Noise-Induced Bistable States and Their Mean Switching Time in Foraging Colonies

We investigate a type of bistability occurring in population systems where noise not only causes transitions between stable states, but also constructs the states themselves. We focus on the experimentally well-studied system of ants choosing between two food sources to illustrate the essential points, but the ideas are more general. The mean time for switching between the two bistable states of the system is calculated. This suggests a procedure for estimating, in a real system, the critical population size above which bistability ceases to occur.

Figure 1

Snapshot of the time series for $z$, obtained with stochastic simulations of the scheme of reactions (1). Parameter values are $\epsilon =1/500$ and $N=250$. Time is expressed in units of $\tau =2\epsilon t/N$.

Figure 2

Equation (5) (solid colors) is compared against simulations of scheme (1) (symbols). Simulations are obtained by taking the normalized histogram of a time series of length $\tau =2.5\times 10^9$. We have used $\epsilon =10^{-3}$ and $N=1500$ (blue line, triangles), $N=N_c\equiv 1000$ (red line, squares), and $N=500$ (purple line, circles).

Figure 3

Equation (6) (solid lines) is compared against stochastic simulations (symbols). Parameters used are $\epsilon =1/50$ (blue line, triangles), $\epsilon =1/100$ (green line, squares), and $\epsilon =1/2000$ (red line, circles). Each symbol has been obtained by averaging over 500 simulations.

Figure 4

Equation (6) (solid lines) is compared against stochastic simulations of the mean time for $\epsilon =1/100$ (circles). Each circle has been obtained from 2500 averages. Parameters used for the analytical formulas are $\epsilon =1/2000$ (red line), $\epsilon =10^{-5}$ (purple line), $\epsilon =10^{-10}$ (green line), and $\epsilon =10^{-15}$ (blue line).

Figure 5

Solid lines: Eq. (7) (red), Eq. (6) (blue) with $\epsilon =1/500$. Circles refer to stochastic simulations of the mean time (as in Fig. 3) with $\epsilon =1/500$.