L-6

Avalanches and Bienaymé-Galton-Watson process

Goal: We solve the mean field version of the cellular automaton, derive its avalanche statistics and make a connection with the Bienaymé-Galton-Watson process used to describe an epidemic outbreak.

Fully connected (mean field) model for the cellular automaton

Let's study the mean field version of the cellular automata introduced in the previous lecture. We introduce two approximations:

• Replace the Laplacian, which is short range, with a mean field fully connected interction

${\sigma }_i=h_{CM}-h_i+m^2(w-h_i),$

.

• The local threshold are all equal. In particular we set

${\sigma }_i^{th}=1,\mathrm{\forall }i$

.

As a consequence, in the limit $L\to \mathrm{\infty }$, the statistical properties of the system are described by the distribution of the local stresses ${\sigma }_i$. For simplicity, instead of the stresses, we study the distance from threshold

$x_i=1-{\sigma }_i$

Our goal is thus to determine their distribution $P_w(x)$, given their intial distribution, $P_0(x)$, and a value of $w$.

Dynamics

Let's rewrite the dynamics with the new variables

• Drive: Increasing $w\to w+dw$ each point decreases its distance to threshold

$x_i\to x_i-m^{2}dw$

.

As a consequence

$P_{w+dw}(x)=P_w(x+dw)\sim P_w(x)+m^{2}dw{\partial }_x{P}_w(x)$

• Instability 1: Stress drop The instability occurs when a point is at $x_i=0$. Then, the point is stabilized (stress drop):

$x_i=0\to x_i=\mathrm{\Delta }$

Increasing $w\to w+dw$, a fraction $m^{2}dw{P}_w(0)$ of the blocks is unstable. Due to the stress drop, their distance to threshold becomes $m^{2}dw{P}_w(0)g(x)$. Hence, one writes

${\partial }_w{P}_w(x)\sim m^2[{\partial }_x{P}_w(x)+P_w(0)g(x)]$

• Instability 2: Stress redistribution The stress drop of a single block induces a stress redistribution where all blocks approach threshold.

$x_i\to =x_i+\frac{1}{L}\frac{\mathrm{\Delta }}{1+m^2}$

The total stress drop is $m^{2}dw{P}_w(0)\int dxxg(x)=m^{2}dw{P}_w(0)\bar{x}$ hence all points move to the origin of

$m^{2}dw{P}_w(0)\frac{\bar{x}}{1+m^2}$

part of them shifts, part of them become unstable... we can write

${\partial }_w{P}_w(x)=m^2[{\partial }_x{P}_w(x)+P_w(0)g(x)][1+P_w(0)\frac{\bar{x}}{1+m^2}+(P_w(0)\frac{\bar{x}}{1+m^2}{)}^2+\dots ]$

and finally:

${\partial }_w{P}_w(x)=\frac{m^2}{1-P_w(0)\frac{\bar{x}}{1+m^2}}[{\partial }_x{P}_w(x)+P_w(0)g(x)]$

Stationary solution

Increasing the drive the distribution converge to the fixed point:

$0={\partial }_x{P}_{\text{stat}}(x)+P_{\text{stat}}(0)g(x)$

• Determne $P_{\text{stat}}(0)=\frac{1}{\bar{x}}$ using

$1=\underset{0}{\overset{\mathrm{\infty }}{\int }}dx{P}_{\text{stat}}(x)=-\underset{0}{\overset{\mathrm{\infty }}{\int }}dxx{\partial }_x{P}_{\text{stat}}(x)$

• Show

$P_{\text{stat}}(x)=\frac{1}{\bar{x}}\underset{x}{\overset{\mathrm{\infty }}{\int }}g(z)dz$