Multiple equilibria and cycles: An example
To give the basic idea for dealing with models with several equilibria, here is an example with two locally stable equilibria.
We take b(φ) = aφ to simplify algebra.
We also let r/a be denoted by r, that is, we normalize both b and r by a. Suppose that there are two possible costs: 0 = c1 ≤ c2 - that is, the distribution function G (c) is a step function; G (c1) = p > 0; G(c2) = 1. The right-hand side of the dynamics for the aggregate equation is either Φ1(φ) = (1 - φ)p - φ2, or Φ2(φ) = 1 - φ - φ2, depending on the range of the argument φ. We show below that Φ1 applies when φ is not greater than ψ defined below, and Φ2 applies otherwise.There are thus two critical points. They are the roots of Φi (φ) = 0, i = 1, 2, and are given by
and
We see that Φi (φi) are negative for i = 1, 2, that is, the critical points are locally stable.
From the optimality condition,
is determined by
or
The two basins of attractions are separated at
where we assume that υ > c2, that is, the value of Φ undergoes a discontinuous change at this value:
and
Therefore, if there is a large positive disturbance near φ1 that makes the variable φ cross the boundary at ψ, then the derivative is posit^^ve and tlιc disturbance is amplified, and φ is attracted to φ2.
Likewise, a large negative disturbance near φ2 will cause the state variable to be attracted to φ1. Figure 1 in Aoki and ShIrai (2000, p. 497) ptots one example of Φ(φ).
and
9.6.1 Asymmetrical cycles
This section presents several examples to show how we approximately evaluate the mean transition times from one basin of attraction to the other, and calculate the equilibrium probabilities for the model to stay in each of the two basins of attractions, as in the example with two locally stable equilibria.
9.6.1.1 Approximate analysis
First, we recognize that we need to calculate only the event from one of the equilibrium state to the boundary between the two basins of attraction, ψ, which is introduced in connection with the example of the multiple equilibria above. The reason is the same one given by van Kampen (1992), as quoted in Aoki p. 151). Hie Ihue needed Ior φ to reach its equilibrium value (φ1, or φ2, depending on the initial value) is much shorter than the time needed to go from one basin of attraction to the other.
A quick way to see this is to solve the deviational equation for φ.[12] To be definite, suppose that φ is in the domain of attraction to φ1, and let x = φ - φ1. Then x is governed by
with the initial condition x(0) = φ(0) — φ1.
and
Suppose that N = 50.
Then W↑g = 0.00665 and W2j1 = 0.01375. We thus estimate the mean first passage times as τ1,2 = 150.4, r2,-1 = 72.7, π1 = 0.67, and π2 = 0.33. The ratio of the mean first-passage times is 1.9. If the number of agents is N = 100, then Wu = 0.00025, W21 = 0.0009, τt2 = 4000, τ2χ = 1100, and
This simple example shows that the model can have several locally stable equilibria and that the fraction of the employed agents may fluctuate between the equilibria. We have shown that this leads to a simpler explanations of asymmetrical cycles, among other things. Aoki and Shirai (2000, p. 499) has another example with a similar set of results.
9.7