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Homework Help: Vector identity proof

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [itex] f(x,y,z), g(x,y,z), h(x,y,z) [/itex] be any [itex] C^2 [/itex] scalar functions. Using the standard identities of vector analysis (provided in section 2 below), prove that

    [itex] \nabla \cdot ( f \nabla g \times \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h) [/itex]


    2. Relevant equations

    Note: The identities below require [itex] f,g,F,G [/itex] to be suitable differentiable, either order [itex] C^1 [/itex] or [itex] C^2 [/itex].

    [itex] 1. \nabla (f+g) = \nabla f + \nabla g [/itex]
    [itex] 2. \nabla (\lambda f) = \lambda \nabla f [/itex], where [itex] \lambda [/itex] is a constant
    [itex] 3. \nabla (fg) = f \nabla g + g \nabla f [/itex]
    [itex] 4. \nabla (\frac{f}{g}) = \frac{g \nabla f - f \nabla g}{g^2} [/itex]
    [itex] 5. \nabla \cdot (F+G) = \nabla \cdot F + \nabla \cdot G [/itex]
    [itex] 6. \nabla \times (F+G) = \nabla \times F + \nabla \times G [/itex]
    [itex] 7. \nabla \cdot (fF) = f \nabla \cdot F + F \cdot \nabla f [/itex]
    [itex] 8. \nabla \cdot (F \times G) = G \cdot (\nabla \times F ) - F \cdot (\nabla \times G) [/itex]
    [itex] 9. \nabla \cdot (\nabla \times F) = 0 [/itex]
    [itex] 10. \nabla \times (fF) = f \nabla \times F + \nabla f \times F [/itex]
    [itex] 11. \nabla \times (\nabla f) = 0 [/itex]
    [itex] 12. {\nabla}^2 (fg) = f{\nabla}^2 g + g{\nabla}^2 f + 2 \nabla f \cdot \nabla g [/itex]
    [itex] 13. \nabla \cdot (\nabla f \times \nabla g) = 0 [/itex]
    [itex] 14. \nabla (f \nabla g - g \nabla f) = f {\nabla}^2 g - g {\nabla}^2 f [/itex]


    3. The attempt at a solution

    Using identity 8,

    [itex] \nabla \cdot ( f \nabla g \times \nabla h ) = \nabla h \cdot (\nabla \times (f \nabla g)) - (f \nabla g) \cdot (\nabla \times (\nabla h)) [/itex]

    One of the terms on RHS, [itex] \nabla \times (\nabla h) = 0 [/itex] by identity 11.

    So the equation reduces to

    [itex] \nabla \cdot ( f \nabla g \times \nabla h ) = \nabla h \cdot (\nabla \times (f \nabla g)) [/itex]

    I'm stuck here. There is no identity that I can use to further simplify this to the one required. (from what I can see, or am I wrong?) How do we proceed?

    Thanks!
     
  2. jcsd
  3. Apr 16, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi madachi! Welcome to PF! :wink:

    Looking at the RHS of the answer, you need to separate the f from the g and h, and 8. won't do that.

    But 7. will, so try 7. :smile:
     
  4. Apr 16, 2010 #3

    Mark44

    Staff: Mentor

    Might be a naive question, but is there an f missing on the left side in this equation?
    [tex] \nabla \cdot ( f \nabla g \times \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h) [/tex]

    If so, here's the corrected version.
    [tex] \nabla \cdot ( f \nabla g \times \bold{f} \nabla h ) = \nabla f \cdot ( \nabla g \times \nabla h) [/tex]
     
  5. Apr 16, 2010 #4
    How do we use 7 ? Because the original equation doesn't look like "the form" of identity 7. Could you show me the first step?

    Thanks!

    No there isn't an f missing on the left side in the equation.

    Thanks.
     
  6. Apr 16, 2010 #5

    tiny-tim

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    Hi madachi! :smile:

    (have a nabla: ∇ :wink:)

    f is f, and F is ∇g x ∇h :smile:
     
  7. Apr 16, 2010 #6
    Thanks! Using what you suggest,

    So

    [itex] \nabla \cdot ( f \nabla g \times \nabla h) [/itex]
    = [itex] f \nabla \cdot (\nabla g \times \nabla h ) + ( \nabla g \times \nabla h ) \cdot \nabla f [/itex]

    Then we use identity 8 to show that the vector field on the first term equal to 0 right? And the second term on RHS is just the identity that we are required to show?

    Thanks!
     
  8. Apr 17, 2010 #7

    tiny-tim

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    Hi madachi! :smile:

    (just got up :zzz: …)
    Not sure how you get it from 8 :redface:

    how about 13? :smile:
     
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