INTRODUCTION

In elementary calculus you learned that differentiation and integration are transforms—this means, roughly speaking, that these operations transform a function into another function. For example, the function f(x) = x² is transformed, in turn, into a linear function, a family of cubic polynomial functions, and a constant by the operations of differentiation, indefinite integration, and definite integration:

Moreover, these two transforms possess the linearity property; this means the transform of a linear combination of functions is a linear combination of the transforms. For α and β constants,

and

provided each derivative and integral exists. In this section we will examine a special type of integral transform called the Laplace transform. In addition to possessing the linearity property, the Laplace transform has many other interesting properties that make it very useful in solving linear initial-value problems.

Integral Transform

If f(x, y) is a function of two variables, then a definite integral of f with respect to one of the variables leads to a function of the other variable. For example, by holding y constant we see that Similarly, a definite integral such as transforms a function f of the variable t into a function F of the variable s. We are particularly interested in an integral transform, where the interval of integration is the unbounded interval If is defined for then the improper integral is defined as a limit

(1)

If the limit in (1) exists, we say that the integral exists or is convergent; if the limit does not exist, the integral does not exist and is divergent. The limit in (1) will, in general, exist for only certain values of the variable s.

A Definition

The function in (1) is said to be the kernel of the transform. The choice as the kernel gives us an especially important integral transform.

The Laplace transform of a function f is also denoted by In general discussion and applications, we shall use a lowercase letter to denote the function being transformed and the corresponding capital letter to denote its Laplace transform. For example,

The Laplace transform is named in honor of the French mathematician, astronomer, and physicist Pierre-Simon Marquis de Laplace (1749–1827), who used the transform in his investigations of probability theory. Most likely the Laplace transform was invented by Leonard Euler. For his services to the Bourbons in restoring the monarchy following the fall of Napoleon Bonaparte in 1814, King Louis XVIII awarded Laplace the title of Marquis.

In a general setting, the variable s could be a complex number. But throughout this chapter we shall assume that s is real. As the next four examples show, the domain of F(s) depends on the function f(t).

EXAMPLE 1 Using Definition 4.1.1

EXAMPLE 2 Using Definition 4.1.1

EXAMPLE 3 Using Definition 4.1.1

EXAMPLE 4 Using Definition 4.1.1

Note that the integral in (4) of Example 4 is the Laplace transform of So (3) and (4) show that Laplace transform of is a multiple of the Laplace transform of :

Thus, with no further work we obtain the additional result

Linearity of the Laplace Transform

Suppose the functions f and g possess Laplace transforms for and , respectively. If c denotes the maximum of the two numbers c₁ and c₂, then for and constants and we can write

In other words,

(5)

Because of the property given in (5), is said to be a linear transform. Furthermore, by the properties of the definite integral, the transform of any finite linear combination of functions is the sum of the transforms provided each transform exists on some common interval of the s-axis.

EXAMPLE 5 Linearity of the Laplace Transform

EXAMPLE 6 Linearity of the Laplace Transform

We state the generalization of some of the preceding examples by means of the next theorem. From this point on we shall also refrain from stating any restrictions on s; it is understood that s is sufficiently restricted to guarantee the convergence of the appropriate Laplace transform.

The result in (b) of Theorem 4.1.1 can be formally justified for n a positive integer using integration by parts to first show that

Then for n = 1, 2, and 3, we have, respectively,

If we carry on in this manner, you should be convinced that

A more extensive list of functions and their Laplace transforms is given in Appendix C.

The Laplace transforms of trigonometric functions can often be obtained with the help of trigonometric identities and (d) and (e) of Theorem 4.1.1.

EXAMPLE 7 Using a Trigonometric Identity

Sufficient Conditions for Existence of {f(t)}

The integral that defines the Laplace transform does not have to converge. For example, neither {1/t} nor {} exists. Sufficient conditions guaranteeing the existence of {f(t)} are that f be piecewise continuous on [0, ∞) and that f be of exponential order for t > T. Recall that a function f is piecewise continuous on [0, ∞) if, in any interval defined by 0 ≤ a ≤ t ≤ b, there are at most a finite number of points t_k, k = 1, 2, …, n(t_k−1 < t_k), at which f has finite discontinuities and is continuous on each open interval defined by t_k−1 < t < t_k. See FIGURE 4.1.1. The concept of exponential order is defined in the following manner.

If f is an increasing function, then the condition |f(t)| ≤ Me^ct, t > T, simply states that the graph of f on the interval (T, ∞) does not grow faster than the graph of the exponential function Me^ct, where c is a positive constant. See FIGURE 4.1.2. The functions f(t) = t, f(t) = e^−t, and f(t) = 2 cos t are all of exponential order c = 1 for t > 0 since we have, respectively,

|t|≤ e^t,        |e^−t|≤e^t,         |2 cos t|≤2e^t.

A comparison of the graphs on the interval [0, ∞) is given in FIGURE 4.1.3.

A positive integral power of t is always of exponential order since, for c > 0,

is equivalent to showing that lim_t→∞tⁿ/e^ct is finite for n = 1, 2, 3, . . . . The result follows by n applications of L’Hôpital’s rule. A function such as is not of exponential order since, as shown in FIGURE 4.1.4, grows faster than any positive linear power of e for t > c > 0. This can also be seen from

for any value of c. By the same reasoning, as for any s and so the improper integral dt diverges. In other words, does not exist.

PROOF:

By the additive interval property of definite integrals,

The integral I₁ exists because it can be written as a sum of integrals over intervals on which e^−stf(t) is continuous. Now f is of exponential order, so there exists constants c, M > 0, T > 0 so that |f(t)| ≤ Me^ct for t > T. We can then write

for s > c. Since Me^{−(s − c)t} dt converges, the integral |e^−stf(t)| dt converges by the comparison test for improper integrals. This, in turn, implies that I₂ exists for s > c. The existence of I₁ and I₂ implies that {f(t)} = e^−stf(t) dt exists for s > c. ≡

EXAMPLE 8 Transform of a Piecewise-Continuous Function

Find {f(t)} for f(t) =

SOLUTION

This piecewise-continuous function appears in FIGURE 4.1.5. Since f is defined in two pieces, {f(t)} is expressed as the sum of two integrals:

≡

In the remaining sections of this chapter we shall refrain from stating any restrictions on s in the examples; it is understood that s is sufficiently restricted to guarantee the convergence of the appropriate Laplace transform.

4.1 Exercises Answers to selected odd-numbered problems begin on page ANS-9.

In Problems 1–18, use Definition 4.1.1 to find {f(t)}.

1. 
2. 
3. 
4. 
5. 
6. 

7. 

8. 

9. 

10. 

11. f(t) = e^t+7
12. f(t) = e^−2t−5
13. f(t) = te^4t
14. f(t) = t²e^−2t
15. f(t) = e^−t sin t
16. f(t) = e^t cos t
17. f(t) = t cos t
18. f(t) = t sin t

In Problems 19–38, use Theorem 4.1.1 to find {f(t)}.

19. f(t) = 2t⁴
20. f(t) = t⁵
21. f(t) = 4t − 10
22. f(t) = 7t + 3
23. f(t) = t² + 6t − 3
24. f(t) = −4t² + 16t + 9
25. f(t) = (t + 1)³
26. f(t) = (2t − 1)³
27. f(t) = 1 + e^4t
28. f(t) = t² − e^−9t + 5
29. f(t) = (1 + e^2t)²
30. f(t) = (e^t − e^−t)²
31. f(t) = 4t² − 5 sin 3t
32. f(t) = cos 5t + sin 2t
33. f(t) = sinh kt
34. f(t) = cosh kt
35. f(t) = e^t sinh t
36. f(t) = e^−t cosh t
37. f(t) = cosh² kt
38. f(t) = sinh² kt

In Problems 39–44, find {f(t)} by first using a trigonometric identity.

39. 
40. 
41. 
42. 
43. 
44. 
45. One definition of the gamma function Γ(α) is given by the improper integral
    

    

    Use this definition to show that Γ(α + 1) = αΓ(α).

46. Use Problem 45 to show that
    

    

    This result is a generalization of Theorem 4.1.1(b).

In Problems 47–50, use the results in Problems 45 and 46 and the fact that to find the Laplace transform of the given function.

47. f(t) = t^−1/2
48. f(t) = t^1/2
49. f(t) = t^3/2
50. f(t) = 6t^1/2 − 24t^5/2

Discussion Problems

51. For what values of s is the following true
    

    

52. Figure 4.1.4 suggests, but does not prove, that the function is not of exponential order. How does the observation that t² > ln M + ct, for M > 0 and t sufficiently large, show that > Me^ct for any c?
53. Use part (c) of Theorem 4.1.1 to show that
    

    

    where a and b are real and i² = −1. Show how Euler’s formula (page 126) can then be used to deduce the results

    

    and

    

54. Under what conditions is a linear function
    

    f(x) = mx + b,

    m ≠ 0, a linear transform?

55. The function is not of exponential order. Nevertheless, show that the Laplace transform exists. [Hint: Use integration by parts.]
56. Explain why the function
    

    

    is not piecewise continuous on

57. Show that the function does not possess a Laplace transform. [Hint: Write as two improper integrals:
    

    

    Show that diverges.]

58. If and a > 0 is a constant, show that
    

    

    This result is known as the change of scale theorem.

In Problems 59–62, use the given Laplace transform and the result in Problem 58 to find the indicated Laplace transform. Assume that a and k are positive constants.