# inordinatum

Physics and Mathematics of Disordered Systems

## Extreme avalanches in the ABBM model

Extreme value statistics is becoming a popular and well-studied field. Just like the sums of random variables exhibit universality described by the central limit theorem, maxima of random variables also obey a universal distribution, the generalized extreme-value distribution. This is interesting for studying e.g. extreme events in finance (see this paper by McNeil and Frey) and climate statistics (see e.g. this physics paper in Science and this climatology paper).

Having refereed a related paper recently, I’d like to share some insights on the statistics of extreme avalanches in disordered systems. An example are particularly large jumps of the fracture front when slowly breaking a disordered solid. A simple but reasonable model for such avalanches is the Alessandro-Beatrice-Bertotti-Montorsi (ABBM) model, originally invented for describing Barkhausen noise. I already touched upon it in a previous blog post, and will focus on it again in the following.
I’ll show an exact formula for the distribution of the maximum avalanche size in an interval, and connect this result to the universal extreme value distribution when considering a large number of avalanches.

## Brief review: The ABBM model

To briefly recap (see my previous post for details), the ABBM model consists in a particle at position $u(t)$ pulled on a random landscape by a spring. A key assumption of the model is that the force resulting from the disordered potential is a Brownian Motion in $u$. This allows computing many observables exactly.
For example, when the force due to the spring increases by $f$, it is well-known (see e.g. arxiv:0808.3217 and the citations therein) that the resulting displacement $S$ of the particle follows the distribition

$\displaystyle P_f(S) = \frac{f}{2\sqrt{\pi} S^{3/2}}\exp\left[-\frac{(kS-f)^2}{4S}\right].$    (1)

Here, $k$ is the spring constant. For simplicity I took a Brownian random force landscape with variance $\sigma=1$ here, but the results are straightforward to generalize. This result is basically the distribution of the first-passage time of a Brownian motion with drift $k$ at a given level $f$. In this context it is also known as the Bachelier-Levy formula (see also my post on first-passage times).
For small forces, $f \to 0$, and weak springs, $k \to 0$, (1) becomes a power-law distribution $P(S) \sim S^{-3/2}$.

## The largest avalanche in the ABBM model

Now let us consider particularly large avalanches. When applying a force $f$, the probability to have a total displacement $S \leq S_0$ is

$\displaystyle F^{tot}_f(S_0) := \int_0^{S_0}\mathrm{d}S\,P_f(S).$    (2)

Note, however, that the total displacement in this case is the sum of many avalanches triggered by infinitesimal steps in $f$. So how do we obtain the probability $F_f(S_0)$ that all avalanches triggered during this ramp-up of the force are smaller than $S_0$? We decompose it in $n$ intervals of size $f/n$, and let $n \to \infty$:

$\displaystyle F_f(S_0) = \lim_{n\to\infty} \left[F^{tot}_{f/n}(S_0)\right]^n.$    (3)

Note that we use here the Markovian property of the Brownian Motion, which ensures that avalanches on disjoint intervals are independent. Only this property permits us to take the $n$-th power to obtain the cumulative distribution for the $n$ intervals; on any non-Markovian landscape the avalanches in these intervals would be correlated and things would be much more complicated.

Combining equations (1), (2) and (3) we can compute $F_f(S_0)$ explicitly:

$\begin{array}{rl} F_f(S_0) =& \lim_{n\to\infty} \left[F^{tot}_{f/n}(S_0)\right]^n = \exp\left[-f\partial_f\big|_{f=0}\int_{S_0}^{\infty}\mathrm{d}S\,P_f(S)\right] \\ =& \displaystyle \exp\left[-f\left(\frac{e^{-k^2 S_0/4}}{\sqrt{\pi S_0}} - \frac{k}{2}\text{erfc} \frac{k\sqrt{S_0}}{2}\right)\right]. \end{array}$    (4)

This satisfies the normalization expected of a cumulative distribution function: For $S_0 \to 0$, $F_f(S_0)\to 0$ and for $S_0 \to \infty$, $F_f(S_0)\to 1$.
The probability distribution of the maximal avalanche size $S_{max}$ is correspondingly

$\displaystyle P_f(S_{max}) = \partial_{S_0}\big|_{S_0=S_{max}}F_f(S_0).$    (5)

Eqs. (4) and (5) are a nice closed-form expression for the size distribution of the largest avalanche during the force increase by $f$!

## From few to many avalanches

As one goes to infinitesimal force steps $f \to 0$, only a single avalanche is triggered. Then it is clear from our construction and eqs. (4), (5) that $P_f(S_{max}) \to P_f(S)$ as defined in (1). So, as expected, when considering a single avalanche the maximal avalanche size and the total avalanche size coincide.

On the other hand, for large force steps $f \to \infty$, the ramp-up of the force triggers many independent avalanches. The total displacement $S$ is then the sum of many independent avalanche sizes $S_1...S_n$. Thus, by the central limit theorem, one expects to find a Gaussian distribution for $S$. We can see this explicitly from (1): The expectation value is $\left\langle S \right\rangle = f/k$, and the fluctuations around it are $\delta S := S-\left\langle S \right\rangle$. From (1) one finds that they scale like $\delta S \sim \sqrt{f/k^3}$. The normalized fluctuations $d := \delta S \sqrt{k^3/f}$ have the distribution

$\displaystyle P(d) = \frac{1}{2\sqrt{\pi}}\exp\left(-\frac{d^2}{4}\right) + \mathcal{O}(f^{-1/2}).$     (6)

For large $f$, we are indeed left with a Gaussian distribution for the normalized fluctuations $d$. This is easily checked numerically, see figure 1 below.

Figure 1: Distribution of the total displacement $S$ for increasing force steps $f$, as given in eq. (1). Observe how the distribution converges to the Gaussian limit eq. (6) indicated by black dotted lines.

So what happens with the maximal avalanche size $S_{max}$ for large steps $f \to \infty$? $S_{max}$ is now the maximum of many independent avalanche sizes $S_1...S_n$, and as mentioned in the introduction we expect its distribution to be a universal extreme value distribution.
Since only large $S_0$ are relevant in (4), the exponent can be approximated by

$\displaystyle \frac{e^{-k^2 S_0/4}}{\sqrt{\pi S_0}} - \frac{k}{2}\text{erfc} \frac{k\sqrt{S_0}}{2} \approx \frac{4}{k^2 \sqrt{\pi} S_0^{3/2}}e^{-k^2 S_0/4}.$    (7)

Inserting this back into (4), we see that for large $f$ the distribution $P_f(S_{max})$ is centered around $S_f := \frac{4}{k^2} \log f$. The cumulative distribution function (4) is well approximated by

$\displaystyle F_f(S_{max}) \approx \exp\left[- \frac{2}{k^2 \sqrt{\pi} S_f^{3/2}}e^{-k^2 (S_{max}-S_f)/4} \right].$    (8)

This is, up to rescaling, the cumulative distribution function of the Gumbel extreme-value distribution $e^{-e^{-x}}$. It is easy to check this universal asymptotic form numerically, see figure 2 below. Note that here the convergence here is much slower than for the total displacement shown in figure 1, since the typical scale for $S_{max}$ only grows logarithmically with $f$.

Figure 2: Distribution of the size of the largest avalanche $S_{max}$ for force steps of increasing size $f$, as given by eq. (5). Observe how the distribution converges to the universal Gumbel limit in eq. (8) indicated by black dotted lines.

For some applications, the limit of a very soft spring, $k \to 0$, is important. I leave the details to the reader but the main picture is that the exponential decay for large $S_0$ in eq. (6) is replaced by a power law $S_0^{-1/2}$. Correspondingly, the universal extreme-value distribution observed for large force steps $f$ is no longer the Gumbel distribution (8) but instead a Fréchet distribution.

## Side note: The minimal avalanche size

One may be tempted to approach similarly the problem of the minimal avalanche size for a slow ramp-up of the applied force. However, this is not well-defined: Due to the roughness of the Brownian force landscape, as we increase the force more and more slowly, the size of the smallest avalanche decreases more and more. Hence, its distribution will always be discretization-dependent and will not yield a finite result such as eq. (4) in the continuum limit.

All this gives a consistent picture of the maximal avalanche in the ABBM model. I find it really nice that it is so simple, knowing the avalanche size distribution (1), to express the distribution of the size of the largest avalanche in closed form and understand how it behaves!

Written by inordinatum

August 19, 2014 at 9:53 pm