Initial-value solution of a damped harmonic trap

For distinct roots $r_{\pm }=(-\gamma \pm \sqrt{{\gamma }^2-4mk})/(2m)$, the deterministic response is $X=(v_0-r_{-}{x}_0)e^{r_{+}t}/(r_{+}-r_{-})+(r_{+}{x}_0-v_0)e^{r_{-}t}/(r_{+}-r_{-})$. It satisfies $X(0)=x_0$ and $\dot{X}(0)=v_0$. With zero-mean future noise and prescribed initial data it is also the conditional ensemble mean, not a full stochastic trajectory.

For $ϵ=km/{\gamma }^2\ll 1$, momentum relaxes in ${\tau }_m=m/\gamma$ while the trap relaxes in ${\tau }_k=\gamma /k$. After the fast transient, the slow amplitude is $x_0+{\tau }_m{v}_0+O(ϵ)$. Neglecting the initial-velocity contribution additionally requires $|{\tau }_m{v}_0|\ll |x_0|$; small Reynolds number alone does not prove this condition.

For $m\ddot{x}+\gamma \dot{x}+kx=0$, let $r_s$ and $r_f$ be the slow and fast negative roots. The coefficient of the slow normal mode is exactlyWith $ϵ=km/{\gamma }^2\ll 1$ and ${\tau }_m=m/\gamma$, expansion gives $A_s=x_0+{\tau }_m{v}_0+O(ϵ)(|x_0|+|{\tau }_m{v}_0|)$. Thus the initial velocity can leave a finite displacement after its own fast decay. In particular, $x_0=0$ and $v_0\ne 0$ produce a nonzero slow tail. A first-order overdamped approximation needs an adjusted initial position after the boundary layer; setting it to $x_0$ requires the additional smallness of ${\tau }_m{v}_0$.