Section 7.4: Inverse Laplace Transform

Transcription

1 Section 74: Inverse Laplace Transform A natural question to ask about any function is whether it has an inverse function We now ask this question about the Laplace transform: given a function F (s), will there be a function f such that F (s) = L f(s)? It turns out that there is at most one continuous function f which satisfies this property (there could be infinitely many discontinuous functions with the same Laplace transform, but we prefer to work with continuous functions) Definition Given a function F (s), if there is a function f that is continuous on [0, ) and satisfies L f = F, then we say that f is the inverse Laplace transform of F (s) and employ the notation f = L F This idea has more than theoretical interest, however; we ll see in the next section that finding inverse Laplace transforms is a critical step in solving initial value problems To determine the inverse Laplace transform of a function, we try to match it with the form of an entry in the right-hand column of a Laplace table Example Determine L F for (a) F (s) =, (b) F (s) = s s + 9, (c) F (s) = s s s + 5! Solution (a) L = L = t s s (b) L = L = sin s + 9 s + = L (c) L s s s + 5 s (s ) + = e t cos Of course, very often the transform we are given will not correspond exactly to an entry in the Laplace table One tool we can use in handling more complicated functions is the linearity of the inverse Laplace transform, a property it inherits from the original Laplace transform Theorem Assume that L F, L F, and L F exist and are continuous on [0, ) and let c be any constant Then Example Determine L 5 s 6 Solution By linearity, we have 5 L s 6 6s s s + 8s + 0 L F + F = L F + L F, L cf = c L F 6s s s + 8s + 0 = 5 L s 6 6 L s + s + 9 L s + 4s + 5

2 Rewriting the final term using completing the square, it becomes L (s + ) + Therefore, by the Laplace table we see that = 5e 6t 6 cos + e t sin t 5 L s 6 6s s s + 8s + 0 Example Determine L 5 (s + ) 4 Solution The fourth power in the denominator suggests that the inverse Laplace transform is of the form L n! = e at t n (s a) n+ In this case, a =, n =, so by linearity we have L 5 = 5! (s + ) 4 6 L = 5 (s + ) 4 6 e t t Example 4 Determine L s + s + s + 0 Solution Using completing the square, the denominator can be rewritten as s + s + 0 = s + s = (s + ) + Therefore, the form of F (s) suggests the following two formulas from the Laplace table: L s a = e at cos(bt), (s a) + b L b = e at sin(bt), (s a) + b where we have a =, b = Thus, we want to write s + s + s + 0 = A s + (s + ) + + B (s + ) + for an appropriate choice of constants A, B By clearing the denominator, we have the equation s + = A(s + ) + B = As + (A + B) Equating coefficients gives us A =, B = /, so by linearity, we have s + L = L s + s + s + 0 (s + ) + L = e t cos e t sin (s + ) +

3 In Example 4, we found it easier to take the Laplace transform of the broken up fraction than of the original combined fraction The procedure we used above, which you may recall from Calculus and which will prove very useful in simplifying these functions, is called the method of partial fractions Because of its importance, we now review this method Review of Partial Fractions A rational function P (s)/q(s), where P (s), Q(s) are polynomials with the degree of P less than the degree of Q, has a partial fraction expansion based on the factorization of Q There are three main cases to consider, the easiest being that of nonrepeated linear factors If Q(s) is a product of distinct linear factors, Q(s) = (s r )(s r ) (s r n ), then the partial fraction expansion has the form P (s) Q(s) = A + A + + A n s r s r s r n The constants A,, A n can be derived by clearing denominators and either equating coefficients of the resulting polynomials as in Example 4, or by careful substitution We demonstrate the second way in the following examples Example 5 Determine L F for F (s) = 7s (s + )(s + )(s ) Solution Since the denominator has three distinct linear factors, the partial fraction expansion has the form 7s (s + )(s + )(s ) = A s + + B s + + C s, and upon clearing denominators (multiplying through by (s + )(s + )(s )), we have () 7s = A(s + )(s ) + B(s + )(s ) + C(s + )(s + ) If we substitute s-values which make the linear factors zero, then each time two terms will drop out and it will be simple to solve for each coefficient First let s = ; then () becomes Similarly, letting s = gives 8 = 4A A = 5 = 5B B =, and letting s = yields 0 = 0C C =

4 4 Therefore, we can now use linearity to calculate the inverse Laplace transform: L 7s = L (s + )(s + )(s ) s + s + + s = L L + L s + s + s = e t e t + e t The second scenario involves repeated linear factors If the highest power of s r that divides Q(s) is (s r) m, then the portion of the partial fraction expansion corresponding to (s r) m is A s r + A (s r) + + A m (s r) m Example 6 Determine L s + 9s + (s ) (s + ) Solution The partial fraction expansion has the form s + 9s + (s ) (s + ) = A s + B (s ) + C s + Clearing denominators gives () s + 9s + = A(s )(s + ) + B(s + ) + C(s ) Again, we substitute values which make the linear factors vanish Letting s = in () gives = 4B B =, and letting s = yields 6 = 6C C = To find the third constant, we can plug in any other value for s; a convenient choice is s = 0: = A + B + C = A = A A = Therefore, the inverse Laplace transform is s L + 9s + = L (s ) (s + ) s + (s ) = L s = e t + te t e t s + + L (s ) L s + The most difficult case is that of a quadratic factor If m is the highest power of (s α) + β that divides Q(s), then the partial fraction expansion for that term is A s + B (s α) + β + A s + B [(s α) + β ] + + A m s + B m [(s α) + β ] m

5 However, for looking up Laplace transforms it is more convenient to have numerators of the form A i (s α) + βb i, so we can equivalently write A (s α) + βb + A (s α) + βb + + A m(s α) + βb m (s α) + β [(s α) + β ] [(s α) + β ] m Example 7 Determine L s + 0s (s s + 5)(s + ) Solution Since the quadratic factor in the denominator is irreducible, we rewrite using completing the square: s s + 5 = (s ) + Then the partial fraction expansion has the form s + 0s A(s ) + B = + C (s s + 5)(s + ) (s ) + s + Clearing denominators gives () s + 0s = [A(s ) + B](s + ) + C(s s + 5) There are two linear terms involved in this expression, so pick the two s-values which make them vanish When s =, we get When s =, we get For the final constant, set s = 0: 8 = 8C C = = 4B + 4C = 4B 4 4B = 6 B = 4 0 = A + B + 5C A = (4) + 5( ) = Therefore, we find the inverse Laplace transform to be L s + 0s (s ) + (4) = L (s s + 5)(s + ) (s ) + s + = L s + 4 L (s ) + (s ) + Homework: pp 74-75, #-9 odd = e t cos + 4e t sin e t 5 L s +