= Forward displacement from a translating localized potential
{title2=$v\Delta x=\int\dot x^2dt$}
Consider an overdamped particle obeying <linear drag> in a translating <potential energy> profile: $\zeta\dot x=F(x-vt)$, where $\zeta>0$, $v>0$ and $F(y)=-U'(y)$. Assume a finite passing trajectory starts and ends at equal values of $U$.
With $y=x-vt$, the <chain rule> gives
$$
\frac{dU}{dt}=-F(y)(\dot x-v)=-\zeta\dot x^2+v\zeta\dot x.
$$
Integrating between the equal-potential endpoints yields
$$
v\Delta x=\int\dot x^2\,dt\geq0.
$$
The net displacement is therefore forward, even though the instantaneous <force> changes sign across a well. The result follows from dissipation and the endpoint condition; <compact support> of the <force> alone does not imply equal endpoint potentials.
Writing $u=F/\zeta$ and using $\int u\,dy=0$ gives the equivalent passing-regime identity
$$
\Delta x=\int\frac{u^2}{v(v-u)}\,dy.
$$
For $\lVert u\rVert_\infty<v$, expanding the denominator gives $\Delta x=v^{-2}\int u^2dy+O(v^{-3})$.
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