4
The CHAPTER Laplace Transform
4.1 Introduction
The Laplace transform provides an eective method of solving initial-value problems for linear dierential equations with constant coecients. However, the usefulness of Laplace transforms is by no means restricted to this class of problems. Some understanding of the basic theory is an essential part of the mathematical background of engineers, scientists and mathematicians. The Laplace transform is dened in terms of an integral over the interval [0, ). Integrals over an innite interval are called improper integrals, a topic studied in Calculus II. DEFINITION Let f be a continuous function on [0, ). The Laplace transform of f , denoted by L[(f (x)], or by F (s), is the function given by
L[f (x)] = F (s) =
0
esx f (x) dx.
(1)
The domain of F is the set of all real numbers s for which the improper integral converges. In more advanced treatments of the Laplace transform the parameter s assumes complex values, but the restriction to real values is sucient for our purposes here. Note that L transforms a function f = f (x) into a function F = F (s) of the parameter s. The continuity assumption on f will hold throughout the rst three sections. It is made for convenience in presenting the basic properties of L and for applying the Laplace transform method to solving initial-value problems. In the last two sections of this chapter we extend the denition of L to a larger class of functions, the piecewise continuous functions on [0, ). There we will apply L to the problem of solving nonhomogeneous equations in which the nonhomogeneous term is piecewise continuous. This will involve some extension of our concepts of dierential equation and solution. As indicated above, the primary application of Laplace transforms of interest to us is solving linear dierential equations with constant coecients. Referring to our work in Chapter 3, the functions which arise naturally in the treatment of these equations are: p(x)erx, p(x) cos x, p(x) sin x, p(x)erx cos x, p(x)erx sin x
where p is a polynomial. We begin by calculating the Laplace transforms of some simple cases of these functions.
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Example 1. Let f (x) = 1 e0x 1 on [0, ). By the Denition,
b
L[1] =
0
esx 1 dx = lim esx s
b
b 0
esx dx
=
b
lim
= lim
0
b
esb 1 1 1 + . + = lim sb s s b se s
Now, lim 1/sesb exists if and only if s > 0, and in this case
b
b
lim
1 = 0. sesb > s 0.
Thus, L[1] = 1 , s
Example 2. Let f (x) = erx on [0, ). Then,
b
L[erx ] =
0
esx erx dx = lim
b
b 0
e(sr)x dx e(sr)b 1 + . (s r) sr
e(sr)x = lim b (s r)
0
= lim
b
The limit exists (and has the value 0) if and only if s r > 0. Therefore L[erx ] = 1 , sr s > r.
Note that if r = 0, then we have the result in Example 1. Example 3. Let f (x) = cos x on [0, ). Then,
b
L[cos x] =
0
esx cos x dx = lim
b 0
esx cos x dx
b
=
b
lim
esx [s cos x sin x] s2 + 2
.
0
(Note the integral was calculated using integration by parts; also, it is a standard entry in a table of integrals.) Now, L[cos x] = lim 1
b esb
s cos b + sin b s . + 2 2 + 2 s s + 2
Since [s cos b + sin b]/(s2 + 2 ) is bounded, the limit exists (and has the value 0) if and only if s > 0. Therefore, s L[cos x] = 2 , s > 0. s + 2 116
The following table gives a basic list of the Laplace transforms of functions that we will encounter in this chapter. While the entries in the table can be veried using the Denition, some of the integrations involved are complicated. The properties of the Laplace transform presented in the next section provide a more ecient way to obtain many of the entries in the table. Handbooks of mathematical functions, for example the CRC Standard Mathematical Tables, give extensive tables of Laplace transforms. Table of Laplace Transforms f (x) F (s) = L[f (x)] 1 , s
1 ex cos x sin x ex cos x ...