Equilibrium Values and Long-Term Behavior
Recall our original paradigm
When the DDS stops changing, change equals zero, and so future equals present, and remains there. A value for which change equals 0, if any exist, is the equilibrium value. Change having stopped gives us a context for the concept of the equilibrium value.
We will define the equilibrium value, also called fixed point, as the value where change stops; i.e., where change equals zero. The value A is an equilibrium value for the DDS
if whenever for some N, we have a(N) = A and all future values a(n) = A for n > N.
Formally, we define the equilibrium value as follows:
Definition 2.1. Equilibrium Value.
The number ev is called an equilibrium value or a fixed point for a discrete dynamical system a(n) if a(k) = ev for all values of к when the initial value a(0) is set to ev. That is, a(k) = ev is a constant solution to the recurrence equation for the dynamical system.
Another way of characterizing such values is to note that a number ev is an equilibrium value for a dynamical system a(n +1) = ,f(a(n),a(n — 1),..., n) if and only if ev satisfies the equation ev = ,f(ev, ev,..., ev). This characterization is the genesis of the term fixed point.
Definition 2.1 shows that a linear homogeneous dynamical system of order 1 only has the value 0 as an equilibrium value.
In general, dynamical systems may have no equilibrium values, a single equilibrium value, or multiple equilibrium values. Linear systems have unique equilibrium values. The more nonlinear a dynamical system is, the more equilibrium values it may have.
Not all DDSs have equilibrium values, many DDSs have equilibrium values that the system will never achieve unless o(0) equals that value.
The DDS from the Tower of Hanoi a(n+ 1) = 2a(n) + 1 has an equilibrium value ev = — 1. If we begin with a(0) = — 1 and iterate, we get
etc. (Note that the physical Tower of Hanoi puzzle cannot have a negative number of disks, so for the puzzle, —1 is an unreachable equilibrium value.)
We can use the observation that ev = f(ev, ev,..., ev) to find equilibrium values, and to determine whether or not a given DDS has an equilibrium value. Look again at the DDS o(n + 1) = 2a(n) + 1. Substituting a(n + 1) = ev and a(n) = ev into our DDS yields
Solving for ev gives the equilibrium value for the DDS as ev = — 1.
Now, let’s consider the DDS a(n + 1) = a(n) + 1. Using our observation, we write
This equation has no solution, so the DDS a(n + 1) = 2a(n) + 1 does not have any equilibrium values.
Example 2.6. Finding Equilibrium Values.
Consider the following four DDSs. Find their equilibrium value(s), if any exist.
Above, we observed that DDSs that have equilibrium values may never attain their equilibrium value, given some specific initial condition a(0). In DDS 3. from Example 2.6 above, choose any initial value A not equal to 0, the ev. Then, by iteration, a(k) = 2~k'A ф 0 for any k. This DDS can never achieve its equilibrium when starting from a nonzero value even though linin^oo a(n) = 0, the equilibrium value.
We found the equilibrium value was —1 for the DDS from the Tower of Hanoi puzzle. Suppose that we begin iterating the Tower’s DDS with an initial value a( 0) = 0. Then
etc. The values continue to get larger and will never reach the value of — 1. Since a(n) represents the number of moves of the disks, the value of —1 makes no real sense in the context of the number of disks to move.
We will study equilibrium values in many of the applications of discrete dynamical systems. In general, a linear nonhomogeneous discrete dynamical system, where the nonhomogeneous part is a constant, will have an equilibrium value. Can you find any exceptions? A linear homogeneous discrete dynamical system will always have an equilibrium value of zero. Why?