1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    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


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted