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

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In summary, this question asks what conditions must be met in order to find a function that satisfies a given equation. However, the equation in question is not easily solved, and it may require borrowing results from other disciplines in order to solve it.f
  • #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
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.
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  • #3
well, as i said i only need to prove the existence of such a function (at least one)
  • #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*.
  • #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 going to 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
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  • #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
  • #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.
  • #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
  • #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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