Extended Data Fig. 3: Minimal model of mass cascade with scale selection. | Nature

Extended Data Fig. 3: Minimal model of mass cascade with scale selection.

From: Pattern formation by turbulent cascades

Extended Data Fig. 3

(a) In the model, two clusters Mn of size 2n−1 can merge into a cluster Mn+1 of size 2n. This aggregation process occurs with a rate \({k}_{n}^{+}\). Conversely, a cluster Mn+1 can split into two clusters Mn (except for monomers M1). This fragmentation process occurs with a rate \({k}_{n+1}^{-}\). (b) The rates \({k}_{n}^{\pm }\) of aggregation/fragmentation depend on the size of the cluster, so that (i) large clusters are more likely to fragment that small ones (blue curve) and (ii) small clusters are more likely to aggregate than large ones (red curve). (c-e) Equations (29) are numerically solved starting from the initial condition cn = 0 for all n. We have set \({\kappa }_{0}^{\pm }=2\) and \({\kappa }_{1}^{\pm }=1\), as well as Jin = Jout = 1. The number distribution cn is plotted in panel c, while the mass distribution ρn = 2n−1m0cn (normalized by its maximum value) is plotted in panel d. In panel e, we show the fluxes \({J}_{n}^{+}\) (red) and \({J}_{n}^{-}\) (blue) defined in Eqs. (31)–(32). The inset of panel e shows the total flux \({J}_{n}={J}_{n}^{+}+{J}_{n}^{-}\), which is constant away from the boundaries. (f) Entropy production in the mass cascade. We plot the rate of entropy production (computed within the mean-field model) as a function of the flux Jin through the system, in logarithmic scale. (As the distribution is stationary, there is a constant flux equal to the input flux). We have set \(N=15,{\kappa }_{0}^{\pm }=0.1,{\kappa }_{1}^{\pm }=1,{J}_{{\rm{out}}}=1\) and cn(t = 0) = 0 for all n. (g) Comparison between mean-field and Monte-Carlo simulations. The red curve (labelled ODE) shows the mean-field solution of Eq. (29), while the blue curve (labelled MC) shows the solution of the kinetic Monte-Carlo simulations (average plus or minus half a standard deviation over 1000 samples). We have set \(N=15,{\kappa }_{0}^{\pm }=0.1,{\kappa }_{1}^{\pm }=1,{J}_{{\rm{in}}}={J}_{{\rm{out}}}=1\) and cn(t = 0) = 0 for all n. (h) Equations (29) with no influx and outflux (Jin = Jout = 0) are numerically solved starting from the initial condition with only monomers cn(t = 0) = c0(0)δn,0, with different values of the number of monomers c0(0). We observe that (i) a peak in the steady-state distribution only arises when the initial number of monomers c0(0) is large enough and (ii) the position of the peak moves as c0(0) increases. We have set \({\kappa }_{0}^{\pm }=2\) and \({\kappa }_{1}^{\pm }=1\).

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