Consider the process of periodic 1D crystallization, where multiple sites initiate crystallizing waves at random with speed $v$. The fraction of the crystallized substance, at position $x$ and time $t$, is given by $$ f(x,t)=1-e^{-\int_{V_X(v)}p(Y) \,dY} $$ where $p(x,t)$ is the crystallization initiation rate at position $x$ and time $t$, and $V_X(v)$ is the past light-cone of the spacetime point $X=(x,t)$ (see figure below, where $x_\pm = x\mp vt$). My goal is to find an expression for $p$ in terms of the expected time of crystallization at each space point, which can be given by $$ t_E(x)=\int_0^\infty t \frac{\partial f(x,t')}{\partial t'}|_{t'=t} \,dt $$
Some ideas: From this paper (in the context of DNA replication, analogous to the process of 1D crystallization, see also this publication and chapter II of this thesis), defining the fraction of uncrystallized substance $s(x,t)=1-f(x,t)$, $p$ is shown to satisfy $$ p(x,t)=-\frac{v}{2}\square \log(s(x,t)), $$ where $\square =\frac{1}{v^2}\partial_t^2-\partial_x^2$ is the d'Alembert operator. Could I use this to write it in terms of the expected time of crystallization at each space point? I was thinking that, given the shape of $f$, perhaps a Laplace Transform could be used to rewrite the fraction $s$ in terms of the expected time at each space point, which would be our data.
A simple example: When $p(x,t)=p$ is constant in spacetime, we have $f(x,t)=1-e^{-pvt^2} $ and thus $$ t_E(x)=2pv\int_0^\infty t^2 e^{-pv t^2}\,dt=\frac12\sqrt{\frac{\pi}{pv}} $$ which can easily be inverted. Any ideas for the general case?
This is just the derivation of the top equation, essentially.
Say the nucleation rate density (with dimension $L^{-1} T^{-1}$) is $I(x,t)$. Therefore the rate of a nucleation in $[a,b]$ at time $t$ is $R(a,b,t)=\int_a^b I(y,t) dy$. This has dimension $T^{-1}$.
Now let us consider an easier problem: the probability $Q(a,b,t)$ that a nucleation occurred in a fixed interval $[a,b]$ within time $t$ satisfies $\frac{\partial Q}{\partial t}(a,b,t)=R(a,b,t) (1-Q(a,b,t))$. The derivation of that is simple enough:
$$P(\text{nucleation within time } t+h) \\ =P(\text{nucleation within time } t)+P(\text{nucleation within time } t+h \mid \text{no nucleation within time } t) \\ \cdot P(\text{no nucleation within time } t) \\ Q(a,b,t+h)=Q(a,b,t)+hR(a,b,t)(1-Q(a,b,t))+o(h).$$
Hence
\begin{align}-\ln(1-Q(a,b,t)) & =\int_0^t R(a,b,s) ds \\ Q(a,b,t) & =1-e^{-\int_0^t R(a,b,s) ds} \\ & =1-e^{-\int_0^t \int_a^b I(y,s) dy ds}. \end{align}
To make the adjustment to this more complicated setting, you have to recompute that conditional probability, but everything else is the same. A close enough nucleation within time $t+h$ given that there were no close enough nucleations within time $t$ means that the nucleation occurred at a site $y \in [x-vt-vh,x+vt+vh]$ at a time $s$ such that $s \in [\max \{ 0,t-|y-x|/v \},t+h-|y-x|/v]$. The resulting figure in the spacetime diagram is a pair of trapezoids. If we neglect the two spacetime triangles on the far left and right which have area $O(h^2)$, then the entire region of interest has time width $h$, so the analogue of $hR(a,b,t)$ is $h \int_{x-vt}^{x+vt} I(y,t-|y-x|/v) dy$.
So the analogous equation for this setting is
$$P(x,t)=1-e^{-\int_0^t \int_{x-vs}^{x+vs} I(y,s-|y-x|/v) \, dy ds}$$
and the initial data is $P(x,0) \equiv 0$. You can strip out most of the nonlinearity by replacing the given quantity $P$ with $S:=-\ln(1-P)$, in which case you have
$$S=\int_0^t \int_{x-vs}^{x+vs} I(y,s-|y-x|/v) dy ds$$
and again $S(x,0) \equiv 0$.
From there I suppose you can derive that wave equation for $S$ with $I$ as forcing, which just solves the inverse problem completely, provided you have $S$. But with less information than that, you need to artificially create some additional information through something like an ansatz, otherwise you're stuck.
Within this formalism, if you have bare rates at a discrete set of sites $Y$ and no rate density otherwise, that amounts to having $I(x,t)=\sum_{y \in Y} \delta(x-y) f(y,t)$.