A problem in one-dimensional diffusion-limited aggregation (DLA) and positive recurrence of Markov chains

TL;DR

This paper studies the growth rate of the aggregate in a one-dimensional diffusion-limited aggregation model, showing it grows like the square root of time for low initial density and conjecturing linear growth for higher density.

Contribution

It provides a rigorous analysis of the growth rate of the DLA aggregate in one dimension and establishes the square root growth for low initial particle density, with a conjecture for linear growth at higher densities.

Findings

For initial mean density $<1$, $R(t)$ grows like $\,$ in $$.

The growth rate of $R(t)$ is of order $\,$ when $<1$.

Conjecture: $R(t)$ grows linearly with $t$ for sufficiently large $$.

Abstract

We consider the following problem in one-dimensional diffusion-limited aggregation (DLA). At time $t$, we have an "aggregate" consisting of $\Bbb{Z}\cap[0,R(t)]$ [with $R(t)$ a positive integer]. We also have $N(i,t)$ particles at $i$, $i>R(t)$. All these particles perform independent continuous-time symmetric simple random walks until the first time $t'>t$ at which some particle tries to jump from $R(t)+1$ to $R(t)$. The aggregate is then increased to the integers in $[0,R(t')]=[0,R(t)+1]$ [so that $R(t')=R(t)+1$] and all particles which were at $R(t)+1$ at time $t'{-}$ are removed from the system. The problem is to determine how fast $R(t)$ grows as a function of $t$ if we start at time 0 with $R(0)=0$ and the $N(i,0)$ i.i.d. Poisson variables with mean $\mu>0$. It is shown that if $\mu<1$, then $R(t)$ is of order $\sqrt{t}$, in a sense which is made precise. It is conjectured that $R(t)$ will grow linearly in $t$ if $\mu$ is large enough.