Forward displacement from a translating localized potential ($v\Delta x=\int {\dot{x}}^{2}dt$)

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 $F(y)=-U^{\prime }(y)$. Assume a finite passing trajectory starts and ends at equal values of $U$.

With $y=x-vt$, the chain rule givesIntegrating between the equal-potential endpoints yieldsThe 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 $\int udy=0$ gives the equivalent passing-regime identityFor $\Vert u{\Vert }_{\infty }<v$, expanding the denominator gives $\Delta x=v^{-2}\int u^{2}dy+O(v^{-3})$.