Given *an* expression for the current and *the* expression for a density, can you determine the velocity field? Suppose you a density field $\rho(\vec r)$ and, due to some equation of motion, you have derived that
$$
\partial_t \rho + \nabla \cdot \vec J = 0, \tag{1}
$$
where $\partial_t = \frac{\partial}{\partial t}$, and you have found an explicit expression for $\vec J$ as a function of $\rho$ and $\partial_t \rho$.
This continuity equation for $\rho$ shows that it is globally conserved.
Comparing it to an equation of motion where $\rho$ is being advected by a velocity field $\vec v$
$$
\partial_t \rho + \nabla \cdot (\rho \vec v) = 0, \tag{2}
$$
it is tempting to identify
$$
\vec v = \frac{\vec J}{\rho}.
$$
However, since Eq. (1) only determines $\vec J$ up to a divergence-free contribution $\vec K$, where $\nabla \cdot \vec K=0$, one can really only say that
$$
\vec v = \frac{\vec J + \vec K}{\rho}. \tag{3}
$$
As I understand it, this stems from the ambiguity in a current as derived from the density field alone. E.g., a constant current of $\vec K=(1,0,0)$ would not be visible in the time evolution of $\rho$.
However, are there reasonable assumptions one can make in order to identify the velocity field as in Eq. (3)?
As context, what I am trying to understand is exemplified in this Physical Review Letters paper from 1997, where the author seems to do this identification between Eqs. (11) and (12).
 A: You can go one step further: If you substract equation (1) from equation (2), you get $\vec\nabla\cdot(\rho\vec v-\vec J)=\vec 0$. Because of the Helmholtz decomposition (only if you consider the fields in $\mathbb R^3$ without a boundary, otherwise there will be a boundary term), there exists a vector field $\vec L$ with:
$$\rho\vec v-\vec J
=\vec\nabla\times\vec L
\Leftrightarrow
\vec v
=\frac{\vec J+\vec\nabla\times\vec L}{\rho}.$$
Of course you could also directly apply the Helmholtz decomposition to $\vec K$, which you already mentioned is divergence-free, and get $\vec K=\vec\nabla\times\vec L$. Basically, the same is done in electrodynamics by writing the divergence-free magnetic field $\vec B$ as $\vec B=\vec\nabla\times\vec A$.
Using Stokes' theorem here won't help much as it just transfers a differential equation into an integral equation, but in fluid dynamics an incompressible fluid leads to the additional condition $\vec\nabla\cdot\vec v=\vec 0$, which you could also use here as the interpretation in Ginsburg-Landau theory is probably quite similar. If we put the upper equation into this condition and using a few identities of vector analysis, we get:
\begin{align*}
\vec\nabla\cdot\vec v
=\vec\nabla\cdot\left(\frac{\vec J+\vec\nabla\times\vec L}{\rho}\right)
=-\frac{\vec J+\vec\nabla\times\vec L}{\rho^2}\cdot\vec\nabla\rho
+\frac{\overbrace{\vec\nabla\cdot\vec J}^{=-\partial_t\rho}+\overbrace{\vec\nabla\cdot(\vec\nabla\times\vec L)}^{=0}}{\rho} 
\stackrel ! =0 \\
\Rightarrow
-\vec J\cdot\vec\nabla\rho
+\vec\nabla\cdot(\vec L\times\vec\nabla\rho)
-\rho\partial_t\rho=0 \\
\Rightarrow
\vec\nabla\cdot(\vec L\times\vec\nabla\rho-\rho\vec J)
-2\rho\partial_t\rho=0,
\end{align*}
and therefore another continuity equation:
$$\partial_t(\rho^2)
-\vec\nabla\cdot(\vec L\times\vec\nabla\rho-\rho\vec J)=0.$$
