Is the continuity equation used to **define** the current density? Recall the continuity equation:
$$\frac{\partial}{\partial t}\rho+\boldsymbol{\nabla\cdot J}=0$$
Given $\rho$, there is obviously not a unique solution $\boldsymbol{J}$, but I guess one could choose an additional requirement (e.g. some condition on the rotation) such that there is a unique solution (by the Helmholtz theroem?). The idea may be to pick the most "obvious" solution. So I hope that someone can elaborate on the mathematical aspect and tell us if this is actually done in practice.
 A: The continuity equation is not the definition of the current density.
One way to define the current density $\mathbf J(\mathbf r)$ is via the three components $J_x(\mathbf r), J_y(\mathbf r), J_z(\mathbf r)$ where
\begin{equation}
J_x(\mathbf r) = \lim_{s \to 0} \frac{I_x(\mathbf r, s)}{\pi s^2}
\end{equation}
where the notation $I_x(\mathbf r, s)$ means the current passing in the positive x-direction through a disk that is perpendicular to the x-axis, centered around the point $\mathbf{r}$, with radius $s$. $J_y$ and $J_z$ are defined similarly. If you choose to use this definition, you might have to do a bit of work to show that the $\bf J$ so defined actually transforms as a vector. However, I think it has pedagogical value in the context of this question: it shows that "current density" is fundamentally nothing more than current per unit area.
A: To find the current from knowing only $\dot \rho$ you will need extra information such as the electrical conductivity. If
$$
{\bf J}= \sigma {\bf E}=- \sigma \nabla \phi
$$
then you have
$$
\dot \rho- \sigma \nabla^2 \phi=0,
$$
which can be solved with suitable boundary conditions.
