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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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J=0
-+1 = pyn + b,
and from Eq. (11) its solution is
(12)
yn = p ny0 + (1 + p + p2 + + p - 1)b.
(13)
If p = 1, we can write this solution in the more compact form
(14)
118
Chapter 2. First Order Differential Equations
EXAMPLE
1
where again the two terms on the right side are the effects of the original population and of immigration, respectively. By rewriting Eq. (14) as
y- = p-(y 1p) + T~p (15)
the long-time behavior of yn is more evident. It follows from Eq. (15) that yn ^ b/(1 p) if Ip \ < 1 and that yn has no limit if \p \ > 1 or if p = 1. The quantity b/(1 p) is an equilibrium solution ofEq. (12), as can readily be seen directly from that equation. Of course, Eq. (14) is not valid for p = 1. To deal with that case, we must return to Eq. (13) and let p = 1 there. It follows that
- = 0 + -b (16)
so in this case yn becomes unbounded as n ^.
The same model also provides a framework for solving many problems of a financial character. For such problems yn is the account balance in the nth time period, pn =
1 + rn, where rn is the interest rate for that period, and bn is the amount deposited or withdrawn. The following is a typical example.
A recent college graduate takes out a $10,000 loan to purchase a car. If the interest rate is 12%, what monthly payment is required to pay off the loan in 4 years?
The relevant difference equation is Eq. (12), where yn is the loan balance outstanding in the nth month, p = 1 + r is the interest rate per month, and b is the monthly payment. Note that b must be negative and p = 1.01, corresponding to a monthly interest rate of 1%.
The solution of the difference equation (12) with this value for p and the initial condition y0 = 10,000 is given by Eq. (15), that is,
yn = (1.01)- (10,000 + 100b) 100b. (17)
The payment b needed to pay off the loan in 4 years is found by setting y48 = 0 and solving for b. This gives
(1.01)48
b = 100^-------------= 263.34. (18)
(1.01)48 1
The total amount paid on the loan is 48 times b or $12,640.32. Of this amount $10,000 is repayment of the principal and the remaining $2640.32 is interest.
Nonlinear Equations. Nonlinear difference equations are much more complicated and have much more varied solutions than linear equations. We will restrict our attention to a single equation, the logistic difference equation
-+1 = py- (1 y) , (19)
which is analogous to the logistic differential equation
- = ry( I 4)
dt K)
(20)
2.9 First Order Difference Equations
119
that was discussed in Section 2.5. Note that if the derivative dy/ dt in Eq. (20) is replaced by the difference (yn+1 yn)/h, then Eq. (20) reduces to Eq. (19) with p = 1 + hr and k = (1 + hr)K/ hr. To simplify Eq. (19) a little more, we can scale the variable yn by introducing the new variable un = yn/ k. Then Eq. (19) becomes
Un+1 = P u-(1 u-), (21)
where p is a positive parameter.
We begin our investigation of Eq. (21) by seeking the equilibrium, or constant solutions. These can be found by setting un+1 equal to un in Eq. (21), which corresponds to setting dy/dt equal to zero in Eq. (20). The resulting equation is
un = pun - pu2n, (22)
so it follows that the equilibrium solutions ofEq. (21) are
un = 0 un = . (23)
p
The next question is whether the equilibrium solutions are asymptotically stable or unstable; that is, for an initial condition near one of the equilibrium solutions, does the resulting solution sequence approach or depart from the equilibrium solution? One way to examine this question is by approximating Eq. (21) by a linear equation in the neighborhood of an equilibrium solution. For example, near the equilibrium solution un = 0, the quantity u- is small compared to un itself, so we assume that we can neglect the quadratic term in Eq. (21) in comparison with the linear terms. This leaves us with the linear difference equation
Un+1 = P un, (24)
which is presumably a good approximation to Eq. (21) for un sufficiently near zero. However, Eq. (24) is the same as Eq. (7), and we have already concluded, in Eq. (9), that un ^ 0 as n if and only if Ipl < 1, or since p must be positive, for 0 < p < 1. Thus the equilibrium solution un = 0 is asymptotically stable for the linear approximation (24) for this set of p values, so we conclude that it is also asymptotically stable for the full nonlinear equation (21). This conclusion is correct, although our argument is not complete. What is lacking is a theorem stating that the solutions of the nonlinear equation (21) resemble those of the linear equation (24) near the equilibrium solution un = 0. We will not take time to discuss this issue here; the same question is treated for differential equations in Section 9.3.
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