Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 71 / 4 / b

Take $x(0)=0$. Integrating the velocity solution givesThe initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general integrated Ornstein-Uhlenbeck displacementUse the thermal initial velocity and the fluctuation-dissipation relation for a Langevin particle. With diffusion coefficient $D=k_{B}T/\zeta$, this simplifies toFor $t\ll \tau$, expansion gives $⟨x^2⟩=(k_{B}T/m)t^2+O(t^3)$: displacement is initially $v_{0}t$, so the particle is ballistic. For $t\gg \tau$, $⟨x^2⟩=2Dt-2D\tau +o(1)$, hence the leading behavior is one-dimensional diffusion with $D=k_{B}T/(6\pi \mu a)$.

The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $⟨x^2⟩=A{t}^3/(3m^2)+O(t^4)$ at very short times. A thermal initial velocity, or a nonzero initial velocity variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-velocity initial condition.