# 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

## 2 Answers

An alternative could be to use separable solutions, that will give solutions which are exponentially decaying in time and trigonometric functions in space. The decay time is identified with the eigenvalues of the spatial part, in accordance with the absorbing boundary at x = 0.

Here's one way to solve it analytically:

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 $v>0$, $$G_{absorbing}(t-t',v,v') = G_0(t-t',v,v')-G_0(t-t',-v,v').$$