The Laplace Transform 3

conditions are specified at some point a 6= 0. All that is required is a simple change of independent variable; a translation. For example, the initial-value problem d2 y dy −3 + 2y = x; dx2 dx y(1) = 0, y 0 (1) = 1 is transformed into d2y dy − 3 + 2y = t + 1, 2 dt dt y(0) = 0, y 0(0) = 1 by setting t = x−1. With this transformation, we have: t = 0 when x = 1; x = t+1; and dy dx = d2 y dx2 = dy dt dy dy = ·1 = dt dx dt dt       d dy d dy dt d dy d2 y = = ·1= 2. dx dx dt dx dx dt dt dt 35. Find the solution of the initial-value problem: y 00 − 3y 0 + 2y = x; y(1) = 0, y 0 (1) = 1 by first solving the transformed problem. 36. Use the Laplace transform method to solve the initial-value problem y 0 − 2y = 1 + x; 139 y(2) = −1. 4.4 Applications to Discontinuous Functions In the preceding sections we assumed that the functions under consideration were continuous on [0, ∞). In this section we consider Laplace transforms of certain types of discontinuous functions that occur in various applications. A particular example of the type of discontinuous function that we will be considering in this section is the unit step function, or Heaviside function u. This function is defined on (−∞, ∞) by ( 0 x<0 (1) u(x) = 1 x ≥ 0. The graph of u is y 3 2 1 -10 -5 5 10 x -1 It is clear that u is continuous on (−∞, ∞) except at x = 0. At x = 0, lim u(x) x→0 does not exist. However, note that the left-hand and right-hand limits of u at x = 0 do exist: lim u(x) = 0 and lim u(x) = 1. x→0− x→0+ Recall from calculus that a function f is continuous at a point c if and only if lim f (x) = f (c), and lim f (x) exists if and only if the left-hand and right-hand limits of x→c x→c f at c both exist and are equal. DEFINITION 1. (Jump Discontinuity) Let the function f = f (x) be defined on an interval I and continuous except at a point c ∈ I, c not an endpoint of I. If the left-hand and right-hand limits of f at c both exist but are not equal, then f is said to have a jump (or finite ) discontinuity at c. Example 1. (a) The unit step function u is continuous on (−∞, ∞) except at x = 0 where it has a jump discontinuity. (b) Let f be defined on [−5, 5] by  x+5    1   − f (x) = x   2x    −(x − 5) 140 −5 ≤ x < −2 −2 ≤ x < 0 0≤x≤2 2 λ. Let c be a real number. The translation of the unit step function u by c is the function uc = u(x − c) defined on (−∞, ∞) by ( 0 x 0 is y 2 1 x c -1 Example 4. The Laplace transform of uc , c > 0 is L[u(x − c)] = Proof: By definition L[u(x − c)] = Z e−cs , s > 0. s (3) ∞ e−sx u(x − c) dx 0 = Z c e−sx · 0 dx + 0 = = lim b→∞ ∞ e−sx · 1 dx c Z b b→∞ c lim Z e −sx dx = lim b→∞  e−sx −s b c e−sb e−sc e−cs + = , s > 0. −s s s 143 Note that if c = 0, then u0 = u(x − 0) = u(x) ≡ 1 on [0, ∞), and L[u(x)] = L[1] = 1/s, s > 0 as we saw in Section 4.1.  Example 5. (a) Let f be the function defined on [0, ∞) by ( 1 0≤x<5 f (x) = 2 5 ≤ x < ∞. y 3 2 1 1 5 3 x 7 -1 Then L[f (x)] = Z ∞ e −sx f (x) dx = Z 0 = =  5 e −sx · 1 dx + 0 −e−sx s 5 Z + lim b→∞ 0 Z ∞ e−sx · 2 dx 5 b 2e−sx dx 0  −e−5s 1 2e−sx + + lim s s b→∞ −s b 5 = 2e−sb 2e−5s −e−5s 1 + + lim + s s b→∞ −s s = −e−5s 1 2e−5s 1 e−5s + + = + , s s s s s s > 0. (b) We calculate L[g(x)] where g is the function given in Example 1(c). L[g(x)] = Z =  ∞ e −sx g(x) dx = 0 = Z 1 e −sx · x dx + 0  −sx 1 −xe−sx e − 2 s s 0 + Z 5 e −sx 1   −sx 5 −2e s 1 · 2 dx + Z ∞ e−sx e−(x−5) dx 5 + lim b→∞ Z b e−[x(s+1)−5] dx 0 " #∞ −e−s −e−[x(s+1)−5] 1 −2e−5s 2e−s −e−s + 2 + 2+ + + lim b→∞ s s s s s s+1 5 = − = e−s 1 2e−5s 2e−s e−s e−5s − 2 + 2− + + s s s s s s+1 1 e−s e−s 2e−5s e−5s + − − + . s2 s s2 s s+1 144  The unit step function and its translations can be used to obtain translations of arbitrary functions. For example, if f is defined on [0, ∞) and c > 0, then the function ( 0 x λ. If c > 0, then L[fc (x)] exists for s > λ and is given by L[fc (x)] = L[f (x − c)u(x − c)] = e−cs F (s). Proof: By the definition, L[fc (x)] = L[f (x − c)u(x − c)] = Z ∞ e−sx f (x − c)u(x − c) dx 0 = lim Z b→∞ c b e−sx f (x − c) dx. Now let t = x − c. Then x = t + c, dx = dt, and With this change of variable, Z b lim e−sx f (x − c) dx = b→∞ c lim t = 0 when Z = e = e−cs lim b→∞ Z  145 e−s(t+c) f (t) dt Z b−c e −st  f (t) dt 0 ∞ e−st f (t) dt = e−cs F (s) 0 since b − c → ∞ as b → ∞. b−c b→∞ 0 −cs x = c. This theorem can be expressed in an equivalent manner using the inverse Laplace transform. COROLLARY If L−1 [F (s)] = f (x) and c > 0, then L−1 [e−cs F (s)] = f (x − c)u(x − c) = fc (x). We now illustrate how to use translations of the unit step function and Theorem 2 to calculate Laplace transforms of piecewise continuous functions. Example 6. We calculate L[f (x)] where f is the function given in Example 4: ( 1 0≤x<5 f (x) = 2 5 ≤ x < ∞. Since f has a jump discontinuity at x = 5, we’ll write f in terms of u(x − 5). Define f1 (x) by ( 1 0≤x<5 f1 (x) = = 1 − u(x − 5) 0 5≤x<∞ and f2 (x) by f2 (x) = ( 0 2 0≤x<5 = 2u(x − 5). 5≤x<∞ Then f (x) = f1 (x) + f2 (x) = 1 − u(x − 5) + 2u(x − 5) = 1 + u(x − 5) and L[f (x)] = L[1 + u(x − 5)] = L[1] + L[u(x − 5)] = Example 7. Let h(x) = ( x2 3 1 e−5s + , s s s > 0. 0≤x<2 . Calculate L[h(x)]. 2 ≤ x < ∞. SOLUTION The graph of h is y 4 3 2 1 1 2 x 3 -1 Let h1 (x) = x2 − x2 u(x − 2) and h2 (x) = 3u(x − 2). Then h(x) = h1 (x) + h2 (x) = x2 − x2 u(x − 2) + 3u(x − 2) 146  and L[h(x)] = L[x2 ] − L[x2 u(x − 2)] + 3L[u(x − 2)] = 2 e−2s 2 − L[x u(x − 2)] + 3 . s3 s Before we can apply Theorem 2 to calculate L[x2 u(x − 2)] we must have the coefficient x2 of u(x − 2) expressed as a function of (x − 2). Since x2 = [(x − 2) + 2]2 = (x − 2)2 + 4(x − 2) + 4 we have   L[x2 u(x − 2)] = L (x − 2)2u(x − 2) + 4(x − 2)u(x − 2) + 4u(x − 2) = L[(x − 2)2u(x − 2)] + 4L[(x − 2)u(x − 2)] + 4L[u(x − 2)] = e−2s 2 1 1 + 4e−2s 2 + 4e−2s . s3 s s It now follows that   2 2e−2s 4e−2s e−2s e−2s 2 −2s 2 −2s 1 −2s 1 L[h(x)] = 3 − e + 4e + 4e − − 2 − + 3 = . s s3 s2 s s s3 s3 s s Example 8. Let    2 g(x) = x   4 0≤x<2 2≤x<4 . x≥4 The graph of g is y 4 3 2 1 2 x 4 Set g1(x) = 2 − 2 u(x − 2) g2(x) = x u(x − 2) − x u(x − 4) = [(x − 2) + 2)]u(x − 2) − [(x − 4) + 4]u(x − 4) = (x − 2)u(x − 2) + 2u(x − 2) − (x − 4)u(x − 4) − 4u(x − 4) g3(x) = 4 u(x − 4). Then g(x) = g1 (x) + g2 (x) + g3(x) = 2 + (x − 2)u(x − 2) − (x − 4)u(x − 4). 147  Therefore, L[g(x)] = 2 1 1 + e−2s 2 − e−4s 2 s s s  The Corollary to Theorem 2 is used to calculate inverse Laplace transforms. Example 9. Suppose 1 e−4s . + s2 s − 2 F (s) = Then     −4s   1 e−4s −1 1 −1 e L +L + =L s2 s − 2 s2 s−2     From the Table of Laplace transforms L−1 1/s2 = x. To calculate L−1 e−4s /(s − 2) , let F (s) = 1/(s − 2). Then L−1 [F (s)] = e2x . Therefore, by the Corollary to Theorem 2,  −4s  −1 e L = e2(x−4) u(x − 4). s−2 −1 It now follows that L −1 [F (s)] = x + e 2(x−4) u(x − 4) = ( x x+ e2(x−4) 0≤x<4 . 4≤x<∞  Example 10. Suppose F (s) = Then L −1 [F (s)] = L −1 e−s e−πs + . (s + 1)2 s2 + 4    −πs  e−s e −1 +L . 2 (s + 1) s2 + 4 To calculate L−1 [e−s /(s + 1)2], let F1 (s) = 1/(s + 1)2. Then, from the Table of Laplace transforms, L−1 [F1 (s)] = xe−x . Therefore,   e−s −1 L = (x − 1)e−(x−1) u(x − 1). (s + 1)2 To calculate L−1 [e−πs /(s2 + 4)], let F2 (s) = s2 1 1 2 = . +4 2 s2 + 4 From the Table of Laplace transforms, L−1 [F2(s)] = 12 sin 2x. Therefore,  −πs  e L−1 2 = 12 sin 2(x − π) u(x − π) = 12 sin 2x u(x − π). (sin 2(x − π) = sin 2x) s +4 Finally, L−1 [F (s)] = (x − 1)e−(x−1) u(x − 1) + 12 sin   0  −(x−1) = (x − 1)e   −(x−1) (x − 1)e + 12 sin 2x 148 2x u(x − π) 0≤x<1 1≤x<π x≥π Exercises 4.4 Use the definition of the Laplace transform to find L[f ]. 1. 2. 3. 4. 5. 6.   x 0≤x<1   f (x) = x−1 1≤x<2 .    0 x≥2 ( −1 0≤x<1 f (x) = . x−1 x≥1 ( 0 0≤x<5 f (x) = . 2 x≥5 ( 2x 0≤x<3 f (x) = . 1 x≥3  1 0≤x<2    f (x) = x−2 2≤x<4 .    −(x−4) e x≥4 ( 1 0≤x<1 f (x) = −1 1≤x<2 and extended periodically on [2, ∞). See Exercises 4.1, Problem 10. ( 2x 0≤x<2 7. f (x) = 2 2≤x<4 and extended periodically on [4, ∞). (Sketch the graph of f .) 8. f (x) = x − n on [n, n + 1), n = 0, 1, 2, . . .. (Sketch the graph of f .) Express the function f in terms of the unit step function and its translations. Then find L[f ]. 9. f (x) = ( x2 0≤x<3 3x x≥3   sin x   10. f (x) = sin 2x    sin 3x . 0≤x<π π ≤ x < 2π . Sketch the graph of f . x ≥ 2π 149   0 0≤x<π   11. f (x) = 1 + cos x π ≤ x < 2π .    2 cos x x ≥ 2π   x 0≤x<2   12. f (x) = 0 2≤x<4 .    x≥4 (x − 4)2   x 0≤x<1      2−x 1≤x<3 . Sketch the graph of f . 13. f (x) =  x−4 3≤x<4      0 x≥4 Calculate the inverse Laplace transform of F (s). 14. F (s) = e−s 2e−3s 6e−4s + − . s s s 15. F (s) = 1 + e−πs . s2 + 1 16. F (s) = 1 e−2s − . s+1 s+1 17. F (s) = s + (s − 1)e−πs . s2 + 1 18. F (s) = 2 e−2s 2e−2s 4e−3s + + − . s2 s2 s3 s 19. F (s) = e−2s . s(s + 1) 20. F (s) = 1 − e−2s . s2 + π 2 150