 Solution

The first part of $\Delta U$ is the change of the quadratic energy associated with the active Ornstein–Uhlenbeck force. The second is minus the change in the particle's potential energy, measured in thermal-noise units. The time integral is the work done by active propulsion, again divided by its noise scale; the full log ratio is the trajectory's time-reversal asymmetry or entropy production.

In a stationary confining state the two endpoint terms remain $O(1)$ as $T\to \infty$ and have zero mean, while the mean active work and the mean log ratio grow proportionally to $T$. Endpoint-term distributions approach time-independent distributions with positive and negative fluctuations. Under mixing assumptions, the time-integrated work has a large-deviation distribution: its central part becomes approximately Gaussian with mean and variance proportional to $T$, while its far tails scale exponentially in $T$. The log-ratio distribution obeys the corresponding fluctuation theorem symmetry.