# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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(a) Calculate 0l(t), ..., 04(t), or (if necessary) Taylor approximations to these iterates. Keep terms up to order six.

(b) Plot the functions you found in part (a) and observe whether they appear to be converging.

? 11. y = — sin y + 1, y(0) = 0

? 12. y = (3t2 + 4t + 2)/2(y — 1), y(0) = 0

13. Let 0n (x) = xn for 0 < x < 1 and show that

r a / ^ (0, 0 < x < 1,

lim 0n(x) = 1 ,

n—TO 11 I 1, x = 1 .

114

Chapter 2. First Order Differential Equations

This example shows that a sequence of continuous functions may converge to a limit function that is discontinuous.

14. Consider the sequence 0n(x) = 2nxe , 0 < x < 1.

(a) Show that lim 0 (x) = 0 for 0 < x < 1; hence

n^^ n

f lim 0 (x) dx = 0.

Jo n^

(b) Show that f 2nxe—nx dx = 1 — e—n; hence

J0

lim [ $ (

lim I $„ (x) dx = 1.

Thus, in this example,

n b n b

lim I $ (x) dx = I lim $ (x) dx,

n^xJa n Ja n^x

even though lim $n (x) exists and is continuous.

n^<^ n

In Problems 15 through 18 we indicate how to prove that the sequence {$n (f)}, defined by Eqs. (4) through (7), converges.

15. If d f/dy is continuous in the rectangle D, show that there is a positive constant K such that

I f(f,yi) - f(f,y2)l < KIyi - y2|,

where (f,y1) and (f,y2) are any two points in Dhaving the same f coordinate. This inequality is known as a Lipschitz condition.

Hinf: Hold f fixed and use the mean value theorem on f as a function of y only. Choose K to be the maximum value of |9 f/dyl in D.

16. If $ i (f) and $n (f) are members of the sequence {$n (f)}, use the result of Problem 15 to show that

I f[f, $n(f)] - f[f, $„-1(f)]I < Kl$n(f) - $n-1(f)l

17. (a) Show that if If I < h, then

l$1(f)l <M If I,

where Mis chosen so that I f(f,y)I < Mfor (f,y) in D.

(b) Use the results of Problem 16 and part (a) of Problem 17 to show that

MKIf I2

I$2(f) - $1(f)I .

(c) Show, by mathematical induction, that

MKn-1\t In MKn-1hn

I$n(f) - $n-1(f)I < -----------------< :----.

n! n!

18. Note that

$n(f) = $1 (f) + [$2(f) - $1(f)] + + [$n(f) - $n-1 (f)].

(a) Show that

I$n(f)I < \$1(f)I + I$2(f) - $1(f)I + ••• + I$n(f) - $n_1(f)I.

2.9 First Order Difference Equations

115

(b) Use the results of Problem 17 to show that

M

I0n() ^ K

(Kh)2 ( Kh)n

Kh + (-------- + ••• + ( )

2! n!

(c) Show that the sum in part (b) converges as n and, hence, the sum in part (a) also

converges as n ^?. Conclude therefore that the sequence [0n(f)} converges since it is the sequence of partial sums of a convergent infinite series.

19. In this problem we deal with the question of uniqueness of the solution of the integral equation (3),

0(f) = f f[s,0(s)] ds.

Jo

(a) Suppose that 0 and 0 are two solutions of Eq. (3). Show that, for f > 0,

0(t) - f(t) = I {f[s,0(s)] - f[s,^(s)]} ds. /0

(b) Show that

I0(t) — 0(t)I < I I f[s, 0(s)] — f[s, 0(s)]I ds.

0

(c) Use the result of Problem 15 to show that

f{ f[s,<

0

f ‘ I f [s,<

0

I0(t) — 0(t)I< k( I0(s) — $(s)I ds,

0

where K is an upper bound for I d f/dyI in D. This is the same as Eq. (30), and the rest of the proof may be constructed as indicated in the text.

2.9 First Order Difference Equations

® While a continuous model leading to a differential equation is reasonable and attractive

for many problems, there are some cases in which a discrete model may be more natural. For instance, the continuous model of compound interest used in Section 2.3 is only an approximation to the actual discrete process. Similarly, sometimes population growth may be described more accurately by a discrete than by a continuous model. This is true, for example, of species whose generations do not overlap and that propagate at regular intervals, such as at particular times of the calendar year. Then the population yn+1 of the species in the year n + 1 is some function of n and the population yn in the preceding year, that is,

yn+1 = f (n, yn), n = °> 1, 2>---- (1)

Equation (1) is called a first order difference equation. It is first order because the value of yn+ j 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

116

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

y0 = 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 n. In this case

yn+1 = f (yn)' n = 0 1 2_.. (3)

If y0 is given, then successive terms of the solution can be found from Eq. (3). Thus

yx = f O0)>

and

y2 = f y) = f [ f (y0)].

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

y3 = f (y2) = f{ f[ f(y0)]}= f3(!0),

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