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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary Chapter 7: The Laplace Transform Part 1 Department of Electrical Engineering National Taiwan University

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Chapter 7: The Laplace Transform – Part 1

王奕翔

Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw

November 26, 2013

1 / 34 王奕翔 DE Lecture 10

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Solving an initial value problem associated with a linear differential equation:

1 General solution = complimentary solution + particular solution. 2 Plug in the initial conditions to specify the undetermined coefficients.

Question: Is there a faster way?

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

In Chapter 4, 5, and 6, we majorly deal with linear differential equations with continuous, differentiable, or analytic coefficients. But in real applications, sometimes this is not true. For example: E(t) L C R E(t) t Square voltage input: Periodic, Discontinuous. Question: How to solve the current? How to deal with discontinuity?

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

In this lecture we introduce a powerful tool:

Laplace Transform

Invented by Pierre-Simon Laplace (1749 - 1827).

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Overview of the Method

Apply Laplace transform Apply inverse Laplace transform Find unknown y(t) that satisfies DE and initial conditions Transformed DE becomes an algebraic equation in Y(s) Solve transformed equation for Y(s) Solution y(t)

  • f original IVP

−1

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

1 Laplace and Inverse Laplace Transform: Definitions and Basics 2 Solve Initial Value Problems using Laplace Transforms 3 Summary

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Definition of the Laplace Transform

Definition For a function f(t) defined for t ≥ 0, its Laplace Transfrom is defined as F(s) := L {f(t)} := ∫ ∞ e−stf(t)dt, given that the improper integral converges. Note: Use capital letters to denote transforms. f(t)

L

− → F(s), g(t)

L

− → G(s), y(t)

L

− → Y(s), etc. Note: The domain of the Laplace transform F(s) (that is, where the improper integral converges) depends on the function f(t)

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Examples of Laplace Transform

Example Evaluate L {1}. L {1} = ∫ ∞ e−st(1)dt = lim

T→∞

∫ T e−stdt = lim

T→∞

[−e−st s ]T = lim

T→∞

1 − e−sT s . When does the above converge? s > 0! Hence, the domain of L {1} is s > 0, and L {1} = 1 s .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Examples of Laplace Transform

Example Evaluate L {t}. L {t} = ∫ ∞ te−stdt = lim

T→∞

∫ T td (−e−st s ) = lim

T→∞

[−te−st s ]T + ∫ T 1 s e−stdt = lim

T→∞

−Te−sT s + 1 s L {1} . When does the above converge? s > 0! Hence, the domain of L {t} is s > 0, and L {t} = 1 s2 .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of tn

L {tn} = n! sn+1 , n = 0, 1, 2, . . . , s > 0 Proof: One way is to prove it by induction. We will show another proof after discussing the Laplace transform of the derivative of a function.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of eat

L { eat} = 1 s − a, s > a Proof: L { eat} = ∫ ∞ eate−stdt = lim

T→∞

∫ T e−(s−a)tdt = lim

T→∞

[−e−(s−a)t s − a ]T = lim

T→∞

1 − e−(s−a)T s − a When does the above converge? s − a > 0! Hence, the domain of L {eat} is s > a, and L {eat} =

1 s−a.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of sin(kt) and cos(kt)

L {sin(kt)} = k s2 + k2 , L {cos(kt)} = s s2 + k2 , s > 0 Proof: L {sin(kt)} = ∫ ∞ sin(kt)e−stdt = ∫ ∞ sin(kt)d (−e−st s ) = [− sin(kt)e−st s ]∞ + k s ∫ ∞ cos(kt)e−stdt = [− sin(kt)e−st s ]∞ + k sL {cos(kt)} When does the above converge? s > 0! = ⇒ [

− sin(kt)e−st s

]∞

0 = 0

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of sin(kt) and cos(kt)

L {sin(kt)} = k s2 + k2 , L {cos(kt)} = s s2 + k2 , s > 0 Proof: L {cos(kt)} = ∫ ∞ cos(kt)e−stdt = ∫ ∞ cos(kt)d (−e−st s ) = [− cos(kt)e−st s ]∞ − k s ∫ ∞ sin(kt)e−stdt = [− cos(kt)e−st s ]∞ − k sL {sin(kt)} When does the above converge? s > 0! = ⇒ [

− cos(kt)e−st s

]∞

0 = 1 s.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of sin(kt) and cos(kt)

L {sin(kt)} = k s2 + k2 , L {cos(kt)} = s s2 + k2 , s > 0 Proof: { L {sin(kt)} = k

sL {cos(kt)}

L {cos(kt)} = 1

s − k sL {sin(kt)}

Solve the above, we get the result: L {sin(kt)} = k sL {cos(kt)} = k s2 − k2 s2 L {sin(kt)} = ⇒ s2 + k2 s2 L {sin(kt)} = k s2 = ⇒ L {sin(kt)} = k s2 + k2 L {cos(kt)} = s kL {sin(kt)} = s s2 + k2 .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform is Linear

Theorem For any α, β, f(t)

L

− → F(s), g(t)

L

− → G(s), L {αf(t) + βg(t)} = αF(s) + βG(s) Proof: It can be proved by the linearity of integral. Example Evaluate L {sinh(kt)} and L {cosh(kt)}. A: sinh(kt) = 1

2

( ekt − e−kt) , cosh(kt) = 1

2

( ekt + e−kt) . Hence sinh(kt)

L

− → 1 2 ( 1 s − k − 1 s + k ) = k s2 − k2 , s > |k| cosh(kt)

L

− → 1 2 ( 1 s − k + 1 s + k ) = s s2 − k2 , s > |k| .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transforms of Some Basic Functions

f(t) F(s) Domain of F(s) tn n! sn+1 s > 0 eat 1 s − a s > a sin(kt) k s2 + k2 s > 0 cos(kt) s s2 + k2 s > 0 sinh(kt) k s2 − k2 s > |k| cosh(kt) s s2 − k2 s > |k|

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Existence of Laplace Transform

Theorem (Sufficient Conditions for the Existence of Laplace Transform) If a function f(t) is piecewise continuous on [0, ∞), and

  • f exponential order,

then L {f(t)} exists for s > c for some constant c. Definition A function f(t) is of exponential order if ∃ c ∈ R, M > 0, τ > 0 such that |f(t)| ≤ Mect, ∀ t > τ. Note: If f(t) is of exponential order, then ∃ c ∈ R such that for s > c, lim

t→∞ f(t)e−st = 0.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Existence of Laplace Transform

Theorem (Sufficient Conditions for the Existence of Laplace Transform) If a function f(t) is piecewise continuous on [0, ∞), and

  • f exponential order,

then L {f(t)} exists for s > c for some constant c. Proof: For sufficiently large T > τ, we split the following integral: ∫ T f(t)dt = ∫ τ f(t)e−stdt

  • I1

+ ∫ T

τ

f(t)e−stdt

  • I2

. We only need to prove that I2 converges as T → ∞: |I2| ≤ ∫ T

τ

|f(t)e−st|dt = ∫ T

τ

|f(t)|e−stdt ≤ ∫ T

τ

Mecte−stdt, which converges as T → ∞ for s > c since L { ect} exists.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

In this lecture, we focus on functions that are piecewise continuous on [0, ∞), and

  • f exponential order

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of Derivatives

Suppose f(t) is continuous on [0, ∞) and of exponential order, and f′(t) is also continuous on [0, ∞), then the Laplace transform of f′(t) can be found as follows: L {f′(t)} = ∫ ∞ e−stf′(t)dt = ∫ ∞ e−std (f(t)) = [ f(t)e−st]∞

0 + s

∫ ∞ e−stf(t)dt = sL {f(t)} − f(0) , s > c Note: since f(t) is of exponential order, f(t)e−st → 0 as t → ∞ for s > c for some constant c. Similarly, if f′(t) is also of exponential order, we can find L {f′′(t)} = sL {f′(t)} − f′(0) = s2L {f(t)} − sf(0) − f′(0) .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Laplace Transform of Derivatives

Theorem If f, f′, . . . , f(n−1) are continuous on [0, ∞) and are of exponential order, and if f(n)(t) is piecewise continuous on [0, ∞), then L { f(n)(t) } = snF(s) − sn−1f(0) − sn−2f′(0) − · · · − f(n−1)(0), where F(s) := L {f(t)}. Example Evaluate L {tn}. A: Let f(t) = tn. Since f(n)(t) = n!, f(k)(0) = 0 for any 0 ≤ k ≤ n − 1, using the above theorem we get L {n!} = snF(s) = n! s = ⇒ F(s) = n! sn+1 .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Inverse Laplace Transform

L {f(t)} = F(s) ⇐ ⇒ L −1 {F(s)} = f(t) F(s): Laplace transform of f(t) ⇐ ⇒ f(t): inverse Laplace transform of F(s)

Note: Inverse Laplace transform is also linear: L −1 {αF(s) + βG(s)} = αf(t) + βg(t)

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Some Inverse Laplace Transforms

F(s) f(t) Domain of F(s) n! sn+1 tn s > 0 1 s − a eat s > a k s2 + k2 sin(kt) s > 0 s s2 + k2 cos(kt) s > 0 k s2 − k2 sinh(kt) s > |k| s s2 − k2 cosh(kt) s > |k|

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Examples

Example Evaluate L −1 {−2s + 6 s2 + 4 } . Step 1: Decompose −2s + 6 s2 + 4 = −2 s s2 + 4 + 3 2 s2 + 4. Step 2: By the linearity of inverse Laplace transform, L −1 {−2s + 6 s2 + 4 } = −2L −1 { s s2 + 4 } + 3L −1 { 2 s2 + 4 } = −2 cos 2t + 3 sin 2t .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Examples

Example Evaluate L −1 { (s + 3)2 (s − 1)(s − 2)(s + 4) } . Step 1: Decompose into partial fractions: F(s) := (s + 3)2 (s − 1)(s − 2)(s + 4) = A s − 1 + B s − 2 + C s + 4. We can find A, B, C by the following: A = [ (s + 3)2

✘✘✘

(s − 1)(s − 2)(s + 4) ]

s=1

= 16 −5, B = [ (s + 3)2 (s − 1)✘✘✘ (s − 2)(s + 4) ]

s=2

= 25 6 C = [ (s + 3)2 (s − 1)(s − 2)✘✘✘ (s + 4) ]

s=−4

= 1 30 Step 2: Linearity = ⇒ f(t) = −16 5 et + 25 6 e2t + 1 30e−4t .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Solving a First-Order IVP with Laplace Transform

Example Solve y′ + 3y = 13 sin 2t, y(0) = 6. Step 1: Laplace-transform both sides: L { y′} + 3L {y} = 13L {sin 2t} = ⇒ (sY(s) − y(0)) + 3Y(s) = 13 2 s2 + 4 = ⇒ (s + 3)Y(s) = 6 + 26 s2 + 4

Note: Use initial condition y(0) = 6 to compute L {y′} = sY(s) − y(0) = sY(s) − 6.

Step 2: Solve Y(s): Y(s) = 6 s + 3 + 26 (s + 3)(s2 + 4). Step 3: Compute the inverse Laplace transform of Y(s): Y(s) = 8 s + 3 + −2s s2 + 4 + 6 s2 + 4 = ⇒ y(t) = 8e−3t − cos 2t + 3 sin 2t .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Partial fraction decomposition: 26 (s + 3)(s2 + 4) = A s + 3 + Bs + C s2 + 4 A = [ 26

✘✘✘

(s + 3)(s2 + 4) ]

s=−3

= 26 9 + 4 = 2 26 = (Bs + C)(s + 3) + A(s2 + 4) = ⇒ B = −A = −2, C = −3B = 6.

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Solving a Second-Order IVP with Laplace Transform

Example Solve y′′ − 3y′ + 2y = e−4t, y(0) = 1, y′(0) = 5. Step 1: Laplace-transform both sides: L { y′′} − 3L { y′} + 2L {y} = L { e−4t} = ⇒ ( s2Y(s) − sy(0) − y′(0) ) − 3 (sY(s) − y(0)) + 2Y(s) = 1 s + 4 = ⇒ (s2 − 3s + 2)Y(s) = s + 2 + 1 s + 4 Step 2: Solve Y(s): Y(s) = s + 2 (s − 1)(s − 2) + 1 (s + 4)(s − 1)(s − 2). Step 3: Compute the inverse Laplace transform of Y(s): Y(s) = −3 − 1

5

s − 1 + 4 + 1

6

s − 2 +

1 30

s + 4 = ⇒ y(t) = −16 5 e−3t + 25 6 e2t + 1 30e−4t .

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Partial fraction decomposition: s + 2 (s − 1)(s − 2) + 1 (s + 4)(s − 1)(s − 2) = A s − 1 + B s − 2 + C s + 4 A = [ s + 2

✘✘✘

(s − 1)(s − 2) + 1 (s + 4)✘✘✘ (s − 1)(s − 2) ]

s=1

= −3 − 1 5 B = [ s + 2 (s − 1)✘✘✘ (s − 2) + 1 (s + 4)(s − 1)✘✘✘ (s − 2) ]

s=2

= 4 + 1 6 C = [ 1

✘✘✘

(s + 4)(s − 1)(s − 2) ]

s=−4

= 1 30

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

General Procedure of Solving IVP with Laplace Transform

any(n) + an1y(n1) + · · · + a1y0 + a0 = g(t) y(0) = y0, y0(0) = y1, . . . , y(n1)(0) = yn1 Initial Value Problem P(s)Y (s) = Q(s) + G(s) P(s) :=

n

X

k=0

aksk, Q(s) =

n

X

k=0

ak

k−1

X

j=0

yk−1−jsj An Algebraic Equation

Laplace Transform L n y(k)(t)

  • = skY (s) −
  • sk−1y0 + sk−2y1 + · · · + yk−1
  • Solve the

Algebraic Equation

Y (s) = Q(s) P(s) + G(s) P(s)

Partial Fraction Decomposition Inverse Laplace Transform

y(t)

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Short Recap

Definition of Laplace Transform and Inverse Laplace Transform Laplace Transform of some Basic Functions Exponential Order Laplace Transform of Derivatives Solving IVP with Laplace Transforms

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Laplace and Inverse Laplace Transform: Definitions and Basics Solve Initial Value Problems using Laplace Transforms Summary

Self-Practice Exercises

7-1: 3, 5, 13, 15, 29, 35, 43, 50, 53, 54, 55 7-2: 1, 3, 13, 15, 19, 21, 29, 35, 43

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