What happens if you apply Jefimenko's equations to a non-conserved current? Maxwell's equations (in differential form) and Jefimenko's equations are in some sense inverses of each other: Maxwell's equations tell you how to use the EM fields $F$ to calculate the current (and charge) density $J$, and Jefimenko's equations tell you how to use the current density $J$ to calculate the EM fields $F$.
Maxwell's equations are only internally consistent if the current is conserved. Put another way: given any EM field $F$, the corresponding current density $J$ that comes out of Maxwell's equations will automatically be conserved.
Q1. What happens if you just blindly plug a non-conserved current $J$ into Jefimenko's equations? Do you still get a seemingly sensible EM field $F$?
Q2. If the answer to Q1 is yes, then what happens if you take that EM field $F$ and plug it into Maxwell's equations to try to get back a current density? The resulting current density $J'$ must be conserved if it comes out of Maxwell's equations, so clearly you can't end up with the same current density $J$ that you started with. This seems to suggest that composing together Maxwell's and Jefimenko's equations gets you some kind of projection map from the space of all possible four-current fields $J$ to the subspace of conserved four-current fields. What is that nature of this projection map? (My guess is that there's some natural decomposition of a general differential form $J$ as the sum of a closed part and a non-closed part, and this projection map simply discards the non-closed part.)
 A: If we use arbitrary charge/current densities in Jefimenko's equations, then the resulting fields generally do not satisfy Maxwell's equations. The derivation of Jefimenko's equations assumes that the current is conserved. Here's a quick review of the derivation, which I'll refer back to when answering Q2.
Half of Maxwell's equations (the metric-independent ones) are satisfied by the definition $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, with no constraints on $A_\mu$. The remaining Maxwell's equations (the metric-dependent ones) are
$$
\newcommand{\bfx}{\mathbf{x}}
\newcommand{\pl}{\partial}
 \pl^2 A_\mu - \pl_\mu \pl\cdot A = J_\mu.
\tag{1}
$$
According to equation (6.66) in ref 1, the distribution
$$
 G(\bfx'-\bfx,t'-t)\equiv\frac{\delta(t-(t-|\bfx'-\bfx|))}{|\bfx'-\bfx|}
$$
satisfies
\begin{align}
 \pl^2 G(\bfx'-\bfx,t'-t)
 &\equiv
 (\pl_t^2-\pl_\bfx^2)G(\bfx'-\bfx,t'-t)
\\
 &=4\pi
 \delta(t'-t)\delta(\bfx'-\bfx).
\tag{2}
\end{align}
Now consider the gauge
$$
 \pl\cdot A=0.
\tag{3}
$$
In this gauge, equation (1) reduces to $\pl^2 A_\mu\propto J_\mu$, which is satisfied by
$$
 A_\mu(\bfx,t)=\frac{1}{4\pi}\int dt'\,d^3\bfx'\ 
 G(\bfx'-\bfx,t'-t) J_\mu(\bfx',t').
\tag{4}
$$
Equation (4) immediately gives Jefimenko's equations as shown on the Wikipedia page (the version written in terms of potentials).
We're not done yet, because we need to check that the condition (3) is consistent with (4). In general, it isn't: they are consistent only if the current is conserved. To see this, apply $\pl^\mu$ to the equation $\pl^2 A_\mu\propto J_\mu$ and then use (3). The result is $0=\pl\cdot J$. This shows that current conservation is a necessary condition for the derivation shown above to be valid. To see that it is also a sufficient condition, start with (4) and use the fact that $G$ depends only on the differences $\bfx'-\bfx$ and $t'-t$ to get
$$
 \pl^\mu A_\mu=\frac{1}{4\pi}\int dt'\,d^3\bfx'\  
 G(\bfx'-\bfx,t'-t) \partial^\mu J_\mu(\bfx',t').
\tag{5}
$$
This shows that current conservation implies (3). Altogether, the derivation that leads to Jefimenko's equations (4) is valid if and only if $\pl\cdot J=0$.
Now we have the material we need to answer Q2. If we use an arbitrary current in equation (4) and put this back into Maxwell's equation (1), then the identity (2) impiles
$$
 \pl^2 A_\mu - \pl_\mu \pl\cdot A = \tilde J_\mu
$$
with
$$
 \tilde J_\mu\equiv J_\mu - \frac{1}{4\pi} \pl_\mu \int G\pl\cdot J.
\tag{6}
$$
The question is whether the modified current (6) is conserved. Apply $\pl^\mu$ to (6) and use (2) to confirm that the answer is yes.


*

*Jackson (1975), Classical Electrodynamics, 2nd edition

