The Laplace Transform. Background: Improper Integrals
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1 The Laplace Transform Background: Improper Integrals Recall: Definite Integral: a, b real numbers, a b; f continuous on [a, b] b a f(x) dx 1
2 Improper integrals: Type I Infinite interval of integration a f(x) dx, b f(x) dx, f(x) dx 2
3 Integrals of the form a f(x) dx Def. a b f(x) dx = lim b a f(x) dx a b 3
4 a f(x) dx converges if b lim b a exists, in which case f(x) dx = L a f(x) dx = L a f(x) dx diverges if b lim b a does not exist. f(x) dx 4
5 The Laplace Transform Definition: Let f be continuous function on [0, ). The Laplace transform of f, denoted L[f(x)], or F(s), is given by L[f(x)] = F(s) = 0 e sx f(x) dx. The domain of F is the set of real numbers s for which the improper integral converges. 5
6 Applications of the Laplace Transform A method for solving initial-value problems for first and second order linear differential equations with constant coefficients. 6
7 Laplace transforms of elementary functions Examples: 1. f(x) c, constant, on [0, ): L[1] = 0 e sx b c dx = lim b 0 e sx c dx 7
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9 2. f(x) = e αx on [0, ): L[e αx ] = 0 e sx e αx dx b = lim b 0 e sx e αx dx = 8
10 3. f(x) = cos βx on [0, ): L[cos βx] = 0 e sx cos βx dx b = lim b 0 e sx cos βx dx Recall, from Calculus II, e sx cos βx dx = e sx s 2 + β2( scos βx β sin βx) Therefore 9
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12 What functions have Laplace Transforms? Definition : f is of exponential order λ if there exists a positive number M and a nonnegative number A such that f(x) Me λx on [A, ). Examples: (a) Bounded functions, e.g., sin x, cos x exponential order 0 10
13 (b) Powers of x: f(x) = x k. exp. order λ for any λ > 0 (c) Exponential fcns: f(x) = e ax. exp. order λ for any λ a 11
14 (d) Complex exponentials : f(x) = e ax cos bx, e ax sin bx, etc. exp. order λ for any λ a (e) f(x) = e x2 is not of exponential order.
15 Theorem: 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. 12
![f(x) F(s) = L[f(x)] c c s, s > 0 e αx 1 s α, s > α cos βx sin βx e αx cos βx e αx sin βx x n, n = 1,2,... x n e αx, n = 1,2,.](/docs-images/79/79316456/images/16-0.jpg ".. x cos βx x sin βx s s 2 + β 2, s > 0 β s 2 + β 2, s > 0 s α (s α) 2 + β 2, s > α β (s α) 2 + β 2, s > α n! s n+1, s > 0 n!")
16 f(x) F(s) = L[f(x)] c c s, s > 0 e αx 1 s α, s > α cos βx sin βx e αx cos βx e αx sin βx x n, n = 1,2,... x n e αx, n = 1,2,... x cos βx x sin βx s s 2 + β 2, s > 0 β s 2 + β 2, s > 0 s α (s α) 2 + β 2, s > α β (s α) 2 + β 2, s > α n! s n+1, s > 0 n! (s α) n+1, s > α s 2 β 2 (s 2 + β 2 ) 2, s > 0 2βs (s 2 + β 2 ) 2, s > 0 13
17 Properties of the Laplace Transform I. L is a linear operator. That is: L[f 1 (x) + f 2 (x)] = L[f 1 (x)] + L[f 2 (x)] L[cf(x)] = cl[f(x)]. 14
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19 II. Laplace transform of derivatives. If f is continuously differentiable and of exponential order λ, then L[f (x)] exists for s > λ and L[f (x)] = sl[f(x)] f(0). 15
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21 Example: L[sin βx] 16
22 If f is twice continuously differentiable with f and f of exponential order λ, then L[f (x)] exists for s > λ and L[f (x)] = s 2 L[f(x)] sf(0) f (0). 17
23 In general, if f, f,, f (n 1) are of exponential order λ, then L[f (n) (x)] exists for s > λ and L[f (n) (x)] = s n L[f(x)] s n 1 f(0) s n 2 f (0) f (n 1) (0). 18
24 III. L[x n f(x)] If L[f(x)] = F(s), then and, in general, L[xf(x)] = df ds, L[x 2 f(x)] = d2 F ds 2 L[x n f(x)] = ( 1) ndn F ds n. 19
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26 Examples: 1. L[1] = 1 s, s > 0 ( ) 1 L[x] = L[x 1] = d ds s = 1 s 2 L[x 2 ] = L[x 2 1] = d2 ds 2 ( ) 1 s = 2 s 3 In general, L[x n ] = n! s n+1 20
27 2. L[e 2x ] = 1 s 2, s > 2 L[xe 2x ] = d ds ( 1 s 2 ) = 1 (s 2) 2 L[x 2 e 2x ] = d2 ds 2 ( 1 s 2 ) = 2 (s 2) 3 L[x n e 2x ] = n! (s 2) n+1 In general: L[x n e αx ] = n! (s α) n+1 21
28 IV. Translation If L[f(x)] = F(s), then L[e rx f(x)] = F(s r). 22
29 Examples: 1. L[cos βx] = F(s) = s s 2 + β 2 L[e αx cos βx] = F(s α) = s α (s α) 2 + β 2 2. L[sin βx] = F(s) = β s 2 + β 2 L[e αx sin βx] = F(s α) = β (s α) 2 + β 2 23
30 Examples: 1. Find the Laplace transform of f(x) = 3 + 4e 3x 2cos 2x. 24
31 1. Find the Laplace transform of f(x) = 3 + 4e 3x 2cos 2x. Answer: F(s) = 3 s + 4 s 3 2s s
32 2. Find the Laplace transform of f(x) = 5x 2 2e 3x sin 2x + 5xe 4x 26
33 2. Find the Laplace transform of f(x) = 5x 2 2e 3x sin 2x + 5xe 4x Answer: F(s) = 10 s 3 4 s 2 + 6s (s 4) 2 27
34 3. Find the Laplace transform of the solution of the initial-value problem: y 2y = 4x; y(0) = 3. 28
35 3. Find the Laplace transform of the solution of the initial-value problem: y 2y = 4x; y(0) = 3. Answer: Y (s) = 3s2 + 4 s 2 (s 2) 29
36 4. Find the Laplace transform of the solution of the initial-value problem: y 2y + 5y = 2x + e x ; y(0) = 2, y (0) = 0. 30
37
38 Ans: Y (s) = s 2 + 2s + 2 s 2 (s + 1)(s 2 2s + 5) + 4 2s s 2 2s
39 f(x) F(s) = L[f(x)] c c s, s > 0 e αx 1 s α, s > α cos βx sin βx e αx cos βx e αx sin βx x n, n = 1,2,... x n e αx, n = 1,2,... x cos βx x sin βx s s 2 + β 2, s > 0 β s 2 + β 2, s > 0 s α (s α) 2 + β 2, s > α β (s α) 2 + β 2, s > α n! s n+1, s > 0 n! (s α) n+1, s > α s 2 β 2 (s 2 + β 2 ) 2, s > 0 2βs (s 2 + β 2 ) 2, s > 0 32
40 Section 4.3. Inverse Laplace transforms Theorem: If f and g are continuous functions on [0, ), and if L[f(x)] = L[g(x)], then f g; that is f(x) = g(x) for all x [0, ). (L is a one-to-one operator.) 33
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42 Def. If F(s) is a given transform and if the function f, continuous on [0, ), has the property that L[f(x)] = F(s), then f is called the inverse Laplace transform of F(s), and is denoted by f(x) = L 1 [F(s)]. The operator L 1 is called the inverse operator of L. 34
43 The operator L 1 is linear; that is L 1 [F(s)+G(s)] = L 1 [F(s)]+L 1 [G(s)] L 1 [c F(s)] = c L 1 [F(s)] Finding inverse Laplace transforms: Crucial step: Partial fraction decomposition! 35
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46 Examples: 1. Given the initial-value problem y + 2y = 3e x, y(0) = 4. (a) Find the Laplace transform of the solution. (b) Find the solution by finding the inverse transform of the result in (a). 36
47 (a) y(x) = e x + 3e 2x 37
48
49 2. F(s) = 2s 1 s 4 + 4s 2 Find f(x) = L 1 [F(s)]. 38
50 Answer: f(x) = x 1 2 cos 2x sin 2x 39
51 3. F(s) = s2 3s 1 (s 2) 2 (s + 4) Find f(x) = L 1 [F(s)]. 40
52 Answer: f(x) = 1 4 e2x 1 2 xe2x e 4x 41
53 4. F(s) = 2s 1 (s + 1)(s 2 4s + 13) Find f(x) = L 1 [F(s)]. 42
54 Answer: f(x) = 1 6 e x e2x cos 3x+ 1 2 e2x sin 3x 43
55 5. Given the initial-value problem y y 2y = sin 2x, y(0) = 1, y (0) = 2. (a) Find the Laplace transform of the solution. (b) Find the solution by finding the inverse transform of the result in (a). 44
56 Answer: y = e2x 2 15 e x cos 2x sin 2x.
57
58 6. Let y = y(x) be the solution of the initial-value problem: y y 6y = 0, y(0) = α, y (0) = 4. Find α such that y = y(x) satisfies lim y(x) = 0. x Answer: α = 2 46
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61 f(x) F(s) = L[f(x)] c c s, s > 0 e αx 1 s α, s > α cos βx sin βx e αx cos βx e αx sin βx x n, n = 1,2,... x n e αx, n = 1,2,... x cos βx x sin βx s s 2 + β 2, s > 0 β s 2 + β 2, s > 0 s α (s α) 2 + β 2, s > α β (s α) 2 + β 2, s > α n! s n+1, s > 0 n! (s α) n+1, s > α s 2 β 2 (s 2 + β 2 ) 2, s > 0 2βs (s 2 + β 2 ) 2, s > 0 47
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