$s=t-t_0$

s=t-t_0

$\displaystyle\frac{\partial P}{\partial s}dv\tau=-\displaystyle\int_{-\infty}^{\infty}du P(v,s|v_0)P(u,\tau|v)+\displaystyle\int_{-\infty}^{\infty}du P(u,s|v_0)P(v,\tau|u)$

\displaystyle\frac{\partial P}{\partial s}dv\tau=-\displaystyle\int_{-\infty}^{\infty}du P(v,s|v_0)P(u,\tau|v)+\displaystyle\int_{-\infty}^{\infty}du P(u,s|v_0)P(v,\tau|u)

$\displaystyle\int_{-\infty}^{\infty}du P(u,s|v)=1$

\displaystyle\int_{-\infty}^{\infty}du P(u,s|v)=1

$\displaystyle\frac{\partial P}{\partial s}\tau=- P(v,s|v_0)+\displaystyle\int_{-\infty}^{\infty}d\xi P(v-\xi,s|v_0)P(v,\tau|v-\xi)$

\displaystyle\frac{\partial P}{\partial s}\tau=- P(v,s|v_0)+\displaystyle\int_{-\infty}^{\infty}d\xi P(v-\xi,s|v_0)P(v,\tau|v-\xi)

$\displaystyle\frac{\partial}{\partial s}P(v,s|v_0)=\sum_{n=1}^{\infty}\displaystyle\frac{(-1)^n}{n!}\displaystyle\frac{\partial^n}{\partial v^n}[M_nP(v,s|v_0)]$

\displaystyle\frac{\partial}{\partial s}P(v,s|v_0)=\sum_{n=1}^{\infty}\displaystyle\frac{(-1)^n}{n!}\displaystyle\frac{\partial^n}{\partial v^n}[M_nP(v,s|v_0)]

$M_n=\displaystyle\frac{1}{\tau}\displaystyle\int_{-\infty}^{\infty}d\xi P(v+\xi,\tau|v)\xi^n=\displaystyle\frac{1}{\tau}\langle (v(\tau)-v(0))^n\rangle$

M_n=\displaystyle\frac{1}{\tau}\displaystyle\int_{-\infty}^{\infty}d\xi P(v+\xi,\tau|v)\xi^n=\displaystyle\frac{1}{\tau}\langle (v(\tau)-v(0))^n\rangle

$\displaystyle\frac{\partial P}{\partial s} = \displaystyle\frac{\partial}{\partial v}(M_1P)+\displaystyle\frac{1}{2}\displaystyle\frac{\partial^2}{\partial v^2}(M_2P)$

\displaystyle\frac{\partial P}{\partial s} = \displaystyle\frac{\partial}{\partial v}(M_1P)+\displaystyle\frac{1}{2}\displaystyle\frac{\partial^2}{\partial v^2}(M_2P)

$M_1=\displaystyle\frac{1}{\tau}(v(\tau)-v(0)) = -\displaystyle\frac{1}{\tau}v$

M_1=\displaystyle\frac{1}{\tau}(v(\tau)-v(0)) = -\displaystyle\frac{1}{\tau}v

$M_2=\displaystyle\frac{1}{\tau}\langle (v(\tau)-v(0))^2\rangle = \displaystyle\frac{1}{\tau}\displaystyle\frac{6k_BT}{m}$

M_2=\displaystyle\frac{1}{\tau}\langle (v(\tau)-v(0))^2\rangle = \displaystyle\frac{1}{\tau}\displaystyle\frac{6k_BT}{m}

$\tau\displaystyle\frac{\partial P}{\partial s} = \displaystyle\frac{\partial}{\partial v}(vP)+\displaystyle\frac{3k_BT}{m}\displaystyle\frac{\partial^2}{\partial v^2}P$

\tau\displaystyle\frac{\partial P}{\partial s} = \displaystyle\frac{\partial}{\partial v}(vP)+\displaystyle\frac{3k_BT}{m}\displaystyle\frac{\partial^2}{\partial v^2}P

$P(v,s|v_0)=\sqrt{\displaystyle\frac{m}{2\pi k_BT(1-e^{-2s/\tau})}}e^{-m(v-v_0e^{-s/\tau})^2/2k_BT(1-e^{-2s/\tau})}$

P(v,s|v_0)=\sqrt{\displaystyle\frac{m}{2\pi k_BT(1-e^{-2s/\tau})}}e^{-m(v-v_0e^{-s/\tau})^2/2k_BT(1-e^{-2s/\tau})}