LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA MINGFENG ZHAO October 28, 2015 The Laplace transform Definition 1. Let f (t) be a function on [0, ∞), then I. The Laplace transform of f , denoted by L[f ](s), is defined as: Z ∞ L[f (t)](s) = f (t)e−st dt, for all s > 0. 0 II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as: L−1 [F (s)](t) = f (t), for all t > 0. Proposition 1. The followings hold: I. Transforms of derivatives: L[f 0 ](s) = sL[f ](s) − f (0) L[f 00 ](s) = s2 L[f ](s) − sf (0) − f 0 (0). II. First Shifting Property: L e−at f (t) (s) = L[f ](s + a) L−1 [F (s + a)](t) = e−at L−1 [F (s)](t). III. Second Shifting Property: L[u(t − a)f (t − a)](s) = e−as L[f (t)](s) L−1 e−as F (s) = u(t − a)L−1 [F (s)](t − a). 1 2 MINGFENG ZHAO Remark 1. In practice, we can use the second shifting property in the following way: L[u(t − a)f (t)](s) = e−as L[f (t + a)](s). Remark 2. In general, let a0 = 0 < a1 < a2 < a3 < · · · < an < ∞, if f1 (t), if 0 ≤ t < a1 , f2 (t), if a1 ≤ t < a2 , f3 (t), if a2 ≤ t < a3 , f (t) = . .. . . . fn (t), if an−1 ≤ t < an , fn+1 (t), if t ≥ an , then, f (t) = n X fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 [u(t − an ) − u(t − ∞)] k=1 = f1 (t)[1 − u(t − a1 )] + n X fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 (t)u(t − an ). k=2 Notice that when a = 0, we know that u(t − 0) = 1 for all t ≥ 0; when a = ∞, we define u(t − ∞) := 0 for all t ≥ 0. Transform of integrals Proposition 2 (Transform of integrals). There holds that Z t L[f (t)](s) . L f (τ ) dτ (s) = s 0 That is, L−1 Z Proof. Let g(t) = Z t F (s) (t) = L−1 [F (s)](τ ) dτ. s 0 t f (τ ) dτ , by the Fundamental Theorem of Calculus, we know that 0 g(0) = 0, and g 0 (t) = f (t). By the Laplace transform of derivative, we have L[g 0 (t)] = sL[g(t)] − sg(0). LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA 3 Then we get t Z L[f (t)] = sL f (τ ) dτ . 0 Hence t Z L 0 L[f (t)](s) f (τ ) dτ (s) = . s 1 Example 1. Find L . s(s2 + 1) By looking up the table, we have −1 L−1 1 = sin(t). s2 + 1 By the transform of integrals, we have L −1 1 2 s(s + 1) " −1 1 s2 +1 t s L = Z L−1 = 0 # 1 (τ ) dτ s2 + 1 t Z = sin(τ ) dτ 0 1 − cos(t). = So L−1 Example 2. Solve t2 = Z 1 (t) = 1 − cos(t) . s(s2 + 1) t eτ x(τ ) dτ . 0 2 Z Let X(s) = L[x(t)], apply the Laplace transform on the both sides of t = t eτ x(τ ) dτ , we have 0 L[t2 ] = L Z t eτ x(τ ) dτ . 0 By the transform of integrals, then Z L t τ e x(τ ) dτ = 0 = = L [et x(t)] s L[x(t)](s − 1) s X(s − 1) . s By the First Shifting Property 4 MINGFENG ZHAO By looking up the table, we have L[t2 ] = 2 , then s3 X(s − 1) 2 = . 3 s s Then X(s − 1) = 2 , that is, s2 X(s) = 2 . (s + 1)2 By the First Shifting Property, we have x(t) = L−1 [X(s)] = 2e−t t. 2 Z Therefore, the solution to t = t eτ x(τ ) dτ is: 0 x(t) = 2e−t t . 2 Z Remark 3. Another way to solve the integral equation t = t eτ x(τ ) dτ , taking the derivative on the both sides, we 0 get 2t = et x(t), then we have x(t) = 2te−t . Transform of Convolution Definition 2. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as: Z (f ∗ g)(t) = t f (τ )g(t − τ ) dτ. 0 Remark 4. It’s easy to see that the followings are true: f ∗ g = g ∗ f, (cf ) ∗ g = f ∗ (cg) = c(f ∗ g), and (f ∗ g) ∗ h = f ∗ (g ∗ h). Proposition 3. There holds that L[(f ∗ g)(t)] = L[f (t)](s) · L[g(t)](s). That is, L−1 [F (s) · G(s)] = L−1 [F (s)] ∗ L−1 [G(s)](t). LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA 5 Proof. In fact, we have ∞ Z L[f ∗ g] (f ∗ g)(t)e−st dt = 0 ∞ Z e−st = ∞ Z ∞ e−st f (τ )g(t − τ ) dtdτ τ 0 ∞ = ∞ Z g(t − τ )e f (τ ) ∞ ∞ Z dt dτ 0 g(t0 )e−s(t +τ ) dt0 f (τ ) = dτ Let t0 = t − τ 0 0 ∞ Z −st τ 0 Z dt 0 = Z f (τ )g(t − τ ) dτ 0 Z t Z = ∞ Z −sτ 0 f (τ )e g(t )e 0 −st0 0 dt dτ 0 Z ∞ f (τ )e−sτ dτ = L[g(t)] · 0 = L[f (t)] · L[g(t)]. Remark 5. The Convolution Property implies Integral Property: let g(t) = 1, then Z t (f ∗ g)(t) = f (τ ) dτ. 0 By the transform of convolution, we have Z t L[f (t)](s) L f (τ ) (s) = L[f (t)](s) · L[1](s) = . s 0 Example 3. Let f (t) = et and g(t) = t for t ≥ 0, find f ∗ g and L[f ∗ g]. By the definition of f ∗ g, we have Z (f ∗ g)(t) t f (τ )g(t − τ ) dτ = 0 Z t eτ (t − τ ) dτ = 0 Z = t τ t τ eτ dτ e dτ − t 0 = Z 0 t t(et − 1) − τ eτ |0 + Z t 0 = t(et − 1) − tet + et − 1 = et − t − 1. eτ dτ Use the integration by parts 6 MINGFENG ZHAO So we know that the convolution of f (t) = et and g(t) = t is: (f ∗ g)(t) = et − t − 1 . By looking up the table, we have 1 , s−1 L[f (t)] = L[et ] = and L[g(t)] = L[t] = 1 . s2 By the transform of convolution, we have L[f ∗ g](s) = L[f (t)] · L[g(t)] = 1 . s2 (s − 1) 1 Example 4. Find L . s2 (s + 1) By looking up the table, we have −1 −1 L 1 = t, s2 −1 and L 1 = e−t . s+1 By the transform of convolution, then L−1 1 s2 (s + 1) 1 1 · s2 s + 1 1 1 −1 −1 = L ∗L s2 s+1 Z t = τ e−(t−τ ) dτ = L−1 0 −t = e +t − 1. Therefore, we have L−1 1 = e−t +t − 1 . s2 (s + 1) Dirac delta and impulse response Definition 3. For any continuous function f (t) on (−∞, ∞), we have Z ∞ δ(t)f (t) dt = f (0). −∞ LECTURE 20: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA 7 Proposition 4. For any constant a > 0, we have L[δ(t − a)] = e−as . That is, L−1 [e−as ] = δ(t − a). Proof. By the definition of the Laplace transform, we have Z ∞ L[δ(t − a)] = e−st δ(t − a) dt 0 Z ∞ = 0 e−s(t +a) δ(t0 ) dt0 Let t0 = t − a −a = e −as , Since a > 0, by the definition of δ. Problems you can do: Lebl’s Book [2]: All exercises on Page 267, Page 268, Page 274 and Page 275. Braun’s Book [1]: All exercises on Page 250, Page 251, Page 256 and Page 257. Read all materials in Section 2.12 and Section 2.13. References [1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992. [2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014. Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: mingfeng@math.ubc.ca