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

For a translating potential energy profile with equal values at the two ends and linear drag force $-\zeta \dot{x}$, a finite passing trajectory satisfies $v\Delta x=\int {\dot{x}}^{2}dt\ge 0$. Equivalently, since $\int Fdy=0$, $\Delta x=\int u^2/[v(v-u)]dy$. Compact support of the force alone is insufficient. At high speed the displacement is $v^{-2}\int (F/\zeta {)}^{2}dy+O(v^{-3})$.