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Getting g such that del dot g = f (given)

  1. Apr 10, 2009 #1
    Hi, it's not quite a homework question, altought this question came up to mind when i was trying to solve a homework problem. sorry if this shouldn't be here...
    The thing is this, what are the conditions i should impose to f: R^n -> R in order to be able to find a g, such that [divergence of g = f] ?
    I'm not sure how to handle this, since it seems to involve some sort of differential equation (does it?) or at least the existence of an integral for f.. well, i don't really need to get the g, i just need to know if it exists to solve this problem the way i want to, which is to transform a volume integral into a surface integral (why would somebody want a divergence if not? :P)
    thanks, and mods: please move the post if this belongs to the homework category..
  2. jcsd
  3. Apr 10, 2009 #2
    Unfortunately, there is no unique g satisfying that equation. Ie., suppose f(x, y, z) = x + y + z.

    Then g1(x, y, z) = (z + C1, z + C2, z(x + y) + z2/2 + C3) and g2(x, y, z) = (x2/2 + D1, y2/2 + D2, z2/2 + D3) both satisfy the equation, but are not equivalent.
    Last edited: Apr 10, 2009
  4. Apr 10, 2009 #3
    well, as i said i only need to prove the existence of such a function (at least one)
  5. Apr 10, 2009 #4
    Wouldn't this work?

    Let all components of g equal zero, except for the first component, where the first component is any antiderivative of f with respect to the variable corresponding to the first component, taking the other variables as constants?


    f(x,y,z) = x + 2y + zexp(z)

    g = ( (1/2)x^2 + x(2y + zexp(z)) + C, 0, 0)

    The the divergence of g is:

    x + 2y + zexp(z) = f(x,y,z).

    You may not always get an elementary function this way, but it is certainly a *function*.
  6. Apr 10, 2009 #5
    i think that may work.. in which case f would have to have an integral with respect to at least one of the variables? .. would a differentiable f be enough?
    it kinda confuses me, since the functions don't always have an antiderivative.. i'm gonna think about it, though that answer feels a bit odd, nevertheless it works for a lot of functions (maybe all of them, as you said it would still be a *function*).
    by the way, if it helps, f = J dot E where J is the current and E an electric field. Both continuous and differentiable functions R^n -> R^n
    Last edited: Apr 10, 2009
  7. Apr 10, 2009 #6
    oh wait.. i actually need that function to make a pass from the volume integral to the surface one using the divergence theorem... so g would need to be bounded, and C^1.. mmm... dammit, i'm getting confused again.. v.v i'm going for a walk
  8. Apr 10, 2009 #7
    I would say that most functions have antiderivatives. Conspicuous counterexamples come to mind, but they're a little contrived.

    Whether or not the antiderivative can be expressed in terms of elementary functions or not... now that's another story. For instance,

    f(x) = exp(x^2)

    Has infinitely many antiderivatives: F(x) + C, for any C, where F(x) = S f(x) dx

    It just so happens that you can't write F(x) using what are called "elementary functions". That doesn't mean that F(x) doesn't define a perfectly good function.
  9. Apr 10, 2009 #8
    and i guess that if the function is made out of antiderivatives, they're continuosly differentiable themselves.. if that's correct then it's solved :D
  10. Apr 10, 2009 #9
    If n = 3, you could borrow a result from electrostatics:

    Suppose you want to find [itex]\vec E[/itex] such that [itex]\nabla \cdot \vec E = \rho[/itex]. Let
    [tex]V(\vec r) = \frac{1}{4\pi} \int_{\mathbb{R}^3} \frac{\rho(\vec r')}{\lvert \vec r - \vec r' \rvert} \,d^3 \vec r',[/tex]
    and take [itex]\vec E = -\nabla V[/itex]. This will also guarantee that [itex]\nabla \times \vec E = 0[/itex].

    I don't know if a similar approach works in other dimensions.
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