The Laplace Transform 1

CHAPTER 4 The Laplace Transform 4.1 Introduction The Laplace transform provides an effective method of solving initial-value problems for linear differential equations with constant coefficients. 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 defined in terms of an integral over the interval [0, ∞). Integrals over an infinite 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 Z ∞ L[f (x)] = F (s) = e−sx f (x) dx. (1) 0 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 sufficient 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 first 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 definition 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 differential equation and solution. As indicated above, the primary application of Laplace transforms of interest to us is solving linear differential equations with constant coefficients. 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 βx, p(x) sin βx, p(x)erx cos βx, p(x)erx sin βx where p is a polynomial. We begin by calculating the Laplace transforms of some simple cases of these functions. 115 Example 1. Let f (x) = 1 · e0x ≡ 1 on [0, ∞). By the Definition, Z ∞ Z b −sx L[1] = e · 1 dx = lim e−sx dx b→∞ 0 0 = lim " b→∞ e−sx −s b # = lim  b→∞ 0    e−sb −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→∞ lim b→∞ −1 = 0. sesb Thus, L[1] = 1 , s s > 0.  Example 2. Let f (x) = erx on [0, ∞). Then, Z b Z ∞ rx −sx rx L[e ] = e · e dx = lim e−(s−r)x dx b→∞ 0 0  e−(s−r)x = lim  b→∞ −(s − r) b 0   = lim b→∞ " # e−(s−r)b 1 + . −(s − r) s−r The limit exists (and has the value 0) if and only if s − r > 0. Therefore L[erx ] = 1 , s−r s > r. Note that if r = 0, then we have the result in Example 1.  Example 3. Let f (x) = cos βx on [0, ∞). Then, Z ∞ Z b −sx L[cos βx] = e · cos βx dx = lim e−sx cos βx dx b→∞ 0 0 = lim b→∞ e−sx [−s cos βx − β sin βx] s2 + β 2 b . 0 (Note the integral was calculated using integration by parts; also, it is a standard entry in a table of integrals.) Now,  L[cos βx] = − lim b→∞ 1 e−sb  s cos βb + β sin βb s . · + 2 s2 + β 2 s + β2 Since [s cos βb + β sin βb]/(s2 + β 2 ) is bounded, the limit exists (and has the value 0) if and only if s > 0. Therefore, s L[cos βx] = 2 , s > 0.  s + β2 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 verified using the Definition, some of the integrations involved are complicated. The properties of the Laplace transform presented in the next section provide a more efficient 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 1 , s 1 , s−α s , s2 + β 2 eαx cos βx sin βx eαx cos βx eαx sin βx xn , n = 1, 2, . . . xn erx , s>0 n = 1, 2, . . . x cos βx x sin βx s>α s>0 β , s>0 s2 + β 2 s−α , s>α (s − α)2 + β 2 β , (s − α)2 + β 2 n! sn+1 , s>α s>0 n! , (s − r)n+1 s>r s2 − β 2 , (s2 + β 2 )2 s>0 2βs , + β 2 )2 s>0 (s2 Exercises 4.1 Use the definition of the Laplace transform to find the Laplace transform of the given function. 1. f (x) = x. 2. f (x) = x2. 117 3. f (x) = sin x. 4. f (x) = xerx . 5. f (x) = sinh x. 6. f (x) = cosh x. 7. Use the fact that Z eax cos bx dx = eax [a cos bx + b sin bx] to find L[erx cos βx] a2 + b 2 Z eax sin bx dx = eax [a sin bx − b cos bx] to find L[erx sin βx] a2 + b 2 by the definition. 8. Use the fact that by the definition. Z 9. Show that L[sin x] = 2π e−sx sin x dx 0 1 − e−2πs . 10. Let f be a continuous function on [0, ∞) and suppose that f is periodic with period p. That is f is continuous and f (x + p) = f (x), p > 0, for all x. Show that Z p e−sx f (x) dx L[f (x)] = 0 . 1 − e−ps 118 4.2 Basic Properties of the Laplace Transform In the preceding section we defined the Laplace transform and calculated the Laplace transforms of some of the functions that occur in solving linear differential equations with constant coefficients. In this section we consider the basic question of the existence of the Laplace transform of a function f , and we develop the properties of the Laplace transform that will be used in solving initial value problems. To motivate the material in this section, consider the differential equation y 00 + ay 0 + by = f (x) (2) where a and b are constants and f is a continuous function on [0, ∞). If we assume that y = y(x) is a solution of (1) and formally apply L, we obtain   L y 00(x) + ay 0 (x) + by(x) = L[f (x)]. (3) The right-hand side of this equation suggests the basic question of the existence of L[f (x)]. That is, for what functions f does L[f ] exist? DEFINITION A function f , continuous on [0, ∞), is said to be of exponential order λ, λ a real number, if there exists a positive number M and a nonnegative number A such that | f (x)| ≤ M eλx on [A, ∞). Example 1. (a) If f is a bounded function on [0, ∞) [for example, f (x) = cos βx or f (x) = sin βx], then f is of exponential order 0. f bounded implies that there exists a positive number M such that | f (x)| ≤ M for all x ∈ [0, ∞). Therefore, | f (x)| ≤ M = M e0x on [0, ∞). [Note: if f (x) = cos βx or f (x) = sin βx, then we could take M = 1.] (b) Let f (x) = x on [0, ∞). For any positive number λ, x lim λx = 0 x→∞ e by L’Hôpital’s rule. Therefore, there exists a nonnegative number A such that x ≤ 1 on [A, ∞). eλx This implies that x ≤ eλx = 1 · eλx on [A, ∞) and f (x) = x is of exponential order λ. The same argument can be used to show that f (x) = xα , α any real number, is of exponential order λ for any positive number λ. In general, if p = p(x) is a polynomial, then p is of exponential order λ for any positive number λ. 119 (c) If f (x) = erx , then f is of exponential order λ for any λ ≥ r. 2 (d) Consider the function f (x) = ex . If f is of exponential order λ for some λ, then there exists a positive number M and a nonnegative number A such that 2 ex ≤ M eλx 2 which implies e−λx ex ≤ M on [A, ∞) on [A, ∞). But, 2 lim e−λx ex = lim ex(x−λ) = ∞, x→∞ a contradiction. Thus f (x) = e number λ.  x→∞ x2 is not of exponential order λ for any positive Our first property of L is a sufficient condition for L[f (x)] to exist. We shall omit the proof. THEOREM 1. Let f be a continuous function on [0, ∞). If f is of exponential order λ, then the Laplace transform L[f (x)] = F (s) exists for s > λ. We now turn to the left-hand side of equation (2) where we have L applied to the linear combination y 00 (x) + ay 0 (x) + by(x). THEOREM 2. The operator L is a linear operator. That is, if g and h are continuous functions on [0, ∞), and if each of L[g(x)] and L[h(x)] exists for s > λ, then L[g(x) + h(x)] and L[c g(x)], c constant, each exist for s > λ, and L[g(x) + h(x)] = L[g(x)] + L[h(x)] L[c g(x)] = cL[g(x)]. Proof: L[g(x) + h(x)] = Z ∞ e −sx [g(x) + h(x)] dx = 0 = = = lim Z b→∞ lim b e−sx [g(x) + h(x)] dx 0 Z b→∞ lim b→∞ Z b e −sx g(x) dx + 0 Z e−sx g(x) dx + lim −sx e−sx g(x) dx + L[c g(x)] = cL[g(x)] for any constant c Z b→∞ 0 0 ∞ The proof that 120 e b = L[g(x)] + L[h(x)]   b h(x) dx 0 0 is left as an exercise. Z Z b e−sx h(x) dx ∞ e−sx h(x) dx 0 COROLLARY Let g1(x), g2(x), . . . , gn (x) be continuous functions on [0, ∞). If L[g1(x)], L[g2(x)], . . . , L[gn (x)] all exist for s > λ, and if c1, c2, . . . , cn are real numbers, then L [c1g1(x) + c2 g2(x) + · · · + cn gn (x)] exists for s > λ and L [c1 g1(x) + c2g2 (x) + · · · + cn gn (x)] = c1 L[g1(x)] + c2 L[g2(x)] + · · · + cn L[gn (x)]. According to the Corollary, if L[y 00], L[y 0 ], and L[y] all exist for s > λ for some λ, then L[y 00 + ay 0 + by] = L[y 00] + aL[y 0] + bL[y]. The next property gives a relationship between the Laplace transform of the derivative of a function and the Laplace transform of the function itself. THEOREM 3. Let g be a continuously differentiable function on [0, ∞). If g is of exponential order λ, then L[g 0(x)] exists for s > λ and L[g 0(x)] = sL[g(x)] − g(0). Proof: By the definition of L, we have Z Z ∞ 0 −sx 0 L[g (x)] = e g (x) dx = lim b→∞ 0 Now, using integration by parts, Z b e−sx g 0(x) dx = 0 b e−sx g(x) 0 + s = e −sb b e−sx g 0(x) dx. 0 Z b e−sx g(x) dx 0 g(b) − g(0) + s Z b e−sx g(x) dx 0 Therefore,  0 L[g (x)] = = = lim e b→∞ −sb g(b) − g(0) + s Z b e −sx Z b  g(x) dx 0 lim e−sb g(b) − g(0) + s lim b→∞ b→∞ e−sx g(x) dx 0 lim e−sb g(b) − g(0) + sL[g(x)] b→∞ provided lim e−sb g(b) exists. b→∞ Since g is of exponential order λ, there exist constants M and A such that | g(x)| ≤ M eλx 121 on [A, ∞). Therefore, e−sb g(b) ≤ e−sb M eλb = M e−(s−λ)b for all b > A, and it follows that lim e−sb g(b) = 0 for s > λ. b→∞ We can now conclude that L[g 0(x)] = sL[g(x)] − g(0).  COROLLARY Let g be a function which is n-times continuously differentiable on [0, ∞). If each of the functions g, g 0, . . . , g (n−1) is of exponential order λ, then L[g (n)] exists for s > λ and L[g (n)(x)] = sn L[g(x)] − sn−1 g(0) − sn−2 g 0(0) − · · · − g (n−1)(0). In particular, if g 00 is continuous on [0, ∞), and if g and g 0 are of exponential order λ, then L[g 00(x)] exists for s > λ and L[g 00(x)] = s2 L[g(x)] − sg(0) − g 0(0). The proof of the Corollary follows by induction. Application to Initial-Value Problems Consider the first order initial-value problem y 0 + ay = f (x); y(0) = α where α is a real number and the function f = f (x) has Laplace transform F = F (s). Let y = y(x) be the solution of the problem. Applying the properties of L established above, we have   L y 0 (x) + ay(x) = L[f (x)] L[y 0 (x)] + aL[y(x)] = F (s) sL[y(x)] − y(0) + a L[y(x)] = F (s) (by linearity) (by Theorem 3) (s + a)L[y(x)] − y(0) = F (s) Applying the initial condition y(0) = α, and solving for L[y(x)] = Y (s), we get Y (s) = F (s) + α . s+a 122 Next consider the second order initial-value problem y 00 + ay 0 + by = f (x); y(0) = α, y 0(0) = β. where α and β are real numbers, and f is a function with Laplace transform L[f (x)] = F (s). Let y = y(x) be the solution of this problem. Then   L y 00(x) + ay 0 (x) + by(x) = L[f (x)] (equation (2)) L[y 00(x)] + aL[y 0(x)] + bL[y(x)] = F (s) (by linearity) s2 L[y(x)] − sy(0) − y 0 (0) + a {sL[y(x)] − y(0)} + bL[y(x)] = F (s) (by the Corollary to Theorem 3) (s2 + as + b)L[y(x)] − sy(0) − y 0 (0) − ay(0) = F (s) Applying the initial conditions y(0) = α, y 0(0) = β and solving for L[y(x)] = Y (s), we get Y (s) = F (s) + αs + β + aα . s2 + as + b Implicit in these derivations is the assumption that the Laplace transform of y and its derivatives exist. Assuming that this is the case, the importance of these results is that it gives us the Laplace transform of the solution of an initial-value problem directly. The question now is: Knowing Y (s), what is y(x)? This is the topic of the next section, “inverting” the Laplace transform. Example 2. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value problem y 0 − 2y = 2e−3x ; y(0) = −2. SOLUTION If y = y(x) is the solution, then L[y 0 (x) − 2y(x)] = L[2e−3x ] = 2L[e−3x ] = L[y 0(x)] − 2L[y(x)] = 2 s+3 sL[y(x)] − y(0) − 2L[y(x)] = 2 s+3 (s − 2)L[y(x)] + 2 = 2 s+3 (s − 2)L[y(x)] = 2 −2 s+3 Therefore, L[y(x)] = Y (s) = 2 2 − . (s − 2)(s + 3) s − 2 123  2 s+3 Example 3. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value problem y 00 + 4y = 3e2x; y(0) = 5, y 0 (0) = −2. SOLUTION If y = y(x) is the solution, then L[y 00(x) + 4y(x)] = L[3e2x ] = 3L[e2x] = L[y 00(x)] + 4L[y(x)] = 3 s−2 s2 L[y(x)] − sy(0) − y 0 (0) + 4L[y(x)] = 3 s−2 (s2 + 4)L[y(x)] − 5s − (−2) = 3 s−2 (s2 + 4)L[y(x)] = 3 s−2 3 + 5s − 2 s−2 Therefore, L[y(x)] = Y (s) = 3 5s − 2 + 2 . 2 (s − 2)(s + 4) s + 4  In the next section we will see how to go from Y (s) to y(x). Additional Applications and Properties Although the main use of Theorem 3 and its Corollary is in solving initial-value problems, the results can also be used to determine entries in a table of Laplace transforms. For example, if f is a continuously differentiable function and L[f (x)] is known, then L[f 0 (x)] can be determined: L[f 0(x)] = sL[f (x)] − f (0). Example 4. In Example 3, Section 4.1, we showed that s . s2 + β 2 L[cos βx] = We could use essentially the same calculations to obtain L[sin βx], but recall that the integrations involved are a little “messy.” Here is a simpler way to obtain L[sin βx]. Since [cos βx]0 = −β sin βx, we can write L[−β sin βx] = L[(cos βx)0 ] = sL[cos βx] − cos(0) = −βL[sin βx] = −β 2 . + β2 s2 Therefore, L[sin βx] = s2 β . + β2 124  s2 s2 −1 + β2 Here are two more properties of the Laplace transform that are useful in determining entries in a table of transforms. THEOREM 4. Let f be a continuous function on [0, ∞), and be of exponential order λ. Then F (s) = L[f (x)] has derivatives of all orders, and for s > λ, dn F (s) = (−1)n L [xn f (x)] , dsn n = 1, 2, . . . . Justification By the definition of L[f (x)] = F (s), we have Z ∞ F (s) = e−sx f (x) dx. 0 Differentiation of this equation with respect to s can be justified and yields Z ∞ Z ∞ dF d −sx e−sx [−xf (x)] dx = [e f (x)] dx = ds ds 0 0 = L[−xf (x)] = −L[xf (x)] Therefore, L[xf (x)] = − dF . ds Differentiating a second time, we have   Z ∞
d2F d −sx d dF = = e [−xf (x)] dx 2 ds ds ds ds 0 Z ∞ = e−sx [x2 f (x)] dx = L[x2f (x)] 0 The result stated in the theorem follows by mathematical induction. Example 5. From the table at the end of Section 4.1, L[erx ] =  1 , s > r. Therefore, s−r  1 1 , s > r, = s−r (s − r)2   1 d2 2 2 rx 00 L[x e ] = F (s) = 2 , s > r, = ds s−r (s − r)3   1 d3 6 L[x3erx ] = −F 000(s) = − 3 , s > r, = ds s−r (s − r)4 d L[xe ] = −F (s) = − ds rx and so on. 0   The final property we’ll consider is called the translation property of L. THEOREM 5. If f is a continuous function on [0, ∞), and if L[f (x)] = F (s) exists for s > λ, then for any real number r, L[erx f (x)] = F (s − r) for s > λ + r. 125 Proof: From the definition of the Laplace transform, Z ∞ Z −(s−r)x F (s − r) = e f (x) dx = 0 ∞ e−sx erx f (x) dx 0 = L[erx f (x)].  Example 6. (a) Since L[x] = 1/s2, s > 0, we have L[xerx ] = F (s − r) = (b) Since L[cos βx] = s2 1 , s>r (s − r)2 (c.f. Example 5) s , s > 0, we have + β2 L[erx cos βx] = F (s−r) = s−r , s>r (s − r)2 + β 2 (as indicated in the Table, Section 4.1). Summary of the Properties of L 1. If f is continuous on [0, ∞), and of exponential order λ, then L[f (x)] = F (s) exists for s > λ. (Theorem 1) 2. L is a linear operator (Theorem 2): L[f + g] = L[f ] + L[g]; L[cf ] = cL[f ]. 3. If f is continuously differentiable on [0, ∞), and of exponential order λ, then L[f 0(x)] exists for s > λ and L[f 0 (x)] = sL[f (x)] − f (0). If f is twice continuously differentiable on [0, ∞), and of exponential order λ, then L[f 00(x)] exists for s > λ and L[f 00(x)] = s2 L[f (x)] − sf (0) − f 0 (0). And so on. (Theorem 3) 4. If f is continuous on [0, ∞), and of exponential order λ, then F (s) = L[f (x)] has derivatives of all orders, and for s > λ, dn F (s) = (−1)n L [xn f (x)] , dsn n = 1, 2, . . . . (Theorem 4) 5. If f is continuous on [0, ∞), and if L[f (x)] = F (s) exists for s > λ, then for any real number r, L[erx f (x)] = F (s − r) for s > λ + r. (Theorem 5) 126