When does an oscillating charged particle stop radiating? (classic case) Although self-force is one of the major problem in physics, I think with a few assumptions it's possible to deal with it, at least in Newtonian mechanics. Let's consider a periodic motion, where we have made a long hole in the Earth such that our charged particle can go through it and oscillates there for a long time (But not forever! It's not a perpetual motion machine). We just want to know how long does it oscillate and therefore radiate. See the picture below:

The red circle is a charged particle with mass $m$ and charge $q$. We know that gravity will force the particle to oscillate, because it applies a backward force (check shell theorem, this is a solid sphere with a hole, not a hollow one, or see this video), $F_g=-kr$, where $k$ is a constant and $r$ indicates the position of particle, that's, if charged particle goes to center of Earth (or $r=0$) there will be no force on it. Let's also assume that initial velocity of particle is zero. We know that when particle arrive at $r=0$ its velocity is not zero, so it continues its motion to the other end of hole, when it arrives there, it goes back and this continues for a long time. Along the way, it radiates too, because it's oscillating after all. It can not radiate forever or else we have a system that gives us energy for free, out of nothing and even better, forever! Eventually, it should stop at the center of the earth, but when?! after how much radiation? Ok, someone might want to use equivalence principle to say that there will be no radiation at all, please don't do that. From mathmatical point of view, this problem is already hard, not to mention particle oscillates so it's not really a free falling particle (even it's, let's put it aside for now). Ok we have Newton's second law, 
$$F_{net}=F_g+F_{rad}\rightarrow mr''=-kr+nr'''$$
Where $n$ is a constant for radiation (it depends on  $q$ and $m$). And $F_{rad}$ is the scary self-force. Even with Wolfram, solving this equation with respect to initial values is hard. It seems more logical to use Larmor formula, 
$$p=c*(r'')^2$$
where $p$ is total power radiated by a non relativistic point charge, $c$ is constant depends on particle's charge. We can use Larmor formula because motion is periodic. But $r''$ or acceleration is not constant, and it is not changing slowly enough to put it outside of integral with respect to time. I wanted to subtract it from initial energy of particle, but I can't do that.
TL;DR
Prove that at finite amount of time, an oscillating charged particle will stop, and it radiates finite amount of energy. Only rigorous solution with mathmatical details will be accepted (but I vote up whatever answer I like).
 A: There is a strong argument to be made that the Abraham-Lorentz equation is ill-posed, and that the Landau-Lifschitz equation is the correct equation to use:
$$
m \ddot{x} = F_\text{ext} + \tau_0 \left( \frac{\partial F_\text{ext}}{\partial t} + \dot{x} \frac{\partial F_\text{ext}}{\partial x} \right) + \mathcal{O}(\tau_0^2),
$$
where $\tau_0 = \mu_0 q^2/(6 \pi m c)$.  This equation can be obtained from the Abraham-Lorentz equation via the "reduction of order" procedure.  A basic description of the reduction-of-order procedure can also be found in Zangwill's Modern Electrodynamics, which I summarize here: one treats the $\dot{a}$ term as though $a$ was the acceleration that the charge would feel without radiation reaction (i.e., $a = F_\text{ext}/m$), takes the time derivative of this quantity, and substitutes it into the Abraham-Lorentz equation.  While this procedure may seem ad hoc if one starts from the Abraham-Lorentz equation, the reduced-order equation can actually be obtained in a more rigorous manner as well (see reference below.)
Taking this approach n your case, you have
$$
F_\text{ext}(x,t) = - k x
$$
and so the differential equation becomes
$$
m \ddot{x} = -k x - \tau_0 k \dot{x}.
$$
This is your standard equation for a damped oscillator, with solution
$$
x(t) = x_0 e^{\Omega t},
$$
where 
$$
\Omega = -\frac{1}{2} \tau_0 \omega_0^2 \pm i \omega_0 \sqrt{ 1 - \frac{\tau_0^2 \omega_0^2}{4}} \approx -\frac{1}{2} \tau_0 \omega_0 \pm i \omega_0.
$$
Note that we must discard the extra term in the square root, since the Landau-Lifshitz equation is only reliable to first order in $\tau_0$.
The solution is then
$$
x(t) = A e^{- \tau_0 \omega_0^2 t} e^{i \omega_0 t}
$$
where $A$ is a complex number determined by the initial conditions.  In particular, if we take $A$ to be real, then the initial energy of the body is $E_0 \approx \frac{1}{2} k A^2$ (to first order in $\tau_0$), and the energy radiated is
\begin{align}
E_\text{rad} &= \int_0^\infty P \,dt \\
&= \int_0^\infty \tau_0 m a^2 \, dt  & \text{(Larmor: $ P = \tau_0 m a^2$)}\\
&= \frac{1}{2} \tau_0 m \int_0^\infty (A \Omega^2 e^{\Omega t}) (A \bar{\Omega}^2 e^{\bar{\Omega} t}) dt \\
&= \frac{1}{2} A^2 |\Omega|^4 \tau_0 m \int_0^\infty e^{2 \Re(\Omega) t} \, dt \\
&= \frac{1}{2} A^2 \omega_0^4 \tau_0 m \int_0^\infty e^{- \tau_0 \omega_0^2 t} dt \\
&= \frac{1}{2} A^2 \omega_0^2 m = \frac{1}{2} k A^2, 
\end{align}
as desired.

For details on why the Landau-Lifschitz equation is a better way to think about radiation reaction than the Abraham-Lorentz equation, see

Gralla, Harte, & Wald, "A Rigorous Derivation of Electromagnetic Self-force."  Phys. Rev. D 80, 024031 (2009).

In this work, the authors use a scaling procedure to take the limit of the force on a continuous body as its size approaches zero but its charge remains fixed.  The self-force term arises as a first-order contribution in $\tau_0$, along with spin-dependent forces and dipole-dependent forces.  The contributions of higher-order terms then depend on the internal structure of the body;  in particular, if one wants to carry this derivation to higher order, one needs a detailed description of how it responds to electromagnetic forces (perfect rigidity is a no-no, since this is inherently a relativistic calculation.)  
