= Solution
The moving <point charge> has distributional <charge density> and <current density>
$$
\rho(\mathbf r',t')=q\delta^{(3)}(\mathbf r'-\mathbf r_0(t')),\qquad
\mathbf J(\mathbf r',t')=q\mathbf v_0(t')\delta^{(3)}(\mathbf r'-\mathbf r_0(t')).
$$
Let
$$
K(\mathbf r,t;\mathbf r',t')=
\frac{\delta(t-t'-|\mathbf r-\mathbf r'|/c)}{|\mathbf r-\mathbf r'|}.
$$
The given <retarded fundamental solution> identity immediately implies that the <retarded potentials>
$$
\phi=\frac1{4\pi\epsilon_0}\int K\rho\,d^3r'\,dt',\qquad
\mathbf A=\frac{\mu_0}{4\pi}\int K\mathbf J\,d^3r'\,dt'
$$
solve $\square\phi=-\rho/\epsilon_0$ and $\square\mathbf A=-\mu_0\mathbf J$ in the sense of <distributions>. Integrating the spatial <Dirac delta function> first gives
$$
\phi=\frac q{4\pi\epsilon_0}\int_{-\infty}^{\infty}
\frac{\delta(g(t'))}{R(t')}\,dt',\qquad
g(t')=t-t'-R(t')/c,
$$
and the analogous integral for $\mathbf A$ has numerator $\mu_0q\mathbf v_0(t')/(4\pi)$.
Assume the physical trajectory has <subluminal speed> and the <retarded time> exists. At its root $t_r$, with $\mathbf n_r=\mathbf R(t_r)/R(t_r)$,
$$
\frac{dR}{dt'}=-\frac{\mathbf R\cdot\mathbf v_0}{R},\qquad
g'(t_r)=-\left(1-\frac{\mathbf n_r\cdot\mathbf v_0(t_r)}c\right).
$$
The positive factor $\kappa=1-\mathbf n_r\cdot\mathbf v_0/c$ also proves <uniqueness of retarded time for a subluminal source>. The <Dirac delta function> substitution formula gives $\delta(g(t'))=\delta(t'-t_r)/\kappa$, so
$$
\boxed{\phi(\mathbf r,t)=\frac q{4\pi\epsilon_0}
\frac1{R(t_r)-\mathbf R(t_r)\cdot\mathbf v_0(t_r)/c},\qquad
\mathbf A(\mathbf r,t)=\frac{\mathbf v_0(t_r)}{c^2}\phi(\mathbf r,t).}
$$
These are the <Liénard–Wiechert potentials>. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the <electric potential> and <magnetic vector potential> are ordinary smooth functions; on the trajectory their <wave equations> have the indicated distributional sources.
For completeness, the <charge continuity equation> $\partial_{t'}\rho+\nabla'\cdot\mathbf J=0$ also verifies the <Lorenz gauge>. Using $\nabla K=-\nabla'K$ and $\partial_tK=-\partial_{t'}K$, <integration by parts> in the retarded integrals gives
$$
\nabla\cdot\mathbf A+c^{-2}\partial_t\phi
=\frac{\mu_0}{4\pi}\int K(\nabla'\cdot\mathbf J+\partial_{t'}\rho)\,d^3r'\,dt'=0,
$$
with the retarded boundary prescription. Thus the construction also has the required <Lorenz gauge condition>.
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