Method of images for Green's function

Suppose there is a wall at $x=0$ and a particle undergoing brownian motion (in $1D$) to the left of it can not cross the wall. The Green’s function in the absence of wall is $G_{0}(x,tmid x_{0},t_{0})$. In the presence of the wall at $x_{0}=0$ the probability of finding the particle at the wall is zero, i.e., $G(x=x_{0},tmid x_{0}=0,t_{0})=0$. Hence using the method of images the Green’s function can be written as: $G(x,tmid x_{0}=0,t_{0})=G_{0}(x,tmid-epsilon,t_{0})-G_{0}(x,tmidepsilon,t_{0})$, here $epsilon$ is very small positive quantity. Now if I take the velocity of the particle into consideration and write the free particle Green’s function as $G_{0}(x,v,tmid0,v_{0},t_{0})$ what should be my Green’s function in the presence of the wall?