L-6: Difference between revisions
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* <Strong> Instability : </Strong> The instability occurs when a point is at <math> x_i =0 </math> | * <Strong> Instability : </Strong> The instability occurs when a point is at <math> x_i =0 </math> and is followed by a stabilization: | ||
<center> <math> x_i =0 \to x_i = \Delta </math></center> with <math>\Delta </math> drawn from <math> g(\Delta) </math> | <center> <math> x_i =0 \to x_i = \Delta </math></center> with <math>\Delta </math> drawn from <math> g(\Delta) </math>. | ||
In our case, fraction <math> m^2 d w P_w(0) </math> of the blocks is unstable. The stabilization induces the change <math> m^2 d w P_w(0) g(x) </math>. Hence, one writes | |||
<center> <math> \partial_w P_{w}(x) \sim m^2 \left[\partial_x P_w(x) + P_w(0) g(x) \right] </math> </center> | <center> <math> \partial_w P_{w}(x) \sim m^2 \left[\partial_x P_w(x) + P_w(0) g(x) \right] </math> </center> | ||
* <Strong> Redistribution </Strong> The | * <Strong> Redistribution </Strong> The stabilization of a single block induces a redistribution where all blocks approach threshold. | ||
<center><math> | <center><math> | ||
x_i \to = x_i - \frac{1}{L} \frac{\Delta}{1+m^2} | x_i \to = x_i - \frac{1}{L} \frac{\Delta}{1+m^2} | ||
</math></center> | </math></center> | ||
The total | The total drop on the force acting on the unstable blocks <math> m^2 d w P_w(0) \int d x x g(x) = m^2 d w P_w(0) \overline{\Delta} </math> per unit length. | ||
This drop in partially compensated by the redistribution. The force acting on all points is increased of | |||
<center> <math> m^2 dw P_w(0) \frac{\overline{\Delta}}{1+m^2} </math> </center> | <center> <math> m^2 dw P_w(0) \frac{\overline{\Delta}}{1+m^2} </math> </center> | ||
Again, most of the distribution will be driven to instability while a franction of the blocks become unstable... we can write | |||
<center> <math>\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] </math> </center> | <center> <math>\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] </math> </center> | ||
and finally: | and finally: |
Revision as of 20:18, 8 March 2025
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
.
- The local threshold are all equal. In particular we set
.
As a consequence, in the limit , the statistical properties of the system are described by the distribution of the local stresses . For simplicity, instead of the stresses, we study the distance from threshold
Our goal is thus to determine their distribution , given their intial distribution, , and a value of .
Dynamics
Let's rewrite the dynamics with the new variables
- Drive: Increasing each point decreases its distance to threshold
.
As a consequence
- Instability : The instability occurs when a point is at and is followed by a stabilization:
with drawn from .
In our case, fraction of the blocks is unstable. The stabilization induces the change . Hence, one writes
- Redistribution The stabilization of a single block induces a redistribution where all blocks approach threshold.
The total drop on the force acting on the unstable blocks per unit length. This drop in partially compensated by the redistribution. The force acting on all points is increased of
Again, most of the distribution will be driven to instability while a franction of the blocks become unstable... we can write
and finally:
Stationary solution
Increasing the drive the distribution converge to the fixed point:
- Determine using
- Show
which is well normalized.
Critical Force
The average distance from the threshold gives a simple relation for the critical force, namely . Hence for the automata model we obtain:
Exercise:
Let's assume an exponential distribution of the thresholds and show
Avalanches or instability?
Given the initial condition and , the state of the system is described by . For each unstable block, all the blocks receive a kick. The mean value of the kick is
Is this kick able to destabilize another block? The equation setting the average position of the most unstable block is
Hence, for large systems we have
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 system is subcritical when and critical for
Mapping to the Brownian motion
Let's define the random jumps and the associated random walk
An avalanche is active until 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 . Under these hypothesis the Sparre-Andersen theorem state that the probability that the random walk remains positive for steps is independent on the jump disribution and for a large number of steps becomes . Hence, the distribution avalanche size is
This power law is of Gutenberg–Richter type. The universal exponent is
- 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
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 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 } and branches with a rate 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 } . 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 , 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
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:
. 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)}
which gives
- 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 is
which gives
with
- 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
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
Hence we find back our previous result