6.2/6.3 Laplace Transform, Derivatives, Integrals, ODE

6.2/6.3 Laplace transform, derivatives, integrals, ODE

Yurii Lyubarskii, NTNU

August 24, 2016

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Remind the deﬁnition: Z ∞ F (s) = Lf (s) = f (t)e−st dt 0 Capital letters will always denote the Laplace transforms of functions denoted by the corresponding small letters

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Rules for derivatives:

L{f 0} = sL{f } − f (0), L{f 00} = s2L{f } − sf (0) − f 0(0), L{f (n)} = snL{f } − sn−1f (0) − sn−2f 0(0) − ... − f (n−1)(0) Assuming all derivatives exist and grow at most exponentially

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Proof for the ﬁrst derivative

Just integration by parts:

Z ∞ (Lf 0)(s) = f 0(t)e−st dt = 0 Z ∞ −st ∞ −st 0 f (t)e |0 − f (t) e dt = 0 Z ∞ s f (t)e−st dt − f (0). 0 Exercise: Prove yourself for the second derivative.

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Rules for integrals :

Z t 1 L f (τ)dτ (s) = F (s). 0 s Proof: R t 0 Denote: g(t) = 0 f (τ)dτ. We have g (t) = f (t) and g(0) = 0. Then

F (s) = L(g 0)(s) = sG(s) − g(0) = sG(s). This is what we need

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Example, inverse Laplace transform :

1 Find L−1 s(s2 + 4) ω Solution: Table formula: L(sin ωt) = s2+ω2 ⇒ 1 1 L−1 (t) = sin 2t. (s2 + 4) 2 Integral rule:

Z t −1 1 1 1 1 L 2 = sin 2τdτ = − cos 2t. s(s + 4) 2 0 4 4

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Applications to ODE

Solve the initial value problem

00 0 0 ay + by + cy = g, y(0) = K0, y (0) = K1

Apply the Laplace transform

2 (as + bs + c)Y − (as + b)K0 − aK1 = G

Then G(s) (as + b)K + aK Y (s) = + 0 1 as2 + bs + c as2 + bs + c We can ﬁnd L{(as2 + bs + c)−1} and use the inverse Laplace transform to compute y.

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Physical interpretation

G(s) (as + b)K + aK Y (s) = + 0 1 as2 + bs + c as2 + bs + c | {z } | {z } contribution of the right handside contribution of the initial conditions

Transfer function 1 Q(s) = as2 + bs + c

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Example 1

y 00 − 9y = 1, y(0) = 1, y 0(0) = 0

2 1 I Apply the Laplace transform s Y − s − 9Y = s I Solve for Y and use partial fractions to write down the answer

s2 + 1 1 5 5 Y (s) = = − + + s(s2 − 9) 9s 9(s − 3) 9(s + 3)

I Find y by performing the inverse transform y(t) = −1/9 + 5/9e3t + 5/9e−3t

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Example 2

y 00 + y 0 − 2y = sin t, y(0) = 0, y 0(0) = 1 1. 1 s2Y − 1 + sY − 2Y = s2 + 1 2. s2 + 2 Y (s) = (s2 + 1)(s2 + s − 2) We want to decompose it using partial fractions (see next slide):

s 1 1 1 Y (s) = −0.1 − 0.3 + 0.5 − 0.4 s2 + 1 s2 + 1 s − 1 s + 2 3. Applying the inverse transform we get

y(t) = −0.1 cos t − 0.3 sin t + 0.5et − 0.4e−2t

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Partial fractions: example

We look for the representation as + b c d Y (s) = + + , s2 + 1 s − 1 s + 2 where s2 +2 = (as +b)(s −1)(s +2)+c(s2 +1)(s +2)+d(s2 +1)(s −1). Now we either open the brackets, compare the coeﬃcients and solve a system of linear equations or we substitute s = 1, 2, i and ﬁnd the constants a = −0.1, b = −0.3, c = 0.5, d = −0.4.

−1 1 Exercise: Find L s(s2+4) by applying partial fraction expansion.

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Laplace transform of discontinuous functions

The building block for discontinuous functions is the step function (Heaviside’s function) uc :

1 uc (t) ( 0, t < c uc (t) = t 1, t ≥ c c For c ≥ 0 we compute its Laplace transform: Z ∞ −cs −st e L{uc }(s) = e dt = , s > 0. c s Further, the change of variables gives the second shift rule −cs L{uc (t)f (t − c)}(s) = e L{f }(s), c ≥ 0. We use it to evaluate the Laplace transform of piecewise deﬁned functions. Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Example

Find the Laplace transform of the function  0, t < 1  f (t) = t − 2, 1 ≤ t ≤ 3 0, t > 3 First we look at the graph of And rewrite the function as this function f (t) = (t − 2)(u3(t) − u1(t))

= (t − 3)u3(t) + u3(t) 1 −(t − 1)u1(t) + u1(t)

0 t Then, using the rules above, we 1 2 3 compute -1 e−3s e−3s e−s e−s Lf (s) = + − + s2 s s2 s

Yurii Lyubarskii, NTNU TMA4120, Lecture 2 Inverse Laplace transform: Example

An important step in the application of the Laplace transform to ODE is to ﬁnd the inverse Laplace transform of the given function. Find f (t) such that L{f } = F is e−2s F (s) = s2 + 2s − 3 First, using the partial functions 1 1 1 1 = − . s2 + 2s − 3 4 s − 1 s + 3 Then we write 1 e−2s e−2s F (s) = − 4 s − 1 s + 3 and using the second shift rule and the table to get u (t) L−1(F )(t) = 2 (et−2 − e−3(t−2)) 4

Yurii Lyubarskii, NTNU TMA4120, Lecture 2