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

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 \sigma_i= h_{CM} - h_i + m^2(w-h_i), \quad }

.

• The local threshold are all equal. In particular we 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 \sigma_i^{th}=1, \quad \forall i }

.

As a consequence, 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 L\to \infty} , the statistical properties of the system are described by the distribution of the local stresses 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 \sigma_i } . For simplicity, instead of the stresses, we study the distance from threshold

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_i = 1-\sigma_i }

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

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 \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 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)} of all blocks, given their intial 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_0(x)} , and a value 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 w } . Let's decompose in steps the dynamics

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

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_i \to x_i - m^2 dw }

.

As a consequence

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+dw}(x) = P_w(x+m^2 dw) \sim P_w(x) + m^2 dw \partial_x P_w(x)}

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

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 \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

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^2 d w P_w(0) \int d x x g(x) = m^2 d w P_w(0) \overline{\Delta} }

\

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

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_i \to x_i - m^2 dw P_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

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 \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:

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 \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:

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 = \partial_x P_{\text{stat}}(x) + P_{\text{stat}}(0) g(x) }

• Determine 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}}(0) =\frac{1}{\overline{\Delta}} } using

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 1= \int_0^\infty dx \, P_{\text{stat}}(x)= - \int_0^\infty dx \, x \partial_x P_{\text{stat}}(x) }

• 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)= \frac{1}{\overline{\Delta}} \int_x^\infty g(z) d z }

which is well normalized.

Critical Force

The average distance from the threshold gives a simple relation for the critical force, namely 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 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} }
• 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- \overline{\Delta}}

Avalanches or instability?

We consider an avalanche starting from a single unstable site 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 } and the sequence of sites more close to instabitity 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< x_2<x_3\ldots } . For each unstable block, all the blocks receive a random 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 \frac{\Delta_1}{(1+m^2)L}, \frac{\Delta_2}{(1+m^2)L}, \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, \lodts } 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 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) } . 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

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 }

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 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 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

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 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 ${\frac {1}{LP_{w}(0)}}$ 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 in average $R_{0}=b/a$ 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 $Q_{0}(t)=1$.

• Evolution equation: Consider the evolution up to the time $t+dt$ as a first evolution from $0$ to $dt$ and a following evolution from $dt$ to $t+dt$. 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

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