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A Graphical Approach to Equilibrium ValuesWe can examine a graph of the DDS’s iterations using Maple. If the values reach a specific value and remain constant, the graph levels, then that value is an equilibrium value (change has stopped). Dynamical systems often represent realworld behavior that we are trying to understand. We model reality to be able to predict future behavior and gain deeper insights into the behavior and how to influence or alter that behavior. Thus, we have great interest in the predictions of the model. Understanding how the model changes in the future is our goal. Models of the Form: a(n + 1) = r • a(n) + b with Constant r and b Consider a savings account with 12% APR interest compounded monthly where we deposit $1000 per month. Repeating our earlier analysis, but adding the monthly deposit. (Remember to start with a fresh Maple worksheet, then load the plots package.) The DDS a(n + 1) = 1.01 a(n) + 1000, with r = 1.01 > 1, grows without bound. The graph suggests that there is no equilibrium value a value where the graph levels at a constant value. Analytically, we have ev = 1.0lev + 1000 ev = —100000. There is an equilibrium value ev = —100000, but that value will never be reached in our savings account problem. Without withdrawals, the savings account can never be $100,000 in the hole! What happens if r is less than 0? Proceed as before, but with a negative r. The new DDS is with a(0) = 0. Analytically solve for the equilibrium value.
The definition of equilibrium value implies that if we start at a(0) = 497.5124378, we stay there forever. From the plot of the DDS in Figure 2.7, we note the oscillations between positive and negative numbers, each growing without bound as the oscillations fan out. Although there is an equilibrium value, the solution to our example does not tend toward this fixed point. FIGURE 2.7: A Linear DDS with Negative r Let’s examine an example with a value of r between 0 and 1, that is, 0 < r < 1. A drug concentration model with a constant dosage of 16 mg each time period (4 hours) has the DDS
with an initial dosage of a(0) applied prior to beginning the regime. Figure 2.8 shows the DDS with 4 different initial values, o(0) = 10, 20, 32, and 50 mg. FIGURE 2.8: Drug Concentration DDS with Four Different Initial Values Regardless of the starting value, each graph shows the future terms of a(n) approach 32. Thus, 32 is the equilibrium value. We could have easily solved for the equilibrium value algebraically as well.
In general, finding the equilibrium value for this type of DDS requires solving the equation ev = r ■ ev + b for ev. We find Applying this formula to the drug concentration DDS aboves calculates the equilibrium value as
the value we observed from the graphs. Stability and LongTerm Behavior For a dynamical system a(n) with a specific initial condition o(0) = /1_{()}. we have shown that we can compute a(l), a(2), and so forth, by iteration. Often the particular values are not as important as the longterm behavior. By longterm behavior, we refer to what will eventually happen to a(n) for larger and larger values of n. There are many types of longterm behavior that can occur with DDSs; we will only discuss a few here. If the values of a(n) for a DDS eventually get close^{0} to the equilibrium value ev, for all initial conditions in a range, then the equilibrium is called a stable equilibrium or an attracting fixed point. Example 2.7. A Stable Equilibrium or Attracting Fixed Point. Consider the DDS a(n + 1) = 0.5 a(n) + 64 with initial conditions either a(0) = 0, 50, 100, 150, or 200. In each case, the ev is 128; therefore, the ev = 128 is a stable equilibrium. Examine Figure 2.9. ^{[1]} FIGURE 2.9: Drug Concentration with Varying Initial Values Let’s use rsolve to find a general formula, and then generate a table of values for the DDS. Notice that all the sequences are converging to 128, the attracting fixed point. (To see more than 10 rows, first execute interface(verboseproc=100).) Example 2.8. A Financial Model with an Unstable Equilibrium. Consider a financial model where $100 is deposited each month in an account that pays 12% APR compounded monthly. The DDS is Let’s first determine a formula for B(n)
The algebraic technique easily gives us the equilibrium value.
The tv value is —10000. If the DDS ever achieves an input of —10000. then the system stays at —10000 forever. Since $100 is deposited each month and there are no withdrawals, the account must stay positive—we can’t reach the equilibrium value! We say this equilibrium value is unstable or a repelling fixed point. Figure 2.10 shows a graph of our financial model and ev. FIGURE 2.10: Unstable Equilibrium Value Generate a table of values. The values increase over time and never move toward —10000. Therefore, the ev is unstable or repelling. Try this model with different initial values and different deposit values. Can the equilibrium become attracting? Often, we characterize the longterm behavior of the system in terms of its stability. If a DDS has an equilibrium value, and if the solution tends to the equilibrium value from starting values near the equilibrium value, then the DDS is said to be stable. We summarize the results for the dynamical system a(n + 1) = r ■ a(n) + b, where b ф 0 in Table 2.4.
From the table we see that the DDS o(n+l) = ra(n)+b has an equilibrium 6/(1 — r) whenever г ф 1 and has no equilibrium when r = 1. Further, we see that the equilibrium is stable only when r < 1. Relationship of the Equilibrium to Analytical Solutions If a discrete dynamical system has an equilibrium ev, we can use the value to find the analytical solution. Consider the DDS for a 6.5% APR mortgage having monthly payments of $639.34:
The ev value is analytically found to be 118031.9927. Substituting the ev and r values into B(k) = r^{k}C + ev and then setting к = 0 gives
which yields The solution to our DDS is (Check this with rsolve).) We can also use this method to determine the mortgage payment . The DDS is now
with solution
Build a system of two equations and two unknowns.
Solve the system. t Thus
(Note: there is a little roundoff error creeping into the calculations.) To find the payment P, we know that
Substitute ev = 118033.6650 and solve to find
This value agrees with our original payment other than some roundoff error. Roundoff in financial calculations can be very serious^{0}—always be careful! Equilibrium Values and the Limit Studying the behavior of equilibrium values relies fundamentally on the concept of limit. Elementary calculus is based on limit and includes a careful study of the concept . Advanced calculus or real analysis presents a precise, carefully crafted definition, but using that would overly complicate our investigation of equilibrium values. The informal definition of limit of a sequence
will serve our needs and keep us focused on the behavior of the discrete dynamical system in question. ^{[2]} ^{[3]} We are interested in what happens to a(k) as к gets larger and larger without bound. (In other words, when we iterate many times.) It may happen that for large values of k, a(k) is close or even equal to some number L. If increasing к (doing more iterations) causes a(k) to get even closer to L until for very large values of k, a(k) is essentially equal to L, then L is the limit of a(k). If a(k) has a limit, then we also say а (к) converges to L or simply a(k) converges. It is important to understand that if a(k) converges to L, then further increases in к will never cause а (к) to move “too far” away from L. Also remember that we are looking at large values of k. How large is large depends upon the DDS. Compare how large к must be for the two sequences DDSj: a{k) = e~^{k} and DDS_{2}: b{k) = l/(ln(k) + 1). If increasing к causes a(k) to continue to increase or to oscillate, and a(k) does not begin to converge to a number L. we say a(k) is diverging and the limit does not exist. Example 2.9. Convergent Difference Equations. Consider 1. the difference equation a(n + 1) = 0.9 o(n) + 2 with a(0) = 1. Iterating with an accuracy of 6 decimal places produces the following:
and for all к > 166, a (к) = 20. Thus, for this difference equation, the limit of a(k) is 20. 2. the difference equation b(n + 1) = —0.1 b(n) + 11 with 6(0) = 1. Iterating with an accuracy of 6 decimal places, we see the following: and for all к > 9, 6(6) = 10. Thus, for this difference equation, the limit of 6(6) is 10. Even though 6(4) is less than 10 and 6(5) is greater than 10, 6(5) is closer to 10 than 6(4) is. Increasing 6 caused а (к) to get closer to 10. Thus, the limit as к —> ос of 6(6) is 10. Graph it! If a(k) does not converge to L. then the sequence diverges. There are several ways in which a sequence may diverge. If a (6) gets infinitely large as к gets infinitely large, then a(k) diverges. If a(6) gets infinitely large in the negative direction as к gets infinitely large, then a(k) also diverges. If o(6) oscillates between large positive and large negative values, always getting further from 0 as к gets infinitely large, then a(k) diverges. If ct(6) oscillates in a pattern between two or more fixed values as к gets infinitely large, а (к) diverges. If a(k) shows absolutely no pattern of behavior as к gets infinitely large, then a(k) diverges. Given a random sequence a(k), it will likely diverge. Example 2.10. Divergent Difference Equations. Consider 1. the difference equation a(n + 1) = 4a(n) + 2 with a(0) = 1. Iteration of this difference equation produces
Further increase in к causes increase in a(k). Thus, a(k) diverges. 2. the difference equation a(n+ 1) = —4a(n) + 2 with a(0) = 1. In this case, iteration produces
As к gets larger and larger, a(6) gets further from 0, always oscillating between positive and negative values. Thus, а (к) diverges. We close this section with a very useful theorem. Theorem. Limits are Stable Equilibria. If a dynamical system has a limit, then that limit is a stable equilibrium value; i.e., is an attracting fixed point. Exercises

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