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?