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Grad f is proportional to grad g

  1. Sep 23, 2011 #1
    If the gradients of the two functions f and g are proportional everywhere in Rn, does that mean there is some differentiable function F of two variables such that F(f(x),g(x)) = 0 everywhere?

    The converse is obviously true by the chain rule, so I was just wondering if this was true, too.
  2. jcsd
  3. Sep 23, 2011 #2


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    Yes, take F(x,y) identically zero. But that's not what you wanted, right? :smile:
  4. Sep 23, 2011 #3

    Heh, I didn't think about that. Yeah, I want there to be some more interesting functional dependence between f and g. But I think stating it as g(x) = F(f(x)) is too strong, because the relation between them might be one-to-many. I was reading about barotropic fluid flows, in which isobars are also surfaces of constant density, and I was just wondering if the conditions

    [tex]\nabla p \times \nabla \rho = 0[/tex]


    [tex]p = p(\rho)[/tex]

    are equivalent, or if the first condition is more general.
  5. Oct 2, 2011 #4
    The answer to your question is basically yes. One way to see this is to note that if the functions' gradients are everywhere proportional, then the functions' level surfaces are identical. So (at least in local patches) each must be a function of the other.
  6. Oct 4, 2011 #5
    Okay, thank you. Is there a global version of this, or does it generally fail to hold globally? I think I see that in regions where the gradients do not vanish, you can derive an explicit formula for the value of f as a function of the value of g, or vice-versa.
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