= Sign of radiation damping and comparison with velocity damping
The sign can be checked independently by multiplying the equation by $\dot x$. With $E=(\dot x^2+x^2)/2$,
$$
\frac{d}{dt}(E-\varepsilon\ddot x\dot x)
=f\dot x-\varepsilon\ddot x^2.
$$
For periodic motion the total-derivative near-field term averages to zero, so positive $\varepsilon$ gives a radiative <energy> loss, consistent with the slow poles' damping. Negative $\varepsilon$ acts as <energy> gain in this model.
For the comparison equation \b[as printed], $\ddot x+x-\varepsilon\dot x=f$, the response denominator instead is $1-\omega^2+i\varepsilon\omega$, with exact poles
$$
\omega_\pm=\frac{i\varepsilon}{2}\pm\sqrt{1-\varepsilon^2/4}
=\pm1+\frac{i\varepsilon}{2}+O(\varepsilon^2).
$$
Thus ordinary positive damping in that sign convention requires $\boxed{\varepsilon<0}$, the opposite sign from slow radiation damping in the third-derivative equation. The word “damped” does not override the explicitly printed minus sign. With the usual equation $\ddot x+x+\gamma\dot x=f$, positive $\gamma$ gives the familiar lower-half-plane poles.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2017-q4-radiation-poles.png]
{title=Exact pole locations for small positive and negative radiation-reaction parameters, including the singular fast pole}
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