} + (uu_y + vv_y + ww_y) \hat{j} + ... \hat{k} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...uux can be rewritten as just du/dx correct where d is the partial

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0yes, it's notation that's faster the write and in many ways easier to read.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm just trying to make sure i understand the notation your using

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0\[ u_x = \frac{\partial u}{\partial x} \] hence \[ uu_x = u\frac{\partial u}{\partial x} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so that finishes up that part of the equation...now i gotta work on breaking up the V X( ∇ X V) portion

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just a quick question....is that the entire equation  v cross del cross v

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \nabla \times V = (w_y  v_z) \hat{i} + ... \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i cross product that answer with V or negative V?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0exactly as you have written it at the top of the question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0minus sign is killing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i've shortcircuited

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(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 \[\nabla \times V = (w_yv_z)i + (u_zw_x)j + (v_xu_y)k\] Next we calculate \[V\times [(w_yv_z)i + (u_zw_x)j + (v_xu_y)k]\] \[=[v(v_xu_y)w(u_zw_x)]i + [w(w_yv_z)u(v_xu_y)]j\] \[ + [u(u_zw_x) v(w_yv_z)]k\] and this equals \[V\times (\nabla \times V).\] Now we do the subtraction. In the i component we get: \[[uu_x + vv_x + ww_x  v(v_xu_y)+w(u_zw_x)]i\] which is equal to \[[uu_x +vu_y + wu_z]i\] Looking good? Let's do the j and k component now: \[[uu_y + vv_y + ww_y  w(w_yv_z)+u(v_xu_y)]j\] which is equal to \[[uv_x +vv_y +wv_z]j\] and at long last, the k component... \[[uu_z + vv_z + ww_z  u(u_zw_x) + v(w_yv_z)]k\] 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I shall now shoot myself because entering all that has been a very annoying experience...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so we all seem to be getting to the point of vxdelxv and going mad lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The challenge wasn't to dream up the math, but to enter the damn stuff! :D
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