University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

458

Pages:

9
Author:

Talon Simpson

The Laplace Transform ∞ Z L(f (t)) = e−st f (t) dt = F (s) 0 Example at Z L(e ) = ∞ e 0 −st at Z e dt = ∞ e −(s−a)t 0 1 −(s−a)t dt = − e s −1 1 ,s > a s −a Issues to explore: I For what f does L exist? I Recall how to take the limit (l’Hospital’s Rule). I Recall how to integrate by parts using a table. ∞ = 0 Some Laplace Transforms We showed that L(eat ) = Z L(t) = ∞ te 0 −st dt 1 s−a . ⇒ What is L(1) = 1 s + t e−st − 1 (−1/s)e−st + 0 (1/s 2 )e−st te−st e−st lim − − 2 T →∞ s s T = 0 1 s2 ⇒ Example Use the computation for L(eat ) to compute L(cos(at)) and L(sin(at)) SOLUTION: Complexify cos(at) + i sin(at) = eiat and so L(eiat ) = 1 s + ia = 2 s − ia s + a2 Therefore, L(cos(at)) = s2 s + a2 L(sin(at)) = s2 a + a2 A (Partial) Table of Transforms f (t) F (s) 1. 1 1 s 2. eat 1 s−a 3. t n , n pos int 4. sin(at) 5. cos(at) Notes s>0 n! , s n+1 a , s 2 +a2 s , s 2 +a2 s>a s>0 Sect 6.1 Sect 6.1 Sect 6.1 s > 0 Sect 6.1 s > 0 Sect 6.1 A function is piecewise continuous, it is integrable. Example on the left, not an example on the right: PWC if finite set, t0 , t1 , . . . , tn s.t. I a = t0 < t1 < t2 < · · · < tn = b I f is cont on each ti < t < ti+1 . I lim f (t) exists. +/− t→ti Def: f is of exponential order if it does not grow faster than an exponential function. That is, we can find values of M, a, t0 so that |f (t)| ≤ Meat , for all t > t0 Examples: I All bounded functions (like sin(t), cos(t), etc) If |f (t)| ≤ M for all t, then a = 0 and t0 is any number. I All polynomials. Note that t n = eln(t n) = en ln(t) ≤ ent for all t > 0 (Left as an exercise: ln(t) < t for all t > 0) I “Not an example”: et 2 I Theorem 1: If f is PWC and of exponential order, then L(f (t)) exists. I Theorem 2: L is a linear operator. That is, for any f , g and constants c, L(f (t) + g (t)) = L(f (t)) + L(g (t)) L(c(f (t)) = cL(f (t)) (Easy to prove- These two things are properties of the integral) Example L(3t − 2e3t + 5) = L(3t) + L(−2e3t ) + L(5) = Continue, using linearity and the previous computations: 3L(t) − 2L(e3t ) + 5L(1) = 3 · 1 1 1 −2 +5 2 s s −3 s The Inverse Transform The inverse function to the Laplace transform exists as a complex integral, and so we will not formally define it. Rather, we will compute the inverse transform with the utilization of a few facts: I The inverse Laplace transform exists. I The inverse Laplace transform is (also) linear. I A table of transforms will be provided- use it to invert. Example: If F (s) = s , s 2 +4 what is f (t)? SOLUTION: Looking at our list of transforms, we see that f (t) = cos(2t). For homework examples from 6.1/6.2, go to the next video.

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