The basic idea escape from a domain of attraction

Central to the computation of extinction times and extinction probabilities or the movement from one peak in a fitness landscape to another is the notion of "escape from a domain of attraction.'' This impressive sounding phrase can be understood through a variety of simple metaphors (Figure 8.1). In the most interesting case, the basic idea is that deterministic and stochastic factors are in conflict - with the deterministic ones causing attraction towards steady state (the bottom of the bowl or the stable steady states in Figure 8.1) and the stochastic factors causing perturbations away from this steady state. The cases of the ball


Figure 8.1. Some helpful ways to think about escape from a domain of attraction. (a) The marble in a cup, when slightly perturbed, will return to the bottom of the cup (a domain of attraction). The converse of this is the ball on the hill, in which any small perturbation is going to be magnified and the ball will move either to the right or the left. (b) In one dimension, we could envision a deterministic dynamical system dX/dt = b(X) in which there is a single steady state that is globally stable (as in the Ornstein-Uhlenbeck process), denoted by s. Fluctuations will cause departures from the steady state, but in some sense the stochastic process has nowhere else to go. On the other hand, if the deterministic system has multiple steady states, in which two stable steady states are separated by an unstable one (denoted by u), the situation is much more interesting. Then a starting value near the upper stable steady state might be sufficiently perturbed to cross the unstable steady state and be attracted towards the lower stable steady state. If X(t) were the size of a population, we might think of this as an extinction. (c) For a two dimensional dynamical system of the form dX/dt = f(X, Y), d Y/dt = g(X, Y) the situation can be more complicated. If a steady state is an unstable node, for example, then the situation is like the ball at the top of the hill and perturbations from the steady state will be amplified (of course, now there are many directions in which the phase points might move). Here the circle indicates a domain of interest and escape occurs when we move outside of the circle. If the steady state is a saddle point, then the separatrix creates two domains of attraction so that perturbations from the steady state become amplified in one direction but not the other. If the steady state is a stable node, then the deterministic flow is towards the steady state but the fluctuations may force phase points out of the region of interest. (d) If we conceive that natural selection takes place on a fitness surface (Schluter 2000), then we are interested in transitions from one local peak of fitness to a higher one, through a valley of fitness.

on the top of the hill or the steady state being unstable or a saddle point are also of some interest, but I defer them until Connections.

We have actually encountered this situation in our discussion of the Ornstein-Uhlenbeck process, and that discussion is worth repeating, in simplified version here. Suppose that we had the stochastic differential equation dX =—Xdt + dW and defined u(x, t) = Pr{X(s) stays within [—A, A] for all s, 0 < s < t|X(0) = xg (8.1) We know that u(x, t) satisfies the differential equation ut = 1 uxx — xux (8-2)

so now look at Exercise 8.1.

Derive Eq. (8.2). What is the subtlety about time in this derivation?

Equation (8.2) requires an initial condition and two boundary conditions. For the initial condition, we set u(x, 0) = 1 if —A < x < A and to 0 otherwise. For the boundary conditions, we set u(— A, t) = u(A, t) = 0 since whenever the process reaches A it is no longer in the interval of interest. Now suppose we consider the limit of large time, for which ut ! 0. We then have the equation 0 = (1/2)uxx — xux with the boundary conditions u(—A) = u(A) = 0.

Show that the general solution of the time independent version of Eq. (8.2) is u(x) = kj J*A exp(s2)ds + k2, where kj and k2 are constants. Then apply the boundary conditions to show that these constants must be 0 so that u(x) is identically 0. Conclude from this that with probability equal to 1 the process will escape the interval [—A, A].

We will thus conclude that escape from the domain of attraction is certain, but the question remains: how long does this take. And that is what most of the rest of this chapter is about, in different guises.

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