Table Entries: Derivative Rules

Table Entries: Derivative Rules 1. t-derivative rule This is a course on differential equations. We should try to compute . L( f 0 ). (We use the notation f 0 instead of f simply because we think the dot does not sit nicely over the tall letter f .) As usual, let L( f )(s) = F (s). Let f 0 be the generalized derivative of f . (Recall, this means jumps in f produce delta functions in f 0 .) The tderivative rule is L( f 0 ) = sF (s) − f (0− ) L( f 00 ) = s2 F (s) − s f (0− ) − f 0 (0− ) L( f (n) n ) = s F (s) − s n −1 − f (0 ) − s (1) (2) n −2 0 − f (0 ) + . . . + f ( n −1) − (0 ). (3) Proof: Rule (1) is a simple consequence of the definition of Laplace transform and integration by parts. L( f 0 ) = = Z ∞ 0− f 0 (t)e−st dt f (t)e −st ∞ +s 0− u = e−st u0 = −se−st Z ∞ 0− v0 = f 0 (t) v = f (t) f (t)e−st dt = − f (0− ) + sF (s). The last equality follows from: 1. We assume f (t) has exponential order, so if Re(s) is large enough f (t)e−st is 0 at t = ∞. 2. The integral in the second term is none other than the Laplace transform of f (t). Rule (2) follows by applying rule (1) twice. L( f 00 ) = sL( f 0 ) − f 0 (0− ) = s(L( f ) − f (0− )) − f 0 (0− ) = sF (s) − s f (0− ) − f 0 (0− ). Rule (3) Follows by applying rule (1) n times. Notes: 1. We will call the terms f (0− ), f 0 (0− ) the ’annoying terms’. We will be happiest when our signal f (t) has rest initial conditions, so all of Table Entries: Derivative Rules OCW 18.03SC the annoying terms are 0. 2. A good way to think of the t-derivative rules is L( f ) = F (s) L( f 0 ) = sF (s) + annoying terms at 0− . L( f 00 ) = s2 F (s) + annoying terms at 0− . Roughly speaking, Laplace transforms differentiation in t to multiplication by s. 3. The proof of rule (1) uses integration by parts. This is clearly valid if f 0 (t) is continuous at t = 0. It is also true (although we won’t show this) if f 0 (t) is a generalized function. –See example 2 below. Example 1. Let f (t) = e at . We can compute L( f 0 ) directly and by using rule (1). Directly: f 0 (t) = ae at ⇒ L( f 0 ) = a/(s − a). Rule (1): L( f ) = F (s) = 1/(s − a) ⇒ L( f 0 ) = sF (s) − f (0− ) = s/(s − a) − 1 = a/(s − a). Both methods give the same answer. . Example 2. Let u(t) be the unit step function, so u(t) = δ(t). . Directly: L(u) = L(δ) = 1. . Rule (1): L(u) = sL(u) − u(0− ) = s(1/s) − 0 = 1. Both methods give the same answer. Example 3. Let f (t) = t2 + 2t + 1. Compute L( f 00 ) two ways. Solution. Directly: f 00 (t) = 2 ⇒ L( f 00 ) = 2/s. Using rule (3): L( f 00 ) = s2 F (s) − s f (0− ) − f 0 (0− ) = s2 (2/s3 + 2/s2 + 1/s) − s · 1 − 2 = 2/s. Both methods give the same answer. 2. s-derivative rule There is a certain symmetry in our formulas. If derivatives in time lead to multiplication by s then multiplication by t should lead to derivatives in s. This is true, but, as usual, there are small differences in the details of the formulas. The s-derivative rule is L(t f )(s) = − F 0 (s) L(tn f )(s) = (−1)n F (n) (s) (4) (5) (6) 2 Table Entries: Derivative Rules OCW 18.03SC Proof: Rule (4) is a simple consequence of the definition of Laplace transform. F (s) = L( f ) = ⇒ F 0 (s) = = d ds Z ∞ Z ∞ 0− 0− Z ∞ 0− f (t)e−st dt f (t)e−st dt −t f (t)e−st = −L(t f (t)). Rule (5) is just rule (4) applied n times. Example 4. Use the s-derivative rule to find L(t). Solution. Start with f (t) = 1, then F (s) = 1/s. The s-derivative rule now says L(t) = − F 0 (s) = 1/s2 –which we know to be the answer. Example 5. Use the s-derivative rule to find L(te at and L(tn e at ). Solution. Start with f (t) = e at , then F (s) = 1/(s − a). The s-derivative rule now says L(te at ) = − F 0 (s) = 1/(s − a)2 . Continuing: L(t2 e at ) = F 00 (s) = 2/(s − a)3 , L(t3 e at ) = − F 000 (s) = 3 · 2/(s − a)4 , L(t4 e at ) = F (4) (s) = 4 · 3 · 2/(s − a)5 , L(tn e at ) = (−1)n F (n) (s) = n!/(s − a)n+1 . With Laplace, there is often more than one way to compute. We know L(tn ) = n!/sn+1 . Therefore the s-shift rule also gives the above formula for L(tn e at ). 3. Repeated Quadratic Factors Recall the table entries for repeated quadratic factors   1 1 L (sin(ωt) − ωt cos(ω t)) = 2ω 3 ( s2 + ω 2 )2   t s L sin(ωt) = 2 2ω ( s + ω 2 )2   1 s2 L (sin(ωt) + ωt cos(ωt)) = 2ω ( s2 + ω 2 )2 (7) (8) (9) Previously we proved these formulas using partial fractions and factoring the denominators on the frequency side into complex linear factors. Let’s prove them again using the s-derivative rule. 3 Table Entries: Derivative Rules OCW 18.03SC Proof of (8) using the s-derivative rule. Let f (t) = sin(ωt). We know F (s) = plies s2 ω . The s-derivative rule im+ ω2 L(t sin ωt) = − F 0 (s) = ( s2 2ωs . + ω 2 )2 This formula is (8) with the factor of 2ω moved from one side to the other. The other two formulas can be proved in a similar fashion. We won’t give the proofs here. 4