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Vector Analysis Problem Involving Divergence

  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data

    Let [itex]f[/itex] and [itex]g[/itex] be scalar functions of position. Show that:
    [tex]\nabla f \cdot \nabla(\nabla ^2 g)-\nabla g \cdot \nabla(\nabla ^2f)[/tex]
    Can be written as the divergence of some vector function given in terms of [itex]f[/itex] and [itex]g[/itex].

    2. Relevant equations

    All the identities given at https://en.wikipedia.org/wiki/Vector_calculus_identities, I suppose. Especially relevant would be the second derivative and divergence identites. Also, [itex]\nabla ^2 =\nabla \cdot \nabla[/itex]

    3. The attempt at a solution

    After considerable time messing around with various vector identites, I've been able to show the above is equivalent to:

    [tex]\nabla \cdot (f \nabla (\nabla ^2 g)-g\nabla (\nabla ^2 f))+g(\nabla \cdot \nabla(\nabla ^2 f))-f(\nabla \cdot \nabla(\nabla ^2 g))[/tex]

    This is painfully close to the result I want, but I can't seem to show that the second and third terms either cancel or are themselves a divergence. I'd really like any hints, and can provide more detail as to the specific identities and manipulations I've used thus far if needed, thanks.
     
  2. jcsd
  3. Mar 29, 2017 #2

    jedishrfu

    Staff: Mentor

    Have you tried working it from the other end starting with the answer or is that an unknown?
     
  4. Mar 29, 2017 #3

    pasmith

    User Avatar
    Homework Helper

    Those terms simplify to [itex]g \nabla^4 f - f \nabla^4g[/itex], which doesn't cancel.

    Rather than building a vector field as a linear combination of [itex]\nabla(\nabla^2 f)[/itex] and [itex]\nabla(\nabla^2 g)[/itex], I would have started by building one as a linear combination of [itex]\nabla f[/itex] and [itex]\nabla g[/itex].
     
  5. Mar 30, 2017 #4
    Unfortuntely the end result is unknown, otherwise that'd be a great suggestion, thanks!

    Yeah, I managed to get to that simplification, which like you said defnitely doesn't cancel, so I guess I must be able to somehow write the quantity as a divergence. I'm sorry, but I don't really understand what you mean by building a vector field?

    Thanks for the help so far, I appreciate it.
     
  6. Mar 30, 2017 #5
    I would write,$$\nabla f \cdot\nabla\left ( \nabla^2 g\right )-\nabla g\cdot \nabla \left ( \nabla^2 f \right ) =\nabla \cdot \vec V $$
    Then do the work of expanding the l.h.s. in it's spacial components and compare with the r.h.s.
     
  7. Mar 30, 2017 #6

    pasmith

    User Avatar
    Homework Helper

    Consider the divergence of [itex]D(g, \nabla^2 g)\nabla f - D(f, \nabla^2 f) \nabla g[/itex] for some function [itex]D[/itex].
     
  8. Mar 30, 2017 #7
    That is a great idea, thanks! I managed to solve it using this idea.
     
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