We know that if charged particle is accelerated, it will radiate. According to Larmor formula, the power is $$P = {2 \over 3} \frac{q^2 a^2}{ c^3}$$ in the Gauss units, where $a$ is acceleration.
And accelerated particle will experience radiation damping force(Abraham–Lorentz force) $$\mathbf{F}_\mathrm{rad} = { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}}$$
The equation of motion of particle should be $$m \frac{d \mathbf{v}}{dt}= \mathbf{F}_{ext}+ { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}} \tag{1}$$
But a contradiction arises when the external force is a constant force. In this case, the particle has constant acceleration ($\mathbf{\dot a}=0$). But it will still radiate energy ($P\neq0$). The work done by external force $\mathbf{F}_{ext}$ will be totally translated to particle's kinetic energy(due to uniform acceleration), but the particle will still radiate energy to infinity. Does this violate the energy conservation?
Note1: For constant force, the uniform acceleration solution $\mathbf{\dot a}=0$ is still the new equation(1) 's solution. That is, $\dot a=0$ is a solution for $a=b$. Certainly this solution($\dot a=0$ ) can also be the solution of $a= b+d \dot a =b$.
Note2: Even though you take the relativistic effect into consideration, you still cannot explain this contradiction. The relativistic generalization of radiation damping force is in Landau's The classical theory of fields (76.2) $$F_{rad}^\mu= \frac{2 e^2}{3 c}(\frac{d^2 u^{\mu}}{ds^2}-(u^\mu u^\alpha)\frac{d^2 u_\alpha}{ds^2})$$ So we see with constant external force $F_{ext}^\mu=\text{cosntant}$, equation of motion, $$m\frac{du^{\mu}}{ds}=F^{\mu}_{ext}+F^{\mu}_{rad}\tag{2}$$
Constant $4$-acceleration $\frac{du^{\mu}}{ds}=\frac{F^{\mu}_{ext}}{m}$, $\frac{d^2 u^{\mu}}{ds^2}=0$ is still the equation(2)'s solution. So there is still no radiation damping force.
