A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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)

  • This Question is Closed
  1. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You need to prove this?

  2. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes sir

  3. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when do i use the (1/2)?

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  7. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. ;-)

  12. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright you gonna be on for a while?

  13. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'll post if i get stuck here

  14. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  15. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[u=u(x,y,z), v=v(x,y,z), w=w(x,y,z)\]

  17. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  18. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  20. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  22. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  23. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  24. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

  26. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what happens with the 1/2 now?

  27. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  28. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    agreed?

  29. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    same as what i wrote or not

  31. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  32. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use any of the above notations?

  33. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't understand what you're asking.

  34. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  35. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  36. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I should have included the components.

  37. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  38. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes?

  39. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  40. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that correct?

  41. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  42. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  43. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  44. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright

  45. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  46. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  47. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nevermind

  48. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  49. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  50. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  51. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    minus sign is killing me

  52. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think i've shortcircuited

  53. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'm getting nowhere

  54. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \[\nabla \times V = (w_y-v_z)i + (u_z-w_x)j + (v_x-u_y)k\] 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.

  55. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  56. amistre64
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  57. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  58. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.