# Method of images for Green's function

Suppose there is a wall at $x=0$ and 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 $G_{0}(x,t\mid x_{0},t_{0})$. In the presence of the wall at $x_{0}=0$ the probability of finding the particle at the wall is zero $i.e$ $G(x=x_{0},t\mid x_{0}=0,t_{0})=0$. Hence using the method of images the Green's function can be written as: $G(x,t\mid x_{0}=0,t_{0})=G_{0}(x,t\mid-\epsilon,t_{0})-G_{0}(x,t\mid\epsilon,t_{0})$, 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 $G_{0}(x,v,t\mid0,v_{0},t_{0})$ what should be my Green's function in the presence of the wall?

The green function can be given by: $$G(x,t\mid x_{0}=0,t_{0})=\frac{1}{\sqrt{4\pi Dt}}[e^{{-(x-x_0-vt)}^2/4Dt}-e^{vx_0/D}e^{{-(x+x_0-vt)}^2/4Dt}]$$ where $v$ is the velocity, $D$ diffusion Coefficient and $t$ is time. You can check that at $x=0$ the concentration is zero.