1550078481-Ordinary_Differential_Equations__Roberts_

232 Ordinary Differential Equations

A function which is of exponential order a > 0 as x ----> +oo may become

infinite as x ----> +oo, but it may not become infinite more rapidly than M eax.
It follows from this definition that all bounded functions are of exponential

order 0 as x ----> +oo. Also if a < b and f ( x) is of exponential order a as

x----> +oo, then f(x) is of exponential order bas x----> +oo, since a< b implies

M eax < ]'/[ ebx. It should be noted that there are functions which are not 2 of

exponential order a as x ----> +oo for any a. For instance, the function ex is
not of exponential order a as x ----> +oo for any a. As a matter of fact, it can
be shown that for any positive constants a and Jvl, no matter 2 how large, there

exists an xo- which depends upon M and a- such that ex > M eax for all

X > Xo.

The following theorem provides sufficient conditions for t he existence of

a Laplace transform of a function f ( x). These conditions are not necessary

conditions as we shall show in the example following the theorem.

THEOREM 5.1 AN EXISTENCE THEOREM FOR THE

LAPLACE TRANSFORM

If f(x) is piecewise continuous on [O, b] for all finite b > 0 and if f(x) is of

exponential order a as x----> +oo, then the Laplace transform of f(x) exists

for s >a.

Proof: Since f(x) is assumed to be of exponential order a as x----> + oo, there

exist positive constants M and xo such that If ( x) I < M eax for x > xo. We

rewrite the Laplace transform of f(x) as follows:

(8) L:[f(x)] = r = f(x)e-sx dx = rxo f(x)e-sx dx + 1 = f(x)e-sx dx.

Jo Jo xo

The first integral on the right-hand side of equation (8) exists, since f(x)

is piecewise continuous on [O, x 0 ], which implies f(x)e-sx is piecewise con-

tinuous on [O, xo], because e-sx is continuous on [O, x 0 ]. Since f(x )e-sx is

p iecewise continuous on [O, xo], it is integrable on [O, x 0 ]. The absolute value

of the second integral on the right-hand side of equation (8) satisfies

11

= f(x)e-sx dxl ::; 1 = lf(x)le-sx dx < M 1 = eaxe-sx dx

xo xo xo

= M eCa-s)x1= = -M e(a-s)xo provided s > a.

a - s xo a - s

Since t he improper integral J = XQ eaxe-sx dx converges for s > a, the improper

integral J;: lf(x)le-sx dx converges for s > a by the comparison test for