For a single charge $e$ with position vector $\textbf R$, the charge density $\rho$ and and current density $\textbf{j}$ are given by:
\begin{equation} \rho(\textbf{r},t)= e\,\delta^3(r-\textbf{R}(t)), \end{equation}
\begin{equation} \textbf{j}(\textbf{r},t)=e\frac{d\textbf{R}}{dt}\delta^3(\textbf{r}-\textbf{R}(t)).\end{equation}
Suppose we want to check the equation of continuity
\begin{equation}\frac{\partial \rho}{\partial t}+\nabla \cdot\textbf{j}=0 .\end{equation}
How to do it? How to deal with the derivatives of a delta function?
Best Answer
It is almost no trouble to generalize to a finite number of point charges $q_i$ at positions ${\bf r}_i(t)$. Then the charge density is
$$ \rho({\bf r},t)~ =~ \sum_i q_i \delta^3({\bf r}-{\bf r}_i(t)), $$
and the current density
$$ {\bf j}({\bf r},t)~ =~ \sum_i q_i \dot{\bf r}_i(t)\delta^3({\bf r}-{\bf r}_i(t)). $$
For clarity let us write ${\bf \nabla}\equiv\frac{\partial}{\partial {\bf r}}.$ The chain rule then yields the continuity equation
$$ -\frac{\partial \rho({\bf r},t)}{\partial t} ~=~ -\sum_i q_i \frac{\partial }{\partial t}\delta^3({\bf r}-{\bf r}_i(t)) ~=~ \sum_i q_i \dot{\bf r}_i(t)\cdot \frac{\partial }{\partial {\bf r}}\delta^3({\bf r}-{\bf r}_i(t)) $$ $$~=~ \frac{\partial }{\partial {\bf r}} \cdot \sum_i q_i \dot{\bf r}_i(t)\delta^3({\bf r}-{\bf r}_i(t))~=~\frac{\partial }{\partial {\bf r}} \cdot {\bf j}({\bf r},t).$$
The same calculation can be repeated more carefully with the help of test functions.