Rate of probability loss from absorbing boundary

Question

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?

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.