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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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+1 = f , ) n = > 1. 2>---- (1)
Equation (1) is called a first order difference equation. It is first order because the value of yn+ x depends on the value of yn, but not earlier values yn_j, yn_2, and so forth. As for differential equations, the difference equation (1) is linear if f is a linear function of yn; otherwise, it is nonlinear. A solution of the difference equation (1) is
Chapter 2. First Order Differential Equations
a sequence of numbers y0, y1, y2,... that satisfy the equation for each n. In addition to the difference equation itself, there may also be an initial condition
0 = a (2)
that prescribes the value of the first term of the solution sequence.
We now assume temporarily that the function f in Eq. (1) depends only on yn, but not . In this case
+1 = f (), = 0 1, 2,.... (3)
If y0 is given, then successive terms of the solution can be found from Eq. (3). Thus
1 = f (0),
y2 = f (yx) = f [ f (0)].
The quantity f [ f (y0)] is called the second iterate of the difference equation and is sometimes denoted by f2(y0). Similarly, the third iterate y3 is given by
3 = f (y2) = f{ f[ f(y0)]}= f3(0), and so on. In general, the nth iterate yn is
= f (Jnl) = fn (J0)-
This procedure is referred to as iterating the difference equation. It is often of primary interest to determine the behavior of yn as n ^ , in particular, whether yn approaches a limit, and if so, to find it.
Solutions for which yn has the same value for all n are called equilibrium solutions. They are frequently of special importance, just as in the study of differential equations. If equilibrium solutions exist, one can find them by setting yn+1 equal to yn in Eq. (3) and solving the resulting equation
= f ( ) (4)
for .
Linear Equations. Suppose that the population of a certain species in a given region in year n + 1, denoted by yn+j, is a positive multiple pn of the population yn in year n; that is,
+1 = , = 0 1 2---- (5)
Note that the reproduction rate pn may differ from year to year. The difference equa-
tion (5) is linear and can easily be solved by iteration. We obtain
1 = P0 0>
y2 = P1 y1 = P1P0 y0,
and, in general,
= Pn1 P0 0 > n = !> 2. (6)
Thus, if the initial population y0 is given, then the population of each succeeding gen-
eration is determined by Eq. (6). Although for a population problem pn is intrinsically
2.9 First Order Difference Equations
positive, the solution (6) is also valid if Pn is negative for some or all values of n. Note, however, that if Pn is zero for some n, then yn+l and all succeeding values of y are zero; in other words, the species has become extinct.
If the reproduction rate Pn has the same value p for each n, then the difference equation (5) becomes
Equation (7) also has an equilibrium solution, namely, yn = 0 for all n, corresponding to the initial value y0 = 0. The limiting behavior of yn is easy to determine from Eq. (8). In fact,
In other words, the equilibrium solution yn = 0 is asymptotically stable for IpI < 1 and unstable if Ip | > 1.
effect of immigration or emigration. If bn is the net increase in population in year n due to immigration, then the population in year n + 1 is the sum of those due to natural reproduction and those due to immigration. Thus
where we are now assuming that the reproduction rate P is constant. We can solve Eq. (10) by iteration in the same manner as before. We have
Note that the first term on the right side of Eq. (11) represents the descendants of the original population, while the other terms represent the population in year n resulting from immigration in all preceding years.
In the special case where bn = b for all n, the difference equation is
yn+1 P yn
and its solution is
y- = p 0.
does not exist,
if Ip I < 1; if p = 1; otherwise.
Now we will modify the population model represented by Eq. (5) to include the
-+1 = P - + b-,
1 = P 0 + b0,
2 = Pp0 + b0) + b1 = P2 0 + p b0 + b1>
y3 = p(p2 0 + Pb0 + bl) + b2 = p3 y0 + p 2b0 + p b1 + b2,
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