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 Failed to parse (unknown function "\dd"): {\displaystyle w \to w + \dd w} each point decreases its distance to threshold

Failed to parse (unknown function "\dd"): {\displaystyle x_i \to x_i - m^2 \dd w }

.

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 Failed to parse (unknown function "\dd"): {\displaystyle w \to w + \dd w} , a fraction $m^{2}dw{P}_w(0)$ of the points of the interface is unstable. Due to the stress drop, their distance to instability will be $m^{2}dw{P}_w(0)g(x)$. Hence, one writes

${\partial }_w{P}_w(x)\sim m^2[{\partial }_x{P}_w(x)+P_w(0)g(x)]$ $\{\begin{array}{l}{\sigma }_i={\sigma }_i-\mathrm{\Delta }\text{stress drop}\\ \\ {\sigma }_{i\pm 1}={\sigma }_{i\pm 1}+\frac{1}{2}\frac{\mathrm{\Delta }}{1+m^2}\text{stress redistribution}\\ \end{array}$

Note that $\mathrm{\Delta }$ is a positive random variable drwan from $g(\mathrm{\Delta })$.

Hence, an unstable point, $x_i<0$, is stabilized to a value $x>0$ drawn from $g(x)$. The stress redistribution induced on each bloch is $\frac{1}{L}\frac{x}{1+m^2}$

we define  and write its evolution equation. 

• Drive: Changing $w\to w+dw$ gives
• Instability: This shift is stable far from the origin, however for a fraction $m^{2}dw{P}_w(0)$ of the points of the interface is unstable. Due to the stress drop, their distance to instability will be $m^{2}dw{P}_w(0)g(x)$. Hence, one writes

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

• Stress redistribution: as a consequence all points move to the origin of

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

where $\bar{x}=\int dxxg(x)$

• Avalanche: Let us call 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

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