Problem statement
Suppose I have a 2D or 3D equation of the form:
$\vec{\nabla} \cdot \left[ \vec{\vec{a}}\left(\vec{x}\right) \cdot \vec{\nabla} f\left(\vec{x}\right) \right] = \vec{\nabla} \cdot \left[ \vec{\vec{a}}\left(\vec{x}\right) \cdot \vec{S} \left(\vec{x}\right) \right]$
with boundary conditions
$\left[\vec{\vec{a}}\left(\vec{x}\right) \cdot \left( -\vec{\nabla} f\left(\vec{x}\right) + \vec{S}\left(\vec{x}\right) \right)\right] \cdot \hat{n}\left(\vec{x}\right) =0$
where $\hat{n}$ is the unit vector pointing outward, normal to the simply-connected domain boundary.
Suppose the rank-2 tensor function $\vec{\vec{a}}\left(\vec{x}\right)$ and vector function $\vec{S}\left(\vec{x}\right)$ are known functions and are continuously-differentiable functions throughout the domain. I wish to solve for $f\left(\vec{x}\right)$ everywhere in the domain.
This is equivalent to solving $\vec{\nabla} \cdot \vec{A}\left(\vec{x}\right) = 0$ throughout the domain and $\vec{A}\left(\vec{x}\right) \cdot \hat{n}\left(\vec{x}\right) = 0$ at the boundaries, where $\vec{A}\left(\vec{x}\right) = \vec{\vec{a}}\left(\vec{x}\right) \cdot \left[ -\vec{\nabla} f\left(\vec{x}\right) + \vec{S}\left(\vec{x}\right) \right]$.
Naive solution
The (potentially?) naive solution is $\vec{\nabla} f\left(\vec{x}\right) = \vec{S}\left(\vec{x}\right)$, or $\vec{A}\left(\vec{x}\right)=0$.
My question
Can it be shown with the information provided that the naive solution is the only solution? If not, what conditions would need to be placed on the known variables to make the naive solution unique?
Of course, $f\left(\vec{x}\right)$ is unique only up to an additive constant. I'm satisfied solving for either that or $\vec{\nabla} f\left(\vec{x}\right)$, which should be unique.