Laplace Transforms

Definition Given an integrable function 𝑓: ℝ ≥0 →ℝ we define the Laplace Transform of 𝑓 𝐹=ℒ𝑓 to be the function 𝐹:𝐷→ℝ 𝐹 𝑠 = 0 ∞ 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 Where 𝐷, the domain of ℒ𝑓, is the domain of 𝑠 for which the integral converges.

Example Compute the Laplace Transform of the function 𝑓 𝑡 =1 Solution Let 𝐹=ℒ𝑓 be the Laplace Transform, and by definition we have: 𝐹 𝑠 = 0 ∞ 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 0 ∞ 𝑒 −𝑠𝑡 𝑑𝑡 = − 1 𝑠 𝑒 −𝑠𝑡 𝑡=0 𝑡=∞ = 1 𝑠 Thus we have our first Laplace Transform formula: ℒ 1 = 1 𝑠

Example Compute the Laplace Transform of 𝑓 𝑡 = 𝑒 𝑘𝑡 Solution ℒ 𝑒 𝑘𝑡 = 0 ∞ 𝑒 −𝑠𝑡 𝑒 𝑘𝑡 𝑑𝑡 = 0 ∞ 𝑒 𝑘−𝑠 𝑡 𝑑𝑡 = 1 𝑘−𝑠 𝑒 𝑘−𝑠 𝑡 𝑡=0 𝑡→∞ = 1 𝑠−𝑘 , 𝑠>𝑘 ℒ 𝑒 𝑘𝑡 = 1 𝑠−𝑘

Example Suppose 𝑓 is a differentiable function on ℝ ≥0 whose Laplace Transform 𝐹=ℒ𝑓 is known. Then compute the Laplace Transform of the derivative 𝑓′. Solution Integrate by parts: 𝑢= 𝑒 −𝑠𝑡 𝑑𝑣= 𝑓 ′ 𝑡 𝑑𝑡 𝑑𝑢=−𝑠 𝑒 −𝑠𝑡 𝑑𝑡 𝑣=𝑓(𝑡) ℒ 𝑓 ′ 𝑡 = 0 ∞ 𝑒 −𝑠𝑡 𝑓 ′ 𝑡 𝑑𝑡 = 𝑒 −𝑠𝑡 𝑓 𝑡 𝑡=0 𝑡→∞ + 0 ∞ 𝑠 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 =−𝑓 0 +𝑠 ℒ 𝑓 𝑡 =−𝑓 0 +𝑠 𝐹(𝑠) ℒ 𝑓 ′ 𝑡 =𝑠 𝐹 𝑠 −𝑓(0) where 𝐹=ℒ𝑓

Example Compute the Laplace Transform of the second derivative 𝑓 ′′ 𝑡 Solution Recall from the previous problem that ℒ 𝑔 ′ 𝑡 =𝑔 0 +𝑠 ℒ 𝑔 𝑡 Let 𝑔=𝑓′ and this formula becomes ℒ 𝑓 ′′ 𝑡 =− 𝑓 ′ 0 +𝑠 ℒ 𝑓 ′ 𝑡 =− 𝑓 ′ 0 +𝑠 −𝑓 0 +𝑠 𝐹 𝑠 = 𝑠 2 𝐹 𝑠 −𝑠𝑓 0 −𝑓′(0) ℒ 𝑓 ′′ 𝑡 = 𝑠 2 𝐹 𝑠 −𝑠𝑓 0 −𝑓′(0)

Example Prove that the Laplace Transform ℒ is a linear operator: for constants 𝑎 and 𝑏 and functions 𝑓 and 𝑔: ℒ 𝑎𝑓 𝑡 +𝑏𝑔 𝑡 =𝑎ℒ 𝑓 𝑡 ++𝑏ℒ 𝑔 𝑡 Solution ℒ 𝑎𝑓 𝑡 +𝑏𝑔 𝑡 = 0 ∞ 𝑒 −𝑠𝑡 𝑎𝑓 𝑡 +𝑏𝑔 𝑡 𝑑𝑡 =𝑎 0 ∞ 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 +𝑏 0 ∞ 𝑒 −𝑠𝑡 𝑔 𝑡 𝑑𝑡 =𝑎ℒ 𝑓 𝑡 +𝑏ℒ 𝑔 𝑡 ℒ 𝑎𝑓 𝑡 +𝑏𝑔 𝑡 =𝑎𝐹 𝑠 +𝑏𝐺(𝑠)

Example Compute the Laplace Transforms of 𝑓 𝑡 = sin 𝜔𝑡 and 𝑔 𝑡 = cos 𝜔𝑡 Solution ℒ sin 𝜔𝑡 = 0 ∞ 𝑒 −𝑠𝑡 sin 𝜔𝑡 𝑑𝑡 = − 𝑒 −𝑠𝑡 𝜔 cos 𝜔𝑡 +𝑠 sin 𝜔𝑡 𝑠 2 + 𝜔 2 𝑡=0 𝑡→∞ = 𝜔 𝑠 2 + 𝜔 2 = 𝑒 −𝑠𝑡 −𝑠 cos 𝜔𝑡 +𝜔 sin 𝜔𝑡 𝑠 2 + 𝜔 2 𝑡=0 𝑡→∞ ℒ cos 𝜔𝑡 = 0 ∞ 𝑒 −𝑠𝑡 cos 𝜔𝑡 𝑑𝑡 = 𝑠 𝑠 2 + 𝜔 2 ℒ sin 𝜔𝑡 = 𝜔 𝑠 2 + 𝜔 2 ℒ cos 𝜔𝑡 = 𝑠 𝑠 2 + 𝜔 2

Example (Power Rule for Laplace Transforms) Compute the Laplace Transform of 𝑓 𝑡 = 𝑡 𝑛 for positive integers 𝑛. Solution Integrate by parts: 𝑢= 𝑡 𝑛 𝑑𝑣= 𝑒 −𝑠𝑡 𝑑𝑡 𝑑𝑢=𝑛 𝑡 𝑛−1 𝑣=− 1 𝑠 𝑒 −𝑠𝑡 ℒ𝑓 𝑠 = 0 ∞ 𝑡 𝑛 𝑒 −𝑠𝑡 𝑑𝑡 = − 1 𝑠 𝑡 𝑛 𝑒 −𝑠𝑡 𝑡=0 𝑡=∞ + 𝑛 𝑠 0 ∞ 𝑡 𝑛−1 𝑒 −𝑠𝑡 = 𝑛 𝑠 ℒ 𝑡 𝑛−1 (𝑠) This is a recursive formula for ℒ 𝑡 𝑛 in terms of ℒ 𝑡 𝑛−1 . Since ℒ 1 = 1 𝑠 , we conclude that ℒ 𝑡 𝑛 = 𝑛! 𝑠 𝑛+1 (this formula can be proved by induction)

The last several slides were preparation. Now an application, to give a taste of how the Laplace Transform is used.

Example Use Laplace Transforms to solve the differential equation/initial value problem 𝑦 ′′ +3𝑦= 𝑒 −𝑡 , 𝑦 0 =1, 𝑦 ′ 0 =2 Solution We begin by taking the Laplace Transform of both sides, letting 𝑌=ℒ𝑦 ℒ 𝑦 ′′ +3𝑦 =ℒ 𝑒 −𝑡 ℒ 𝑦 ′′ +3ℒ 𝑦 = 1 𝑠+1 ℒ sin 𝜔𝑡 = 𝜔 𝑠 2 + 𝜔 2 ℒ cos 𝜔𝑡 = 𝑠 𝑠 2 + 𝜔 2 𝑠 2 𝑌 𝑠 −𝑠𝑦 0 −𝑦′(0) +3𝑌 𝑠 = 1 𝑠+1 𝑌(𝑠)= 𝑠 2 +3𝑠+3 ( 𝑠 2 +3)(𝑠+1) Solve for 𝑌(𝑠). Recall 𝑦 0 =1, 𝑦 ′ 0 =2: ℒ 𝑒 𝑘𝑡 = 1 𝑠−𝑘 Break apart the right-hand-side with partial fractions and simplify: 𝑌 𝑠 = 1/4 𝑠+1 + 9/4 𝑠 2 +3 + 3/4 𝑠 𝑠 2 +3 = 1 4 1 𝑠−(−1) + 3 3 4 3 𝑠 2 +3 + 3 4 𝑠 𝑠 2 +3 * 𝑦 𝑡 = 1 4 𝑒 −𝑡 + 3 3 4 sin 3 𝑡 + 3 4 cos 3 𝑡 Conclude that

Laplace Transforms allow us to work more easily with “ugly” piecewise-defined functions.

Example Compute the Laplace Transform of the Heaviside Step Function at 𝑎 𝑢 𝑎 𝑡 = 0 0≤𝑡<𝑎 1 𝑎≤𝑡 Solution ℒ 𝑢 𝑎 𝑡 = 0 ∞ 𝑒 −𝑠𝑡 𝑢 𝑎 𝑡 𝑑𝑡 = 0 𝑎 𝑒 −𝑠𝑡 0 𝑑𝑡 + 𝑎 ∞ 𝑒 −𝑠𝑡 1 𝑑𝑡 = 𝑒 −𝑎𝑠 𝑠 ℒ 𝑢 𝑎 𝑡 = 𝑒 −𝑎𝑠 𝑠

Example (t-Shift-Rule for Laplace Transforms) Suppose we know the Laplace Transform ℒ 𝑓 𝑡 =𝐹(𝑠) of a function 𝑓. Find the Laplace Transform of the shifted function (where 𝑎 is a number 𝑎≥0) 𝜏 𝑎 𝑓 𝑡 = 𝑢 𝑎 𝑡 𝑓 𝑡−𝑎 Solution ℒ 𝜏 𝑎 𝑓(𝑡) = 0 ∞ 𝜏 𝑎 𝑓(𝑡) 𝑒 −𝑠𝑡 𝑑𝑡 = 𝑎 ∞ 𝑓 𝑡−𝑎 𝑒 −𝑠𝑡 𝑑𝑡 Substitute 𝑢=𝑡−𝑎, 𝑑𝑢=𝑑𝑡 = 0 ∞ 𝑓 𝑢 𝑒 −𝑠 𝑢+𝑎 𝑑𝑢 = 𝑒 −𝑎𝑠 0 ∞ 𝑓 𝑢 𝑒 −𝑠𝑢 𝑑𝑢 = 𝑒 −𝑎𝑠 𝐹(𝑠) ℒ 𝜏 𝑎 𝑓 𝑡 = 𝑒 −𝑎𝑠 𝐹(𝑠) Where 𝜏 𝑎 𝑓 𝑡 = 𝑢 𝑎 𝑡 𝑓 𝑡−𝑎

Example Compute the Laplace Transform of the floor function 𝑓 𝑡 = 𝑡 Solution Notice that we can write 𝑓 𝑡 = 𝑛=1 ∞ 𝑢 𝑛 𝑡 Therefore ℒ𝑓 𝑠 = 𝑛=1 ∞ 𝑒 −𝑛𝑠 𝑠 = 1 𝑠 𝑛=0 ∞ 𝑒 −𝑠 𝑒 −𝑠 𝑛 = 1 𝑠 𝑒 −𝑠 1− 𝑒 −𝑠 Here we used the Geometric Series formula 𝑛=0 ∞ 𝑎 𝑟 𝑛 = 𝑎 1−𝑟

Example Solve the differential equation 𝑦 ′ = 𝑡 , 𝑦 0 =0 Solution We take the Laplace Transform of both sides, using the previous problem for the right-hand-side. ℒ 𝑦 ′ =ℒ 𝑡 𝑦 0 +𝑠 𝑌 𝑠 = 𝑛=1 ∞ 𝑒 −𝑛𝑠 𝑠 𝑌 𝑠 = 𝑛=1 ∞ 𝑒 −𝑛𝑠 𝑠 2 We now take the inverse Laplace Transform of both sides: 𝑦 𝑡 = 𝑛=1 ∞ 𝑢 𝑛 𝑡 𝑡−𝑛

Example (Laplace Transform of Periodic Functions) Compute ℒ𝑓 where 𝑓 is a periodic function of period 𝐿 Solution = 𝑛=0 ∞ 𝑛𝐿 𝑛+1 𝐿 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 𝑛=0 ∞ 0 𝐿 𝑓 𝑢 𝑒 −𝑠 𝑢+𝑛𝐿 𝑑𝑢 ℒ𝑓 𝑠 = 0 ∞ 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 𝑛=0 ∞ 𝑒 −𝑠𝑛𝐿 0 𝐿 𝑓 𝑢 𝑒 −𝑠𝑢 𝑑𝑢 = 0 ∞ 𝑒 −𝑠𝑛𝐿 ℒ 𝑓 (𝑠) Where 𝑓 𝑡 = 𝑓(𝑡) 0≤𝑡≤𝐿 0 𝐿<𝑡 Although the last formula appears “cleaner”, the penultimate formula is more useful in practice: = ℒ 𝑓 𝑠 1− 𝑒 −𝑠𝐿 If 𝑓 is 𝐿-periodic, then ℒ𝑓(𝑠)= 𝑛=0 ∞ 𝑒 −𝑠𝑛𝐿 0 𝐿 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡

Example Find the Laplace Transform of the square-wave function of period 2𝑎 Solution Using the periodic function formula with 𝐿=2𝑎 If 𝑓 is 𝐿-periodic, then ℒ𝑓(𝑠)= 𝑛=0 ∞ 𝑒 −𝑠𝑛𝐿 0 𝐿 𝑓 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 Where 0 2𝑎 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 1 𝑠 1− 𝑒 −𝑎𝑠 2 = 1 𝑠 − 2 𝑒 −𝑎𝑠 𝑠 + 𝑒 −2𝑎𝑠 𝑠 Obtain ℒ𝑓 𝑠 = 𝑛=0 ∞ 𝑒 −2𝑎𝑛𝑠 1 𝑠 − 2 𝑒 −𝑎𝑠 𝑠 + 𝑒 −2𝑎𝑠 𝑠 ℒ𝑓 𝑠 = 𝑛=0 ∞ 𝑒 −2𝑛𝑎𝑠 𝑠 − 2 𝑒 − 2𝑛+1 𝑎𝑠 𝑠 + 𝑒 − 2𝑛+2 𝑎𝑠 𝑠 ℒ𝑓 𝑠 =− 1 𝑠 + 2 𝑠 𝑛=0 ∞ 𝑒 −2𝑛𝑎𝑠 − 𝑒 − 2𝑛+1 𝑎𝑠 Where 𝑓 is the unit square-wave of period 2𝑎

Example (s-shift formula) Suppose we know the Laplace transform of 𝑓(𝑡) is 𝐹(𝑠). Find a function whose Laplace transform is 𝐹 𝑠+𝑎 , where 𝑎 is a real number. Solution We are looking for a function 𝑓 𝑡 with ℒ 𝑓 𝑠 =𝐹 𝑠+𝑎 Write out the definition of the Laplace Transform: 0 ∞ 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡 = 0 ∞ 𝑒 − 𝑠+𝑎 𝑡 𝑓 𝑡 𝑑𝑡 Set the integrands equal to each other to find that 𝑒 −𝑠𝑡 𝑓 𝑡 = 𝑒 − 𝑠+𝑎 𝑡 𝑓 𝑡 𝑓 𝑡 = 𝑒 −𝑎𝑡 𝑓 𝑡 Conclude: If ℒ𝑓=𝐹, then ℒ 𝑒 −𝑎𝑡 𝑓 𝑡 =𝐹 𝑠+𝑎

Example A mass-spring system with resistance is acted upon by a alternating force modelled by the square wave function 𝑓(𝑡) with period 2𝐿 and (semi-)amplitude 𝐴. Solve the differential equation assuming the mass starts at rest at 𝑥=0. Assume underdamping ( 𝛽 2 −4𝑚𝑘<0) Solution 𝑚 𝑥 ′′ +𝛽 𝑥 ′ +𝑘𝑥=𝑓 𝑡 Take the Laplace Transform of both sides 𝑚 𝑠 2 𝑋+𝛽𝑠𝑋+𝑘𝑋=− 𝐴 𝑠 + 2𝐴 𝑠 𝑛=0 ∞ 𝑒 −2𝑛𝑎𝑠 − 𝑒 − 2𝑛+1 𝑎𝑠 𝑋=− 𝐴 𝑠 𝑚 𝑠 2 +𝛽𝑠+𝑘 + 2𝐴 𝑠 𝑚 𝑠 2 +𝛽𝑠+𝑘 𝑛=0 ∞ 𝑒 −2𝑛𝑎𝑠 − 𝑒 − 2𝑛+1 𝑎𝑠 𝛾= 𝛽 2𝑚 𝜔= 4𝑚𝑘− 𝛽 2 2𝑚 Write 𝑚 𝑠 2 +𝛽𝑠+𝑘=𝑚 𝑠+𝛾 2 + 𝜔 2 , where 𝛾= 𝛽 2𝑚 and 𝜔= 4𝑘𝑚− 𝛽 2 2𝑚 = −𝐴/𝑚 𝑠 𝑠+𝛾 2 + 𝜔 2 + 2𝐴/𝑚 𝑠 𝑠+𝛾 2 + 𝜔 2 𝑛=0 ∞ 𝑒 −2𝑛𝐿𝑠 − 𝑒 − 2𝑛+1 𝐿𝑠 Apply partial fractions. = 𝐴 𝑚 𝛾 2 + 𝜔 2 1 𝑠 − 𝑠+2𝛾 𝑠+𝛾 2 + 𝜔 2 −1+2 𝑛=0 ∞ 𝑒 −2𝑛𝐿𝑠 − 𝑒 − 2𝑛+1 𝐿𝑠

Example A mass-spring system with resistance is acted upon by a alternating force modelled by the square wave function 𝑓(𝑡) with period 2𝐿 and (semi-)amplitude 𝐴. Solve the differential equation assuming the mass starts at rest at 𝑥=0. Assume underdamping ( 𝛽 2 −4𝑚𝑘<0) Solution 𝑋(𝑠)= 𝐴 𝑚 𝛾 2 + 𝜔 2 1 𝑠 − 𝑠+2𝛾 𝑠+𝛾 2 + 𝜔 2 −1+2 𝑛=0 ∞ 𝑒 −2𝑛𝐿𝑠 − 𝑒 − 2𝑛+1 𝐿𝑠 ℒ sin 𝜔𝑡 = 𝜔 𝑠 2 + 𝜔 2 ℒ cos 𝜔𝑡 = 𝑠 𝑠 2 + 𝜔 2 = 𝐴 𝑚 𝛾 2 + 𝜔 2 1 𝑠 − 𝑠+𝛾 𝑠+𝛾 2 + 𝜔 2 − 𝛾 𝜔 𝜔 𝑠+𝛾 2 + 𝜔 2 −1+2 𝑛=0 ∞ 𝑒 −2𝑛𝐿𝑠 − 𝑒 − 2𝑛+1 𝐿𝑠 Let 𝑔 𝑡 =1− 𝑒 −𝛾𝑡 cos 𝜔𝑡 − 𝛾 𝜔 sin 𝜔𝑡 and 𝐵= 𝐴 𝑚 𝛾 2 + 𝜔 2 If ℒ𝑓=𝐹, then ℒ 𝑒 −𝑎𝑡 𝑓 𝑡 =𝐹 𝑠+𝑎 𝑥(𝑡)=−𝐵𝑔(𝑡)+2𝐵 𝑛=0 ∞ 𝜏 2𝑛𝐿 𝑔 𝑡 − 𝜏 2𝑛+1 𝐿 𝑔(𝑡) ℒ 𝜏 𝑎 𝑓 𝑡 = 𝑒 −𝑎𝑠 𝐹(𝑠) Where 𝜏 𝑎 𝑓 𝑡 = 𝑢 𝑎 𝑡 𝑓 𝑡−𝑎

Example Plot the solution to the previous problem with 𝑚=2.3 kg, 𝛽=0.4 kg/s, 𝑘=25 N/m, 𝐴=20 N, 𝐿=3 s.

Example Same example: 𝑚=0.3 kg, 𝛽=1.2 kg/s, 𝑘=30 N/m, 𝐴=40 N, 𝐿=12 s