Suppose $(M,g)$ is a three dimensional smooth compact simply connected Riemannian manifold with boundary and suppose that $\Sigma$ is a smooth simply connected hypersurface in $M$ with a smooth boundary $\partial \Sigma \subset \partial M$. Let $D\_X Y$ denote the Levi-Civita connection on $(M,g)$. Let $Z$ be a smooth vector field on $\Sigma$ and let $f:\Sigma \to \mathbb R$ be a smooth function on $\Sigma$. Does there exist a smooth function $\phi$ in $M$ such that $\Sigma=\{\phi=0\}$ and such that
$$ (D\_{\nabla \phi} \nabla \phi)\big|\_{\Sigma} = f\, Z.$$
Note that the left hand side is just the restriction of $D\_{\nabla \phi}\nabla \phi$ to $\Sigma$ and that $\nabla \phi$ is the gradient of $\phi$ with respect to $g$.


If the answer is no, would it make a difference if the latter condition is replaced with
$$(D\_{\nabla \phi} \nabla \phi)\big|\_{\Sigma} - f\, Z \in \textrm{Span}\{(\nabla \phi)|\_{\Sigma}\}$$



A question on Levi-Civita connection and a fixed hyper surface
Not a full answer, but there definitely should be some integrability constraint.


Take the simplest case where $M = \mathbb{R}^3$ (the boundary is unimportant for the discussion here) and $\Sigma$ is the $x$-$y$ plane.


If $\phi$ is a defining function of $\Sigma$, then restricted to $\Sigma$ we have $\nabla \phi|\_{\Sigma} = \eta \partial\_z$ for some function $\eta: \Sigma \to \mathbb{R}$.


Symmetry of the Hessian requires then
$$ \nabla^2 \phi|\_{\Sigma} = \begin{pmatrix}
0 & 0 & \partial\_x \eta \\
0 & 0 & \partial\_y \eta \\
\partial\_x \eta & \partial\_y \eta & \* \end{pmatrix} $$
This shows that
$$ D\_{\nabla\phi} \nabla\phi = \eta (\partial\_x \eta, \partial\_y \eta, \*) $$


So a necessary condition for your equation to hold (setting $f \equiv 1$ since as you agreed it is unimportant) in the flat case is that $Z$ is the gradient of some scalar function on $\Sigma$, and that allowing there to be normal components (as in your modified question) has no impact.




---


The conclusion is unchanged in the Lorentzian setup, if you require $\Sigma$ to be non-degenerate. In the case where $\Sigma$ is null, however, since $\nabla\phi$ now lies in the tangent space of $\Sigma$ and $\Sigma$ is two dimensional, it may in fact be possible for the second variant to hold (but I don't have the energy to run through the analysis now).


