The Laplace Transform Inverse of the Laplace Transform Philippe B. Laval KSU Today Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 1 / 12
Outline Introduction Inverse of the Laplace Transform Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 2 / 12
Introduction By now, we should have become effi cient at finding the Laplace transform of a function f. Think of the Laplace transform as a mapping which transforms a given function f (t) into the Laplace transform F (s). What if we wanted to go in the other direction? That is, if we were given the end product, F (s), could we find the function f (t) of which F (s) is the Laplace transform? This is the last step we need before we can apply the Laplace transform to solve differential equations. What we are looking for is the inverse of the Laplace transform. Before we give a formal definition of the inverse of the Laplace transform and look at examples, we explain why such an inverse is useful. Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 3 / 12
Introduction Consider the initial value problem y y = t; y (0) = 0, y (0) = 1. If we take the transform of both sides of our equation, we get L {y y} (s) = L { t} (s). Using the properties of the Laplace transform and letting Y (s) = L {y} (s), we see that L { y y } (s) = L { t} (s) L { y } (s) Y (s) = 1 s 2 s 2 Y (s) 1 Y (s) = 1 s 2 This is an algebraic equation in Y we can solve. The solution is Y (s) = 1 1. Hence, L {y} (s) = Y (s) = s2 s 2. So, we are looking for the function y (t) whose Laplace transform is 1 s 2 in 1 other words, we are looking for the inverse Laplace transform of s 2. From the examples we did in previous sections, we know that L {t} (s) = 1. So, it is natural to think that y (t) = t. s2 Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 4 / 12
Inverse of the Laplace Transform Definition Given a function F (s), if there is a function f (t) that is continuous on [0, ) and satisfies L {f } = F (1) then we say that f (t) is the inverse Laplace transform of F (s). We will use the notation f = L 1 {F }. Remark When we wrote L {f } = F, we omitted the independent variables for clarity. The complete version is L {f (t)} (s) = F (s). Similarly, the complete version of f = L 1 {F } is f (t) = L 1 {F (s)} (t). Remark With the table of Laplace transforms, we already know the inverse Laplace transform of many functions. Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 5 / 12
Inverse of the Laplace Transform Example Using the table of Laplace transforms given in the notes, find the following inverse Laplace transforms. { } 2 1 L 1 s 3 (t) { } 3 2 L 1 s 2 (t) + 9 3 L 1 { s 1 s 2 2s + 5 } (t) Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 6 / 12
In practice, we do not always encounter a Laplace transform that corresponds to an entry in the second column of our table. We will have to use some of the properties of L 1. One such property is linearity. Theorem Suppose that L 1 {F } and L 1 {G} exist and are continuous on [0, ) and let c be any constant. Then L 1 {F + G} = L 1 {F } + L 1 {G} (2) and L 1 {cf } = cl 1 {F } (3) Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 7 / 12
Example { } 10 Find L 1 (s 3) 3 (t) Example { 5 Find L 1 s 6 6s } s 2 + 9 + 3 s 2 + 4s + 5 Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 8 / 12
Looking at the right column of our transform table, we see that the expressions have a very specific format. 1 b They involve fractions of the form (s a) n or s 2 + b 2 or s s 2 + b 2 or b (s a) 2 + b or s a 2 (s a) 2 (up to a constant. Remember if we are + b2 missing a constant, we can always introduce it using linearity). These are precisely the fractions we obtain when we do partial fraction decomposition. We assume the reader is familiar with this technique. We will only comment in the case the denominator has quadratic factors. Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 9 / 12
In a traditional calculus class, [ students are taught that if a fraction contains in its denominator (s α) 2 + β 2] m where (s α) 2 + β 2 is irreducible, that is it cannot be factored further, [ then the portion of the partial fraction expansion that corresponds to (s α) 2 + β 2] m is C 1 s + D 1 (s α) 2 + β 2 + C 2 s + D 2 [ (s α) 2 + β 2] 2 +... + C m s + D m [(s α) 2 + β 2] m However, it is more convenient for us to write it as A 1 (s α) + βb 1 (s α) 2 + β 2 + A 2 (s α) + βb 2 [ (s α) 2 + β 2] 2 +... + A m (s α) + βb m [(s α) 2 + β 2] m This is a better match to some of the entries on the right side of our transform table. Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 10 / 12
Example { Find L 1 2s 2 } + 10s (s 2 2s + 5) (s + 1) 1 What is the form of the partial fraction decomposition? Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 11 / 12
Example { Find L 1 2s 2 } + 10s (s 2 2s + 5) (s + 1) 1 What is the form of the partial fraction decomposition? 2s 2 + 10s A (s 1) + 2B 2 (s 2 = 2s + 5) (s + 1) (s 1) 2 + C + 2 2 s + 1 Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 11 / 12
Example { Find L 1 2s 2 } + 10s (s 2 2s + 5) (s + 1) 1 What is the form of the partial fraction decomposition? 2s 2 + 10s A (s 1) + 2B 2 (s 2 = 2s + 5) (s + 1) (s 1) 2 + C + 2 2 s + 1 3 What is the partial fraction decomposition? Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 11 / 12
Example { Find L 1 2s 2 } + 10s (s 2 2s + 5) (s + 1) 1 What is the form of the partial fraction decomposition? 2s 2 + 10s A (s 1) + 2B 2 (s 2 = 2s + 5) (s + 1) (s 1) 2 + C + 2 2 s + 1 3 What is the partial fraction decomposition? 2s 2 + 10s 3 (s 1) + 2 (4) 4 (s 2 = 2s + 5) (s + 1) (s 1) 2 1 + 2 2 s + 1 Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 11 / 12
Example { Find L 1 2s 2 } + 10s (s 2 2s + 5) (s + 1) 1 What is the form of the partial fraction decomposition? 2s 2 + 10s A (s 1) + 2B 2 (s 2 = 2s + 5) (s + 1) (s 1) 2 + C + 2 2 s + 1 3 What is the partial fraction decomposition? 2s 2 + 10s 3 (s 1) + 2 (4) 4 (s 2 = 2s + 5) (s + 1) (s 1) 2 1 + 2 2 s + 1 { 5 Hence L 1 2s 2 } + 10s (s 2 = 3e t cos 2t + 4e t sin 2t e t 2s + 5) (s + 1) Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 11 / 12
Exercises Do the problems at the end of section 4.3 in my notes on the inverse of the Laplace transform. Philippe B. Laval (KSU) Inverse of the Laplace Transform Today 12 / 12