I'm working with the metric $ds^2=\frac{dxdy}{xy}$ and this applies to the vector field $\vec V.$ Also working with a semi-Riemannian manifold.
I have a vector field $\vec V=\langle x\log x, -y\log y\rangle$ on the $x-y$ plane for $x,y\in (0,1).$ Assume $\vec V$ to be the projection onto $(x,y,0)$. I've found that $f(x,y)=\big(\frac{x^2}{2}\log x - \frac{x^2}{4} - \frac{y^2}{2}\log y + \frac{y^2}{4} + C\big).$ So $\textbf{Grad}\big(f(x,y)\big)=\textbf{Grad}\big(\frac{x^2}{2}\log x - \frac{x^2}{4} - \frac{y^2}{2}\log y + \frac{y^2}{4} + C\big)=\big\langle x\log x,-y \log y \big\rangle.$
Now here is $\vec V$ sitting in the $x-y$ plane, with another "copy" sitting on the plane sliced by $z=1.$
Here, two more "copies" are added. I left out the last two copies because it would make the drawing too confusing.
How can one reconstruct a (possibly unique) vector field in $(0,1)^3$ based on the boundary vector fields?
Just to make it clear the 3-vector field in $(0,1)^3$ projected onto the faces of the cube should yield $\vec V$ up to congruence.
My thought is to find the surfaces whose gradients equal the copies of $\vec V$ on the boundary. But epistemologically, this doesn't get me any closer to finding the vector field in 3-space. I think I need a new idea to solve the problem.
Does anyone have any idea how to do this or is it not possible from the given information?
My constraints are:
$1)$ All the boundary vector fields must be Killing (they preserve the metric(s))
$3)$ The integral curves of each boundary vector fields must be analytic.
$4)$ The boundary vector fields and the interior vector field must be bounded.
Here is the shape that I think should resemble the interior 3-vector field:
