OpenYear of origin: 1992
Posted online: 2018-08-08 16:39:12Z by Henrik Shahgholian99
Cite as: P-180808.2
Assume we have $n$ identical particles at the origin of the lattice $\mathbb Z^d$ ($d\geq 2$). One by one each of these particles starts a simple symmetric random walk from the origin (with walks being independent of each other), until reaching an unoccupied site (vertex) on $\mathbb Z^d$. Then the particle stops, and never moves from there.
All particles follow this process, until we get at most 1 particle hosted by any vertex of $\mathbb Z^d$. One can prove, that this process, which is called internal diffusion limited aggregation (IDLA for short), with probability 1 will reach a stable state.
Let $D_n\subset \mathbb Z^d$ be the set of occupied vertices of $\mathbb Z^d$ for $n$ particles at the origin. It is proved in [1] that as $n \to \infty$, the set $n^{-1/d} D_n$ converges to a ball with probability 1; see [1] for the precise statement on existence of the scaling limit of IDLA. For more general point sources, see [2].
Here I shall describe two problems related to this model, on wedge shaped domains, where the boundary of the domains play the role of absorbents, and particles reaching there will disappear from the system.
Consider again an IDLA on $$D:=\{ x_1 > k |x_2|\} \cap \frac{1}{n} \mathbb Z^2, \qquad \hbox{for some } k > 0.$$ Observe that now $ \mathbf 0 \notin D$.
Governing rules:
i) Initially all sites within $D$ are filled with one particle each.
ii) Particles walking from $z= (1,0)$ are allowed to walk out from $D$ and occupy new sites.
Question:How many particles $N=N(k,n)$ (in average) do we need to run before we can occupy the origin?
The related Hele-Shaw problem (in the continuous setting) says that there is a delay time for the flow (injected at $(1,0$)) before making the flow at the origin to move, see [4].
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Edited: (general update ) at 2018-08-08 18:02:37Z
Created at: 2018-08-08 16:39:12Z View this version
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