L-6

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

$x_{i}=1-\sigma _{i}$

The instability occurs when a block is at $x_{i}=0$ and is followed by its stabilization and a redistribution on all the blocks :

${\begin{cases}x_{i}=0\to x_{i}=\Delta &(stabilization)\\x_{j}\to x_{j}-{\frac {1}{L}}{\frac {\Delta }{1+m^{2}}}&(redistribution)\\\end{cases}}$

Dynamics

Our goal is thus to determine the distribution $P_{w}(x)$ of all blocks, given their intial distribution, $P_{0}(x)$, and a value of $w$. Let's decompose in steps the dynamics

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

$x_{i}\to x_{i}-m^{2}dw$

.

As a consequence

$P_{w+dw}(x)=P_{w}(x+m^{2}dw)\sim P_{w}(x)+m^{2}dw\partial _{x}P_{w}(x)$

• Stabilization : A fraction $m^{2}dwP_{w}(0)$ of the blocks is unstable. The stabilization induces the change $m^{2}dwP_{w}(0)\to m^{2}dwP_{w}(0)g(x)$. Hence, one writes

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

The stabilization of the unstable blocks induce a drop of the force per unit length

$m^{2}dwP_{w}(0)\int dxxg(x)=m^{2}dwP_{w}(0){\overline {\Delta }}$

\

• Redistribution This drop is (partially) compensated by the redistribution. The force acting on all points is increased:

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

Again, most of the distribution will be driven to instability while a fraction of the blocks 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:

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

• Determine $P_{\text{stat}}(0)={\frac {1}{\overline {\Delta }}}$ 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?

We consider an avalanche starting from a single unstable site $x_{0}=0$ and the sequence of sites more close to instabitity $x_{1}<x_{2}<x_{3}\ldots$. For each unstable block, all the blocks receive a random kick:

${\frac {\Delta _{1}}{(1+m^{2})L}},\quad {\frac {\Delta _{2}}{(1+m^{2})L}},\quad {\frac {\Delta _{3}}{(1+m^{2})L}},\ldots$

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 \Delta_1,\Delta_2,\Delta_3, \ldots } drwan from 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 g(\Delta) } Are these kick able to destabilize other blocks?

Given the initial condition and $w$, 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) } . From the extreme values theory we know 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

$x_{1}\sim {\frac {1}{LP_{w}(0)}},\;x_{2}\sim {\frac {2}{LP_{w}(0)}},\;x_{3}\sim {\frac {3}{LP_{w}(0)}},\ldots$

Hence we need to compare the mean value of the kick with the mean gap between nearest unstable sites:

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{\overline{\Delta}}{(1+m^2)L} \quad \text{versus }\quad \frac{1}{ P_w(0) L} }

Note 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 L } simplifies. We expect three possibilities:

• if the mean kick is smaller than the mean gap the system is subcritical and avalanches quickly stops.
• if the mean kick is equal to the mean gap the system is critical and avalanches are power law distributed
• if the mean kick is larger of the mean gap the system is super-critical and avalanches are unstable.

Note that in the stationary regime the ratio between mean kick and mean gap is $1/(1+m^{2})$. Hence, 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

$\eta _{1}={\frac {\Delta _{1}}{(1+m^{2})L}}-x_{1},\;\eta _{2}={\frac {\Delta _{2}}{(1+m^{2})L}}-(x_{2}-x_{1}),\;\eta _{3}={\frac {\Delta _{3}}{(1+m^{2})L}}-(x_{3}-x_{2})\ldots$ 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 $X_{0}=0$. 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

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

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 in average 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 } .

• Evolution equation: Consider the evolution up to the 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+dt} as a first evolution from 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} 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 dt} and a following evolution from 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 dt} 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 t+ dt} . 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}} }