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:

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 f_c= 1- \int_0^\infty d x x P_{\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

• 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_{\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

$x_{\text{kick}}={\frac {\overline {\Delta }}{(1+m^{2})L}}$

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 $\sim 1/P_{w}(0)$, 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 $m>0$ 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 } $X_{n}=\sum _{i=1}^{n}\eta _{i}\quad \quad {\text{with}}\;{\overline {\eta _{i}}}={\frac {\overline {\Delta }}{L(1+m^{2})}}-{\frac {1}{LP_{w}(0)}}$

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 $n$ 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 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 \tau=3/2}

• Stationary regime: Replacing 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 \frac{1}{LP_w(0)}} with 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 \frac{1}{LP_{\text{stat}}(0)} = \frac{\overline{\Delta}}{L} } we get 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 \; \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 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 \tau=3/2} until a cut-off

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

Bienaymé Galton Watson process

A time 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 t=0 } appears as infected individual which dies with a rate $a$ and branches with a rate $b$. On average, each infection generates 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 R_0 = b/a } new ones. Real epidemics corresponds to 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 R_0>1 } .

At time 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 t } , the infected population 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 n(t) } , while the total infected population 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 N(t) = \int_0^t n(t') d t' }

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

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_s(t)=\int_0^\infty P(N) e^{-s N} dN=\left\langle e^{-s\int_0^t n(t') dt'}\right\rangle }

. Note that the normalization imposes 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_0(t)=1 } .

• Backward approach: derive the following equation 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 Q_s(t)}

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_s(t+dt) = (1-(a+b) d t) e^{-s dt} Q_s(t) +a dt + b dt Q_s^2(t) +O(dt^2) }

which gives

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 \frac{d Q_s(t)}{d t}= -(a+b+s) Q_s(t)+a+ b Q_s^2(t) }

• Critical case: the stationary solution: Let's 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 b=a} 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 a=1} to recover the results of the mean field cellular automata. In the limit 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 t \to \infty} the total population coincides with the avalanche size, 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(t\to \infty) =S} . The Laplace transform of 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)} 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 0= -(2+s) Q_s^{\text{stat}}+1+ (Q_s^{\text{stat}})^2 }

which gives

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

with

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^\infty d s P(S) e^{-sS}= Q_s^{\text{stat}} }

• Critical case: Asymptotics: We want to predict the power law tail of the avalanche distribution 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) \sim A \cdot S^{-\tau} } . Taking the derivative with respect to 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 s } 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 A \int_0^\infty d s S^{1-\tau} e^{-sS}=\frac{1}{2 \sqrt{s}} }

and conclude that 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 \tau=3/2 } 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 A =\frac{1}{2 \int_0^\infty d z e^{-z}/\sqrt{z}}= \frac{1}{2 \sqrt{\pi}} }

Hence we find back our previous result

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