The Laplace Transform (Sect. 4.1). The Laplace Transform (Sect. 4.1).

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1 The Laplace Transform (Sect. 4.1). s of Laplace Transforms. The Laplace Transform (Sect. 4.1). s of Laplace Transforms.

2 The definition of the Laplace Transform. Definition The function F : D F R is the Laplace transform of a function f : [, ) R iff for all s D F holds, F (s) = e st f (t) dt, where D F R is the set where the integral converges. Remark: The domain D F of F depends on the function f. Notation: We often denote: F (s) = L[f (t)]. This notation L[ ] emphasizes that the Laplace transform defines a map from a set of functions into a set of functions. Functions are denoted as t f (t). The Laplace transform is also a function: f L[f ]. The Laplace Transform (Sect. 4.1). s of Laplace Transforms.

3 Review: Improper integrals. Recall: Improper integral are defined as a limit. t N g(t) dt = lim g(t) dt. N t The integral converges iff the limit exists. The integral diverges iff the limit does not exist. Compute the improper integral Solution: Since N e at dt = lim N e at dt, with a >. e at dt = lim 1 N a lim N e an = for a >, we conclude ( ) e an 1. e at dt = 1 a. The Laplace Transform (Sect. 4.1). s of Laplace Transforms.

4 s of Laplace Transforms. Compute L[1]. Solution: We have to find the Laplace Transform of f (t) = 1. Following the definition we obtain, But L[1] = e at dt = 1 a e st 1 dt = e st dt for a >, and diverges for a. Therefore L[1] = 1, for s >, and L[1] does not exists for s. s In other words, F (s) = L[1] is the function F : D F R given by f (t) = 1, F (s) = 1 s, D F = (, ). s of Laplace Transforms. Compute L[e at ], where a R. Solution: Following the definition of Laplace Transform, L[e at ] = e st e at dt = e (s a)t dt. We have seen that the improper integral is given by e (s a)t dt = 1 (s a) for (s a) >. We conclude that L[e at ] = 1 s a f (t) = e at, F (s) = for s > a. In other words, 1 (s a), s > a.

5 s of Laplace Transforms. Compute L[sin(at)], where a R. Solution: In this case we need to compute L[sin(at)] = lim N N e st sin(at) dt. Integrating by parts twice it is not difficult to obtain: 1 s [ e st sin(at) ] N This identity implies N e st sin(at) dt = a [ e st s 2 cos(at) ] N a2 s 2 N e st sin(at) dt. (1 + a2 s 2 ) N e st sin(at) dt = 1 s [ e st sin(at) ] N a [ e st s 2 cos(at) ] N. s of Laplace Transforms. Compute L[sin(at)], where a R. Solution: Recall the identity: (1 + a2 s 2 ) N e st sin(at) dt = 1 s [ e st sin(at) ] N a [ e st s 2 cos(at) ] N. Hence, it is not difficult to see that ( s 2 + a 2 ) s 2 which is equivalent to L[sin(at)] = e st sin(at) dt = a s 2, s >, a s 2 + a 2, s >.

6 The Laplace Transform (Sect. 4.1). s of Laplace Transforms. A table of Laplace Transforms. f (t) = 1 F (s) = 1 s >, s f (t) = e at F (s) = 1 s > max{a, }, s a f (t) = t n F (s) = n! s >, s (n+1) a f (t) = sin(at) F (s) = s 2 + a 2 s >, s f (t) = cos(at) F (s) = s 2 + a 2 s >, a f (t) = sinh(at) F (s) = s 2 a 2 s > a, s f (t) = cosh(at) F (s) = s 2 a 2 s > a, f (t) = t n e at n! F (s) = s > max{a, }, (s a) (n+1) f (t) = e at sin(bt) F (s) = b (s a) 2 + b 2 s > max{a, }.

7 The Laplace Transform (Sect. 4.1). s of Laplace Transforms. Properties of the Laplace Transform. Theorem (Sufficient conditions) If the function f : [, ) R is piecewise continuous and there exist positive constants k and a such that f (t) k e at, then the Laplace Transform of f exists for all s > a. Theorem (Linear combination) If the L[f ] and L[g] are well-defined and a, b are constants, then L[af + bg] = a L[f ] + b L[g]. Proof: Integration is a linear operation: [a f (t) + b g(t)] dt = a f (t) dt + b g(t) dt.

8 Properties of the Laplace Transform. Theorem (Derivatives) If the L[f ] and L[f ] are well-defined, then holds, L[f ] = s L[f ] f (). (1) Furthermore, if L[f ] is well-defined, then it also holds L[f ] = s 2 L[f ] s f () f (). (2) Proof of Eq (??): Use Eq. (??) twice: L[f ] = L[(f ) ] = sl[(f )] f () = s ( sl[f ] f () ) f (), that is, L[f ] = s 2 L[f ] s f () f (). Properties of the Laplace Transform. Proof of Eq (??): Recall the definition of the Laplace Transform, L[f ] = e st f (t) dt = lim n n e st f (t) dt Integrating by parts, lim n n [( ) e st f (t) dt = lim e st n f (t) n [ ] L[f ] = lim n e sn f (n) f () +s n ] ( s)e st f (t) dt e st f (t) dt = f ()+s L[f ], where we used that lim n e sn f (n) = for s big enough, and we also used that L[f ] is well-defined. We then conclude that L[f ] = s L[f ] f ().

9 The Laplace Transform (Sect. 4.1). s of Laplace Transforms. Laplace Transform and differential equations. Remark: Laplace Transforms can be used to find solutions to differential equations with constant coefficients. Idea of the method: [ ] Differential Eq. L for y(t). (1) Algebraic Eq. for L[y(t)]. (2) (2) Solve the Algebraic Eq. for L[y(t)]. (3) Transform back to obtain y(t). (Using the table.)

10 Laplace Transform and differential equations. Use the Laplace transform to find the solution y(t) to the IVP y + 2y =, y() = 3. Solution: We know the solution: y(t) = 3e 2t. (1): Compute the Laplace transform of the differential equation, L[y + 2y] = L[] L[y + 2y] =. Find an algebraic equation for L[y]. Recall linearity: L[y ] + 2 L[y] =. Also recall the property: L[y ] = s L[y] y(), that is, [ ] s L[y] y() + 2 L[y] = (s + 2)L[y] = y(). Laplace Transform and differential equations. Use the Laplace transform to find the solution y(t) to the IVP y + 2y =, y() = 3. Solution: Recall: (s + 2)L[y] = y(). (2): Solve the algebraic equation for L[y]. L[y] = y() 3, y() = 3, L[y] = s + 2 s + 2. (3): Transform back to y(t). From the table: L[e at ] = 1 s a 3 s + 2 = 3L[e 2t ] 3 s + 2 = L[3e 2t ]. Hence, L[y] = L[3e 2t ] y(t) = 3e 2t.