Random walk of a fly with a receding electric fence

A similar but mathematically more challenging variant of the question has been asked here. I would like to propose a simpler version of the problem, where we consider a semi-infinite geometry instead of a bounded region, in the hope that it may be solved analytically. By solving the problem, I mean to calculate the probability that the particle hits the boundary.

Again, let's go back to our fly analogy, in which a dumb (meaning it has no sense of its surroundings, as opposed to intelligent walkers) one-dimensional fly acts as a random walker, and starts out at position $x_0 > 0$. At time $t=0$, a train sits still at the origin, and then recedes to the left with position described by $L(t) = -\alpha \sqrt{t}$, for $\alpha > 0$. To solve this problem, we would equivalently have to solve the diffusion equation

$ \frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2}, $

subject to $c(L(t), t) = 0, t>0$ and $c(x,0) = \delta(x-x_0)$. I would like to know if there is a clean solution to this problem.

NOTE 1: The diffusion equation can be solved in the semi-infinite geometry, with a static boundary. The concentration (probability density) of the particle can be derived with the method of images, by placing an anti-Gaussian at the mirrored position:

$ c(x,t) = \frac{1}{\sqrt{4 \pi D t}} \left[\operatorname{exp}\left(- \frac{(x-x_0)^2}{4Dt}\right) - \operatorname{exp}\left(- \frac{(x+x_0)^2}{4Dt}\right)\right], $

and the first-passage probability density $F(t)$ is given by the flux at $x=0$:

$ F(t) = D \left.\frac{\partial c}{\partial x}\right|_{x=0} = \frac{x_0}{\sqrt{4 \pi Dt^3}} \operatorname{exp}\left(- \frac{x_0^2}{4 D t}\right). $

Finally, the probability of first-passage at time $t$ is $ P(T \le t) = \int_0^t \mathrm{d}\tau\, F(\tau) = \operatorname{erfc}\left(\frac{x_0}{\sqrt{4Dt}}\right). $

NOTE 2: If $L(t) = -vt$, in which case the velocity of the train is constant, the problem can be solved rather elegantly by a method of images: for the static boundary, we place an image charge with amplitude $w$. This gives

$ c(x,t) = \frac{1}{\sqrt{4 \pi D t}} \left[\operatorname{exp}\left(- \frac{(x-x_0)^2}{4Dt}\right) - w \operatorname{exp}\left(- \frac{(x+x_0)^2}{4Dt}\right)\right]. $

Notice that for the concentration to vanish at $x=0$, we must have $w = \operatorname{exp} \left(\frac{x x_0}{Dt}\right) $, which can be interpreted as a moving boundary with $v = -x/t$. Then $w = e^{-v x_0/D}$, and we have the concentration for a linearly moving boundary diffusion problem:

$ c(x,t) = \frac{1}{\sqrt{4 \pi D t}} \left[\operatorname{exp}\left(- \frac{(x-x_0)^2}{4Dt}\right) - \operatorname{exp}\left(- \frac{(x+x_0)^2}{4Dt} - \frac{v x_0}{D} \right)\right]. $

Other quantities can then be calculated accordingly. The problem with this seems to be the fact that, for the $\sqrt{t}$ motion, the image would not be static.