I've already got the electric fields and magnetic fields derived from the Lienard-Wiechert potentials:
$${\bf E}=\frac{q}{4\pi\epsilon_0}\frac{R}{(\bf R\cdot u)^3}[(c^2-v^2){\bf u}+\bf R\times(u\times a)]$$
$${\bf B}=\frac{\bf R}{cR}\times\bf E$$
where ${\bf R=r-r'}$ and ${\bf u}=\frac{c\bf R}{R}-\bf v$.
I wonder if they satisfy Maxwell's equations, I've tried to derive Gauss's law, but in vain. So do they? Or is there something wrong in my derivation?
I will take you back to Maxwell's equations in Lorenz Gauge. $$\vec \nabla^2 \varphi-\frac{1}{c^2} \frac{\partial^2\varphi}{\partial t^2}=-\frac{\rho}{\varepsilon_0} $$ $$\vec \nabla^2 \bf{A}-\frac{1}{c^2} \frac{\partial^2\bf A}{\partial t^2}=\mu_0\bf J$$ also $$\bf E=-\vec \nabla \varphi-\frac{\partial \bf A}{\partial t}$$ $$\bf B=\vec \nabla \times \bf A$$ Note that $\bf E$ and $\bf B$ satisfy all Maxwell's Equations.Solution to these can be given as $$\varphi = \frac{1}{4\pi \varepsilon_0}\int \frac{\rho(\bf r',t_r')}{\vert \bf r -\bf r' \vert}d^3\bf r'$$ $$\bf{A}=\frac{\mu_0}{4 \pi} \int \frac{\bf J (\bf r',t_r')}{\vert \bf r- \bf r' \vert}d^3\bf r'$$
Using $\rho =q\delta^3(\bf r-r_0)$ and $ \bf J$ $=$ $q\bf v \delta^3(\bf r- \bf r_0)$ Remember that both $\bf r_0$ and $\bf v$ are functions of retarted time $t_r'$ to simplify we add another delta term $\delta (t' -t_r')$ we get that. $$\varphi=\frac{q}{4 \pi \varepsilon_0}\int \frac{\delta(t'-t_r')}{\vert \bf r - \bf r_0 (t_r') \vert}dt'$$ $$\bf A=\frac {q\mu_0}{4 \pi} \int \bf v \frac{\delta (t'-t_r')}{\vert \bf r-\bf r_0 \vert}dt'$$ Now using $\vert \bf R \vert'=\frac{\bf R}{\vert \bf R \vert} \cdot \bf v$ and $\delta (t'-t_r')=\frac{\delta (t'-t_r)}{ \frac{\partial }{\partial t'}(t'-t_r') \vert _{t'=t_r} } $ we get our potential $$\varphi = \frac{q}{4 \pi \varepsilon_0} \frac{1}{(1-\bf \hat R \cdot \frac{\bf v}{c})\vert \bf R \vert}$$ $$\bf A = \frac{\bf v}{c^2} \varphi$$
So we just derived our Potential starting from the assumption that Maxwell's Equations are valid which includes Gauss's Law
They are derived from Maxwell's equations, so they satisfy Maxwell's equations, but taking vector derivatives is very complicated with retardation.
