# 1D drift-diffusion equation with single absorbing boundary

If we have just the simple diffusion equation (in 1D): $$\frac{\partial P(x,t)}{\partial t} = D \frac{\partial^2 P(x,t)}{\partial x^2}$$ with an absorbing boundary at x=0 and initial condition $P(x,0) = \delta(x-x_0)$, we can use the method of images to get the solution $$P(x,t) = \frac{1}{\sqrt{4 \pi D t}}e^{\frac{-(x-x_0)^2}{4 D t}} - \frac{1}{\sqrt{4 \pi D t}}e^{\frac{-(x+x_0)^2}{4 D t}}.$$ However I am interested in solving this in the case where there is also a drift (ultimately one that is not constant in time, but to start with just a solution with a constant drift velocity would be great). I haven't been able to find anything about this problem, does anyone have any ideas?

• What is causing the drift term? How would the changing charge density impact the drift term? Why would the drift velocity remain constant? – Jon Custer Jul 22 '15 at 23:33
• Sorry to be clear I mean I want to solve the drift-diffusion equation with an absorbing boundary at x=0, so $\frac{\partial P(x,t)}{\partial t} = D \frac{\partial^2 P(x,t)}{\partial x^2} - \frac{\partial (v P(x,t))}{\partial t}$. Ultimately I want a solution for general v(t), but I think just for a constant v would be a good starting point – Henry Jul 23 '15 at 0:06
• You likely will not find an analytic solution, you'll have to do it numerically. – Kyle Kanos Jul 23 '15 at 1:42
• In the unlikely event this is ever useful to anyone: the case where $v(t)=v$, i.e. a constant, can be solved exactly - see chapter 3 of Sidney Redner's book on first passage processes. The general $v(t)$ case is much harder however and I wasn't able to obtain an exact solution, but I was able to derive an integral equation for the first passage density. – Henry Sep 17 '17 at 20:40

The diffusion operator with a drift term is diagonalized by Hermite polynomials. Scaling the drift-diffusion equation gives \begin{equation} {\partial \over \partial t} f(t,v) = {\partial \over \partial v} \left(v+{\partial \over \partial v} \right) f(t,v), \end{equation} Expanding $f(t,v)$ in terms of (probabilist's) Hermite polynomials \begin{equation} f(t,v) = \sum_{n=0}^\infty f_n(t) e^{-v^2/2} \mathit{He}_n(v), \end{equation} and applying the operator ${1 \over \sqrt{2 \pi} n!} \int_{-\infty}^\infty dv\; \mathit{He}_n(v)$, to both sides of the first equation, gives $${\partial \over \partial t} f_n(t) = -n f_n(t), ~~~~~~~ \Rightarrow~~~~~~~ f_n(t) = f_n(0) e^{-nt}.$$ Using $$f_n(t) = {1 \over \sqrt{2 \pi} n!} \int_{-\infty}^\infty dv\; f(t,v) \mathit{He}_n(v),$$ allows one to write \begin{equation} f(t,v) = {e^{-v^2/2} \over \sqrt{2 \pi}} \int_{-\infty}^\infty dv' \; \left[ \sum_{n=0}^\infty {e^{-nt} \over n!} \mathit{He}_n(v) \mathit{He}_n(v') \right] f(0,v'). \end{equation} Now, the Kernel in the above equation can be identified as the Mehler kernel \begin{equation} \sum_{n=0}^\infty \frac{\rho^n}{n!} ~ \mathit{He}_n(x)\mathit{He}_n(y) = \frac 1{\sqrt{1-\rho^2}}\exp\left(-\frac{\rho^2 (x^2+y^2)- 2\rho xy}{2(1-\rho^2)}\right), \end{equation} which allows the propagator solution to the drift-diffusion equation to be written explicitly as $$f(t,v) = \int_{-\infty}^\infty dv' \; \left\lbrace {1 \over \sqrt{2 \pi(1-e^{-2t})}} \exp \left[ - {(v-v'e^{-t})^2 \over 2(1-e^{-2t})} \right] \right\rbrace f(0,v').$$
Now, we can define the free-space Green's function $$G_{0}(t-t',v,v') = {\Theta(t-t') \over \sqrt{2 \pi(1-e^{-2(t-t')})}} \exp \left[ - {(v-v'e^{-(t-t')})^2 \over 2(1-e^{-2(t-t')})}\right].$$
To enforce an absorbing boundary condition at $v = 0$, define the new Green's function for $v>0$, $$G_{absorbing}(t-t',v,v') = G_0(t-t',v,v')-G_0(t-t',-v,v').$$