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Is there a way to construct a unit vector field $V(\textbf{x})=(V_x,V_y,\sqrt{1-V_x^2-V_y^2})$ in $\mathbb{R}^3$ knowing its divergence $\nabla \cdot V$ and Darboux vector $\omega=\kappa(\textbf{x}) (V(\textbf{x})\times V'(\textbf{x})) + \tau(\textbf{x}) V(\textbf{x})$ at every point, where $\kappa(\textbf{x}),\tau(\textbf{x})$ are also known?

How could I go about proving either?

So far I have tried brute force through substitution a component of the divergence into the other scalars $\kappa(\textbf{x}), \tau(\textbf{x})$, but I end up with a differential equation which is very complicated. Is there a known approach for the construction of vector fields from their "internal" scalars?

Michael Paris
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    The divergence of a vector field is a particular case of the exterior derivative. See, for example, https://en.wikipedia.org/wiki/Divergence#Relation_with_the_exterior_derivative and https://en.wikipedia.org/wiki/Exterior_derivative#Divergence. There’s an algorithm for finding, under some fairly mild conditions, an antiderivative of a $k$-form via integration, so it seems like this could be doable Do you have a concrete example in mind? – amd Jul 13 '17 at 20:50
  • Basically, my concrete example is the given case. I know the pseudoscalar $\ast d\ast V^\flat$ and the result of the inner product $<.,\ast d V^\flat>$. Which algorithm do you have in mind? Do you think I should restate the question I terms of exterior algebra? – Michael Paris Jul 14 '17 at 00:46
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    See https://math.stackexchange.com/a/1683675/265466 for an algorithm that finds an antiderivative of a closed differential form defined on a star-shaped region. – amd Jul 14 '17 at 03:18
  • I checked out the link and I believe I understood the method, thank you for that. Unfortunately, I think the method can only be applied for explicit forms and does not give me a result for something general such as, $\omega = s(x,y,z) dx\wedge dy \wedge dz$, because I would need to be able to determine the integral of $s(tx,ty,tz)$ in the given domain, right? – Michael Paris Jul 14 '17 at 16:04
  • Actually, you’ll need to compute $I(x,y,z)=\int_0^1s(tx,ty,tz),t^2,dt$. The algorithm produces $I(x,y,z)(z,dx\wedge dy+x,dy\wedge dz+y,dz\wedge dx)$. – amd Jul 14 '17 at 22:35

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