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

$\sigma _{i}=h_{CM}-h_{i}+m^{2}(w-h_{i}),\quad$

.

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

$\sigma _{i}^{th}=1,\quad \forall i$

.

As a consequence, in the limit $L\to \infty$, 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):

$x_{i}=0\to x_{i}=\Delta$

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

$\partial _{w}P_{w}(x)\sim m^{2}\left[\partial _{x}P_{w}(x)+P_{w}(0)g(x)\right]$

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

$x_{i}\to =x_{i}-{\frac {1}{L}}{\frac {\Delta }{1+m^{2}}}$

The total stress drop is $m^{2}dwP_{w}(0)\int dxxg(x)=m^{2}dwP_{w}(0){\overline {\Delta }}$ hence all points move to the origin of

$m^{2}dwP_{w}(0){\frac {\overline {\Delta }}{1+m^{2}}}$

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

$\partial _{w}P_{w}(x)=m^{2}\left[\partial _{x}P_{w}(x)+P_{w}(0)g(x)\right]\left[1+P_{w}(0){\frac {\overline {\Delta }}{1+m^{2}}}+(P_{w}(0){\frac {\overline {\Delta }}{1+m^{2}}})^{2}+\ldots \right]$

and finally:

$\partial _{w}P_{w}(x)={\frac {m^{2}}{1-P_{w}(0){\frac {\overline {\Delta }}{1+m^{2}}}}}\left[\partial _{x}P_{w}(x)+P_{w}(0)g(x)\right]$

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 $P_{\text{stat}}(0)={\frac {1}{\overline {\Delta }}}$ using

$1=\int _{0}^{\infty }dx\,P_{\text{stat}}(x)=-\int _{0}^{\infty }dx\,x\partial _{x}P_{\text{stat}}(x)$

• Show

$P_{\text{stat}}(x)={\frac {1}{\overline {\Delta }}}\int _{x}^{\infty }g(z)dz$

which is well normalized.

Critical Force

The average distance from the threshold gives a simple relation for the critical force, namely $1-f_{c}={\overline {x}}$. Hence for the automata model we obtain:

$f_{c}=1-\int _{0}^{\infty }dxxP_{\text{stat}}(x)=1-{\frac {1}{2}}{\frac {\overline {\Delta ^{2}}}{\overline {\Delta }}}$

Exercise:

Let's assume an exponential distribution of the thresholds and show

• $P_{\text{stat}}(x)=e^{-x/{\overline {\Delta }}}/{\overline {\Delta }}$
• $f_{c}=1-{\overline {\Delta }}$

Avalanches or instability?

Given the initial condition and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w } , the state of the system is described by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_w(x) } . For each unstable block, all the blocks receive a kick. The mean value of the kick is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{\text{kick}}= \frac{\overline{\Delta}}{(1+m^2)L} }

Is this kick able to destabilize another block? The equation setting the average position of the most unstable block is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{x_1} P_w(t) dt =\frac{1}{L} }

Hence, for large systems we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_1 \sim \frac{1}{L P_w(0)}, \; x_2 \sim \frac{2}{L P_w(0)}, \; x_3 \sim \frac{3}{L P_w(0)}, \ldots }

We expect three possibilities:

• if the mean kick, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim \overline{\Delta}/(1+m^2) } is smaller than the mean gap Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim 1 /P_w(0)} , the system is subcritical and avalanches quickly stops.
• if the mean kick, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim \overline{\Delta}/(1+m^2) } is equal to the mean gap Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim 1 /P_w(0)} , the system is critical and avalanches are power law distributed
• if the mean kick, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim \overline{\Delta}/(1+m^2) } is larger of the mean gap Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim 1 /P_w(0)} , the system is super-critical and avalanches are unstable.

Note that in the stationary regime the system is subcritical when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m>0 } and critical for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=0 }

Mapping to the Brownian motion

Let's define the random jumps and the associated random walk

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta_1 = \frac{\Delta_1}{(1+m^2)L}- x_1, \; \eta_2=\frac{\Delta_2}{(1+m^2)L}- (x_2-x_1), \; \eta_3=\frac{\Delta_3}{(1+m^2)L}- (x_3-x_2) \ldots } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_n= \sum_{i=1}^n \eta_i \quad \quad \text{with} \; \overline{\eta_i} = \frac{\overline{\Delta}}{L(1+m^2)} -\frac{1}{LP_w(0)} }

An avalanche is active until Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_n } is positive. Hence, the size of the avalanche identifies with first passage time of the random walk.

• Critical case : In this case the jump distribution is symmetric and we can set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_0=0} . Under these hypothesis the Sparre-Andersen theorem state that the probability that the random walk remains positive for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} steps is independent on the jump disribution and for a large number of steps becomes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(n) \sim \frac{1}{\sqrt{\pi n}}} . Hence, the distribution avalanche size is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(S)= Q(S)-Q(S+1) \sim \frac{1}{\sqrt{\pi S}} -\frac{1}{\sqrt{\pi (S+1)}} \sim \frac{1}{2 \sqrt{\pi}}\frac{1}{S^{3/2}} }

This power law is of Gutenberg–Richter type. The universal exponent is $\tau =3/2$

• Stationary regime: Replacing ${\frac {1}{LP_{w}(0)}}$ with ${\frac {1}{LP_{\text{stat}}(0)}}={\frac {\overline {\Delta }}{L}}$ we get $\;{\overline {\eta _{i}}}\sim -{\frac {m^{2}}{1+m^{2}}}{\frac {\overline {\Delta }}{L}}$. For small m, the random walk is only sliglty tilted. The avalanche distribution will be power law distributed with $\tau =3/2$ until a cut-off

$S_{\max }\sim m^{-4}$

Bienaymé Galton Watson process

A time $t=0$ appears as infected individual which dies with a rate $a$ and branches with a rate $b$. On average, each infection generates $R_{0}=b/a$ new ones. Real epidemics corresponds to $R_{0}>1$.

At time $t$, the infected population is $n(t)$, while the total infected population is

$N(t)=\int _{0}^{t}n(t')dt'$

Our goal is to compute $P(N(t))$ and we introduce its Laplace Transform:

$Q_{s}(t)=\int _{0}^{\infty }P(N)e^{-sN}dN=\left\langle e^{-s\int _{0}^{t}n(t')dt'}\right\rangle$

. Note that the normalization imposes $Q_{0}(t)=1$.

• Backward approach: derive the following equation for $Q_{s}(t)$

$Q_{s}(t+dt)=(1-(a+b)dt)e^{-sdt}Q_{s}(t)+adt+bdtQ_{s}^{2}(t)+O(dt^{2})$

which gives

${\frac {dQ_{s}(t)}{dt}}=-(a+b+s)Q_{s}(t)+a+bQ_{s}^{2}(t)$

• Critical case: the stationary solution: Let's set $b=a$ and $a=1$ to recover the results of the mean field cellular automata. In the limit $t\to \infty$ the total population coincides with the avalanche size, $N(t\to \infty )=S$. The Laplace transform of $P(S)$ is

$0=-(2+s)Q_{s}^{\text{stat}}+1+(Q_{s}^{\text{stat}})^{2}$

which gives

$Q_{s}^{\text{stat}}={\frac {(2+s)-{\sqrt {s^{2}+4s}}}{2}}\sim 1-{\sqrt {s}}+O(s)$

with

$\int _{0}^{\infty }dsP(S)e^{-sS}=Q_{s}^{\text{stat}}$

• Critical case: Asymptotics: We want to predict the power law tail of the avalanche distribution $P(S)\sim A\cdot S^{-\tau }$. Taking the derivative with respect to $s$ we have

$A\int _{0}^{\infty }dsS^{1-\tau }e^{-sS}={\frac {1}{2{\sqrt {s}}}}$

and conclude that $\tau =3/2$ and

$A={\frac {1}{2\int _{0}^{\infty }dze^{-z}/{\sqrt {z}}}}={\frac {1}{2{\sqrt {\pi }}}}$

Hence we find back our previous result

$P(S)\sim {\frac {1}{2{\sqrt {\pi }}}}{\frac {1}{S^{3/2}}}$