The following is the solution to the 1D diffusion equation with diffusion coefficient D, initial particle position $x_0$, and a perfectly absorbing boundary at $x=0$ (s.t. $P(x=0)=0$).

$$ P(x;t)=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x-x_0)^2}{4 D t}} - \frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x+x_0)^2}{4 D t}} $$

If I understand correctly, for an $x_0>0$.

$$ P(\text{no collision at time $t$})=\int_0^\infty P(x;t) dx $$

In other words, the total probability in allowed space at time $t$ is exactly the probability that the particle never contacted the absorbing boundary at $x=0$ up to time $t$. What I want to compute is the rate of probability loss $k(t)$. From above, it seems that would be:

$$ k(t) = \frac{d}{dt} \int_0^\infty P(x;t) dx $$

evaluating with mathematica reveals: $$ k(t)=-\frac{D x_0}{2 \sqrt{\pi}} \left(\frac{1}{D t}\right)^{3/2} e^{-\frac{x_0^2}{4 D t}} $$

which seems reasonable.

It seems that there should be a way to compute $k(t)$ without computing the spatial integration across $x$, perhaps some computation involving only the boundary. I thought since all the loss occurs at $x=0$, the time derivative of $P(x;t)$ evaluated at $x=0$ should be k(t). However, the result of that calculation is 0.

Question: is there a way to compute k(t) without computing the spatial integral over the $x$ domain?

  • $\begingroup$ Huh, that's interesting, the functional form $P(x;t)=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x-x_0)^2}{4 D t}} - \frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x+x_0)^2}{4 D t}}$ almost reminds me of the method of image charges in electrostatics; it's as if the absorbing boundary is equivalent to there being a negative-valued mirror particle that diffused on the other side of the boundary. This makes me wonder if there's a clever combinatorial argument to show this... $\endgroup$ Mar 16, 2014 at 22:34
  • $\begingroup$ It is a method of images solution. Positive particle at $x_0$ and a negative image at $-x_0$ to impose the P(x=0)=0 boundary. $\endgroup$
    – vector07
    Mar 16, 2014 at 23:05

1 Answer 1


PART 1 (an unrelated alternate derivation of $P(x,t)$:

You can give a purely combinatorial derivation of the form of $P(x,t)$ based on the partitioning of random walks.

Let $x>0$, let the absorbing boundary be located at $x=0$, and let the initial source be located at $x_0>0$.

In the absence of an absorbing boundary, the amplitude $P_1(x,t)=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x-x_0)^2}{4 D t}}$ is due to random walks originating at $x_0$ and terminating at $x$. Let $S$ be the set of these random walks. Then $S$ can be partitioned as the disjoint union $$S=\overset{\infty}{\underset{j=0}{\bigcup}}S_j$$ where $S_j$ is the set of random walk which cross $x=0$ $j$ times.

In the presence of the absorbing boundary, the amplitude $P_2(x,t)$ is solely due to the contribution of paths $s\in S_0$. Thus, if it is possible to obtain an analytic form for $P_3(x,t)$ where $P_3(x,t)$ is the amplitude contributed by paths $s\in\bigcup_{j=1}^\infty S_j$, then we have $$P_2(x,t)=P_1(x,t)-P_3(x,t).$$

However, note that any path $s$ starting from $x_0$ that crosses the boundary $j$ times and terminates at $x$ has an associated path of equal travel distance $\overline{s}$ that crosses the boundary $j-1$ times that started at $-x_0$; to construct $\overline{s}$, simply take the mirror-image of the first portion of $s$ right up to the point it first encounters the boundary, and concatenate the remainder of $s$ to it.

Summing over the paths $s\in\bigcup_{j=1}^\infty S_j$ and replacing $s$ by $\overline{s}$, one then obtains $$P_3(x,t)=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x+x_0)^2}{4 D t}}$$ which then automatically implies that $$P(x;t)=\frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x-x_0)^2}{4 D t}} - \frac{1}{\sqrt{4 \pi D t}}e^{-\frac{(x+x_0)^2}{4 D t}}.$$

This is admittedly a pretty bizarre way of solving the problem, but it has the strange novelty that it involves no integrals or differential equation manipulation.

Part 2 (computing $k(t)$ without doing an integral):

By Fick's law, you have that the flux across the boundary is given by $$J=-D P^{(1,0)}(0,t)=-\frac{D x_0}{2 \sqrt{\pi}} \left(\frac{1}{D t}\right)^{3/2} e^{-\frac{x_0^2}{4 D t}}.$$

Since boundary flux is equal to rate of loss, you have $k(t)=J$ as desired.

  • $\begingroup$ This is very nice. How do you get k(t) from this though? That's what I was asking about. $\endgroup$
    – vector07
    Mar 16, 2014 at 23:43
  • $\begingroup$ Sorry, I accidentally thought it was $P(x,t)$ that you were interested in rather than $k(t)$ (was reading too fast and not paying attention). I'm not sure if it's possible to get that from the boundary though, but let me think about it. By the way, is there some reason why you're looking for an alternate method of computing $k(t)$? Are you attempting to do this for a problem where the integration has no symbolic solution? $\endgroup$ Mar 16, 2014 at 23:48
  • $\begingroup$ Your answer may not answer my specific question but it is a neat alternative explanation for the method of images. In most textbooks the method of images is proposed on geometric grounds. $\endgroup$
    – vector07
    Mar 16, 2014 at 23:51
  • $\begingroup$ You're right on for my intentions. The actual problem is much more complex. Since I have to numerically integrate the space, an alternative formulation where, say, I numerically integrate on the boundary should be much more efficient. $\endgroup$
    – vector07
    Mar 17, 2014 at 0:14
  • $\begingroup$ Apparently you can do it based on the boundary alone. So note that your initial method gives zero since $P(0,t)$ is always zero, and thus the time-derivative always vanishes, so it clearly can't be the right way to go. However, according to Fick's law, you can compute it using the first spatial derivative at the boundary without performing the integral. I'll append the computation to my answer. $\endgroup$ Mar 17, 2014 at 0:14

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