AB2.7: Laplace Transforms. Inverse Transform. Linearity. Shifting

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Valentine Bridges
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1 AB2.7: Laplace Tranform. Invere Tranform. Linearity. Shifting The function F () = L (f) = e t f(t)dt () i called the Laplace tranform of the original function f(t) and i denoted by L (f). The original function f(t) in () i called the invere tranform of F (), and we hall write f(t) = L (F ). EXAMPLE Laplace tranform Let f(t) = for t. Find F (). F () = L (f) = L () = e t dt = e t =, >. The integral in () i called the improper integral and, by definition, i evaluated according to the rule T e t f(t)dt = lim e t f(t)dt. T From thi definition, it follow e t dt = lim [ ] T e t = lim [ T e T + ] e = ( > ). Note that for the limit in the definition of the improper integral doe not exit, and the improper integral itelf doe not exit. EXAMPLE 2 Laplace tranform of an exponential function Let f(t) = e at for t, a = cont. Find F (). F () = L (e at ) = e t e at dt = a e ( a)t = a, > a. Linearity of the Laplace tranform L (af(t) + bg(t)) = al f(t) + bl g(t). EXAMPLE 3 Hyperbolic function Find the Laplace tranform of f(t) = coh at = 2 (eat + e at )

2 . L (coh at) = 2 L (eat ) + 2 L (e at ) = 2 ( a + ) = + a For the Laplace tranform of the hyperbolic ine, we have L (inh at) = 2 L (eat ) 2 L (e at ) = 2 ( a ) = + a 2 a 2, > a. a 2 a 2, > a. EXAMPLE 4 Coine and ine Setting a = iω (i = ), we obtain On the other hand, L (e iωt ) = iω = + iω ( iω)( + iω) = + iω 2 + ω = ω + i ω ω. 2 L (e iωt ) = L (co ωt + i in ωt) = L (co ωt) + il (in ωt), o that L (co ωt) = 2 + ω, L (in ωt) = ω ω. 2 For the Laplace tranform of the hyperbolic ine, we have L (inh at) = 2 L (eat ) 2 L (e at ) = 2 ( a ) = + a a 2 a 2, > a. Shifting THEOREM 5..2 If f(t) ha a tranform F (), where > k, then e at f(t) ha the tranform F ( a), where a > k: L (e at f(t)) = F ( a) PROOF e at f(t) = L (F ( a)). We obtain F ( a) by replacing by a in the integral of the Laplace tranform () of the original function f(t), to obtain F ( a) = F () = L (f) = e ( a)t f(t)dt = e t f(t)dt, e t e at f(t)dt = L (e at f(t)).

3 Taking the invere of boith ide we obtain L (F ( a)) = e at f(t). EXAMPLE 5 Damped vibration Since we have L (co ωt) = L (e at co ωt) = 2 + ω, L (in ωt) = ω ω, 2 a ( a) 2 + ω 2, L (eat in ωt) = For a < the function e at co ωt decribe damped vibration. Important tranform ω ( a) 2 + ω 2. Calculate the Laplace tranform () of the function t n for n =, 2,... uing the induction and integration by part. We have L (t n+ ) = e t t n+ + t= Proceeding further we obtain L (t n+ ) = n + L (t n ) = (n + )n! n The gamma function i defined a e t t n+ dt = e t dt n+ = + n + t n+ d( e t ) = (n + )n L (t n ) =... = Γ(z) = = (n + )n! n+ = e x x z dx. (n + )! n+2. e t t n dt = n + L (t n ). (n + )n! n L () = Uing the gamma function, we obtain the Laplace tranform of t a for arbitrary a > : becaue L (t a ) = e t t a dt = e x x a dx = a+ Γ(z + ) = e x ( x Γ(a + ) a+ e x x z dx. ) a dx =

4 Important Laplace tranform f(t) = : F () = L () = f(t) = t : F () = L (t) = f(t) = t 2 : F () = L (t) = f(t) = t n : F () = L (t n ) = f(t) = t a : F () = L (t a ) = f(t) = e at : F () = L (e at ) = f(t) = co ωt : L (co ωt) = f(t) = in ωt : L (in ωt) = f(t) = coh at : L (coh at) = f(t) = inh at : L (inh at) = f(t) = e at co ωt : L (e at co ωt) = f(t) = e at in ωt : L (e at in ωt) = e t t a dt = f(t) = t n e at : F () = L (t n e at ) = f(t) = te at : F () = L (te at ) = e t dt =. e t tdt = 2. e t t 2 dt = 2! 3. e t t n dt = n! n+. Γ(a + ) a+ (a > ). e t e at dt = a, > a. e t co ωtdt = e t in ωtdt = e t coh atdt = e t inh atdt = e t e at co ωtdt = e t e at in ωtdt = e t t n e at dt = e t te at dt = 2 + ω 2. ω 2 + ω 2, 2 a 2. a 2 a 2. a ( a) 2 + ω 2. ω ( a) 2 + ω 2. n! ( a) n+. ( a) 2.

5 Exitence of Laplace tranform THEOREM 5..3 Let f(t) be a function that i piecewie continuou on every finite interval for t and atifie f(t) Me kt (t ) for ome poitive contant k and M. Then the Laplace tranform of f(t) exit for all > k. PROOF Since f(t) i a piecewie continuou function, e t f(t) i integrable on every finite interval for t. From the condition f(t) Me kt (t ) and auming that > k we obtain L (f) = e t f(t)dt f(t) e t dt Me kt e t dt = M e ( k)t dt = M k, where the condition > k i needed to provide the exitence of the lat integral. Note important particular cae: coh t e t, t > (M = k = ); t n n!e t, t > (M = n!, k = ).

6 PROBLEM 5.. Find the Laplace tranform of f(t) = 2t + 6. L (2t + 6) = 2L (t) + 6L () = PROBLEM 5..2 Find the Laplace tranform of f(t) = a + bt + ct 2. L (a + bt + ct 2 ) = al () + bl (t) + cl (t 2 ) = a + b + 2c 2c + b + a2 = PROBLEM 5..4 Find the Laplace tranform of f(t) = co 2 ωt. L (co 2 ωt) = L ( + co 2ωt) = 2 2 L () + 2 L (co 2ωt) = ω. 2 PROBLEM 5..5 Find the Laplace tranform of f(t) = e a bt. L (e a bt ) = L (e a e bt ) = e a L (e bt ) = ea + b. PROBLEM 5..9 Find the Laplace tranform of f(t) = t for t, f(t) = for t. L (f(t)) = e t ( t)dt = e t dt ( e ) + ( e ) + [ t e t [ e ] ( e ) te t dt = ( e ) ] e t dt = = + e 2. td( e t ) =

7 PROBLEM 5.. Find the Laplace tranform of f(t) = for < t < 4, f(t) = for t > 4 and t <. L (f(t)) = 4 e t dt = e t 4 = e e 4. PROBLEM Find the Laplace tranform of f(t) = t 2 e 3t. f(t) = t 2 : F () = L (t) = e t t 2 dt = 2! 3. Uing the hifting theorem with a = 3 and f(t) = t 2, we obtain the reult L (e at f(t)) = L (e 3t t 2 ) = F ( a) = 2 ( + 3) 3. PROBLEM 5..3 Find the Laplace tranform of f(t) = e αt co βt. Solution. Uing the hifting theorem with a = α and f(t) = co βt, we obtain the reult f(t) = e αt co βt : L (e αt co βt) = e t e αt co βtdt = + α ( + α) 2 + β 2. PROBLEM Find the invere Laplace tranform of F () = ( + ) 2. Solution. By definition of the invere Laplace tranform, f(t) = L (F ). On the other hand, by the hifting theorem with a = and f(t) = t, we have F () = L (te t ) = e t te t dt = So that, by the uniquene of the invere Laplace tranform, f(t) = L (F ) = te t. ( + ) 2.

8 PROBLEM Find the invere Laplace tranform of L (t 3 ) = F () = 2 ( 3) 4. By the hifting theorem with a = 3 and f(t) = 2t 3, F () = 2L (t 3 e 3t ) = 2 e t t 3 dt = 3! 3+ = 6 4. e t t 3 e 3t 6 dt = 2 ( 3). 4 So that, by the uniquene of the invere Laplace tranform, f(t) = L (F ) = 2t 3 e 3t. PROBLEM Find the invere Laplace tranform of F () = = ( + 3) From the table of the Laplace tranform, we have f(t) = in 3t : L (in 3t) = By the hifting theorem with a = 3 and f(t) = in 3t, F () = L (e 3t in 3t) = e t in 3tdt = e t e 3t in 3tdt = So that, by the uniquene of the invere Laplace tranform, f(t) = L (F ) = e 3t in 3t. 3 ( + 3)

Math 22B, Homework #8 1. y 5y + 6y = 2e t

Math 22B, Homework #8 1. y 5y + 6y = 2e t Math 22B, Homework #8 3.7 Problem # We find a particular olution of the ODE y 5y + 6y 2e t uing the method of variation of parameter and then verify the olution uing the method of undetermined coefficient.