The Inverse Problem

If F(s) represents the Laplace transform of a function f(t), that is, {f(t)} = F(s), we then say f(t) is the inverse Laplace transform of F(s) and write f(t) = ⁻¹{F(s)}. For example, from Examples 1, 2, and 3 in Section 4.1 we have, respectively,

1 = ⁻¹, t = ⁻¹, and e^−3t = ⁻¹.

The analogue of Theorem 4.1.1 for the inverse transform is presented next.

When evaluating inverse transforms, it often happens that a function of s under consideration does not match exactly the form of a Laplace transform F(s) given in a table. It may be necessary to “fix up” the function of s by multiplying and dividing by an appropriate constant.

EXAMPLE 1 Applying Theorem 4.2.1

Linearity of the Inverse Laplace Transformation

Analogous to (5) of Section 4.1, the inverse Laplace transform is also a linear transform. Suppose the functions and are piecewise continuous on and of exponential order. Then for constants and we can write

(1)

Like (5) of Section 4.1, (1) extends to any finite linear combination of Laplace transforms.

EXAMPLE 2 Termwise Division and Linearity

Partial Fractions

Partial fractions play an important role in finding inverse Laplace transforms. The decomposition of a rational expression into component fractions can be done quickly by means of a single command on most computer algebra systems. Indeed, some CASs have packages that implement Laplace transform and inverse Laplace transform commands. But for those of you without access to such software, we will review in this and subsequent sections of this chapter some of the basic algebra in the important cases in which the denominator of a Laplace transform F(s) contains (i) distinct linear factors, (ii) repeated linear factors, and (iii) quadratic polynomials with no real factors.

Although we shall examine each of these cases as this chapter develops, it still might be a good idea for you to consult either a calculus text or a current precalculus text for a more comprehensive review of this theory.

The following example illustrates partial fraction decomposition in the case when the denominator of F(s) is factorable into distinct linear factors.

EXAMPLE 3 Partial Fractions and Linearity

Transform of a Derivative

As pointed out in the introduction to this chapter, our immediate goal is to use the Laplace transform to solve differential equations. To that end we need to evaluate quantities such as {dy/dt} and {d²y/dt²}.

To begin, we consider the case when the function f is continuous, f′ is piecewise continuous on [), and both functions are of exponential order with c as specified in Definition 4.1.2. For simplicity, we also assume that f′ has a single point of discontinuity t₀ on the interval [). Then integration by parts gives

Because the function f is of exponential order, the term as for , and the last term is recognized as Using the notation we have shown that

(6)

The result in (6) can be used recursively to find Laplace transforms of higher derivatives. For example, if and in are replaced by and we get

or

(7)

Similarly, yields

(8)

The results in (6), (7), and (8) are special cases of the following theorem.

Solving Linear ODEs

It is apparent from the general result given in Theorem 4.2.2 that {dⁿy/dtⁿ} depends on Y(s) = {y(t)} and the n − 1 derivatives of y(t) evaluated at t = 0. This property makes the Laplace transform ideally suited for solving linear initial-value problems in which the differential equation has constant coefficients. Such a differential equation is simply a linear combination of terms y, y′, y″, ..., y⁽ⁿ⁾:

where the coefficients a_i, i = 0, 1, ..., n and y₀, y₁, ..., y_n−1 are constants. By the linearity property, the Laplace transform of this linear combination is a linear combination of Laplace transforms:

(9)

From Theorem 4.2.2, (9) becomes

(10)

where {y(t)} = Y(s) and {g(t)} = G(s). In other words:

If we solve the general transformed equation (10) for the symbol Y(s), we first obtain P(s)Y(s) = Q(s) + G(s), and then write

(11)

where P(s) = a_nsⁿ + a_n−1sⁿ⁻¹ + … + a₀, Q(s) is a polynomial in s of degree less than or equal to n − 1 consisting of the various products of the coefficients a_i, i = 1, ..., n, and the prescribed initial conditions y₀, y₁, ..., y_n−1, and G(s) is the Laplace transform of g(t).^* Typically we put the two terms in (11) over the least common denominator and then decompose the expression into two or more partial fractions. Finally, the solution y(t) of the original initial-value problem is y(t) = ⁻¹{Y(s)}, where the inverse transform is done term by term.

The procedure is summarized in FIGURE 4.2.1.

The next example illustrates the foregoing method of solving DEs.

EXAMPLE 4 Solving a First-Order IVP

EXAMPLE 5 Solving a Second-Order IVP

Examples 4 and 5 illustrate the basic procedure for using the Laplace transform to solve a linear initial-value problem, but these examples may appear to demonstrate a method that is not much better than the approach to such problems outlined in Sections 2.3 and 3.3–3.6. Don’t draw any negative conclusions from the two examples. Yes, there is a lot of algebra inherent in the use of the Laplace transform, but observe that we do not have to use variation of parameters or worry about the cases and algebra in the method of undetermined coefficients. Moreover, since the method incorporates the prescribed initial conditions directly into the solution, there is no need for the separate operations of applying the initial conditions to the general solution y = c₁y₁ + c₂y₂ + + c_n y_n + y_p of the DE to find specific constants in a particular solution of the IVP.

The Laplace transform has many operational properties. We will examine some of these properties and illustrate how they enable us to solve problems of greater complexity in the sections that follow.

We conclude this section with a little bit of additional theory related to the types of functions of s that we will generally be working with. The next theorem indicates that not every arbitrary function of s is a Laplace transform of a piecewise-continuous function of exponential order.

PROOF:

Since f(t) is piecewise continuous on the closed interval [0, T], it is necessarily bounded on the interval. That is, |f(t)| ≤ M₁ = M₁e^0t. Also, because f is assumed to be of exponential order, there exist constants γ, M₂ > 0, and T > 0, such that |f(t)| ≤ M₂e^γt for t > T. If M denotes the maximum of {M₁, M₂} and c denotes the maximum of {0, γ}, then

for s > c. As s → ∞, we have |{f(t)}| → 0, and so {f(t)} → 0. ≡

As a consequence of Theorem 4.2.3 we can say that functions of s such as F₁(s) = 1 and F₂(s) = s/(s + 1) are not the Laplace transforms of piecewise-continuous functions of exponential order since F₁(s) 0 and F₂(s) 0 as s → ∞. But you should not conclude from this that F₁(s) and F₂(s) are not Laplace transforms. There are other kinds of functions.

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

4.2.1 Inverse Transforms

In Problems 1–30, use Theorem 4.2.1 to find the given inverse transform.

In Problems 31 and 32, find the given inverse Laplace transform by finding the Laplace transform of the indicated function f.

31. ; f(t) = a sin bt − b sin at
32. ; f(t) = cos bt − cos at

4.2.2 Transforms of Derivatives

In Problems 33–46, use the Laplace transform to solve the given initial-value problem.

33. − y = 1, y(0) = 0
34. 2 + y = 0, y(0)= −3
35. y′ + 6y = e^4t, y(0) = 2
36. y′ − y = 2 cos 5t, y(0) = 0
37. y″ + 5y′ + 4y = 0, y(0) = 1, y′(0) = 0
38. y″ − 4y′ = 6e^3t − 3e^−t, y(0) = 1, y′(0) = −1
39. y″ + y = t, y(0) = 10, y′(0) = 0
40. y″ + 9y = e^t, y(0) = 0, y′(0) = 0
41. 2y‴ + 3y″ − 3y′ − 2y = e^−t, y(0) = 0, y′(0) = 0, y″(0) = 1
42. y‴ + 2y″ − y′ − 2y = sin 3t, y(0) = 0, y′(0) = 0, y″(0) = 1
43. 
44. 
45. 
46. 

In Problems 47 and 48, use one of the inverse Laplace transforms found in Problems 31 and 32 to solve the given initial-value problem.

The inverse forms of the results in Problem 53 in Exercises 4.1 are

and

In Problems 49 and 50, use the Laplace transform and these inverses to solve the given initial-value problem.

49. y′ + y = e^−3t cos 2t, y(0) = 0
50. y″ − 2y′ + 5y = 0, y(0) = 1, y′(0) = 3

In Problems 51 and 52, use the table of Laplace transforms in Appendix C to solve the given initial-value problem.

51. y⁽⁴⁾ + 16y = 0, y(0) = 0, y′(0) = 0, y″(0) = 1, y‴(0) = 0
52. y⁽⁴⁾ + 2y″ + y = 0, y(0) = 0, y′(0) = 1, y″(0) = 0, y‴(0) = 0

Discussion Problems

53. Using the transform in (6) is the same as
    

    

    With discuss how this result in conjunction with (c) of Theorem 4.1.1 can be used to find

54. Proceed as in Problem 53, but this time discuss how to use (7) with in conjunction with (d) and (e) of Theorem 4.1.1 to find
55. Suppose is continuous for and of exponential order. If then use (6) to show that where
56. Make up two functions f₁ and f₂ that have the same Laplace transform. Do not think profound thoughts.

^*The numbers 1, 2, and −4 are the zeros of the common denominator (s − 1)(s − 2)(s + 4).

*The polynomial P(s) is the same as the nth degree auxiliary polynomial in (13) in Section 3.3, with the usual symbol m replaced by s.