Laplace Transforms 9781737462460, 173746246X

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Laplace Transforms Laplace Transforms Robert P. Massé Subtitle Second Edition

Lorem Ipsum Dolor Facilisis

Copyright © 2021 Robert P. Massé

Also by Robert P. Massé Physics: Nature of Physical Fields and Forces

All rights reserved.

Second Edition

Physics: Where It Went Wrong Vectors and Tensors of Physical Fields

ISBN: 978-1-7374624-6-0

Prime Numbers and Congruence Theory Complex Variables

Massé, Robert P. Laplace Transforms

i

Dedication This book is dedicated to my beloved wife Kay whose loving and insightful support made it possible

ii

Preface Mathematics is an essential tool in the continuing struggle to understand the physical world.

iii

! The subject of this book is the Laplace transform which is a particular mathematical transform. The Laplace transform provides an extremely powerful method for solving linear differential equations involving initial conditions. It does this by converting differential equations into algebraic equations. Differentiation is thereby replaced with algebraic manipulations. Even many differential equations having impulsive or discontinuous forcing functions can be easily solved using Laplace transforms. These transforms therefore find application in the numerous diverse fields where differential equations are used to describe changes in an entity. ! The material in this book is organized and presented in a manner designed to make this a comprehensive reference work on Laplace transforms. Most of the sources for this book are given in the references at the end of the book. ! The research that forms the foundation of this book could not have been accomplished without the outstanding interlibrary loan programs of Florida made available through the good resources of the Tampa Bay Library Consortium.

!

!

First Edition – January 18, 2021

!

Second Edition – June 29, 2021

Robert P. Massé

iv

4! Derivatives and Integrals of Laplace Transforms

Contents

!

4.1!

Derivatives of Laplace Transforms

!

4.2!

Integrals of Laplace Transforms

!

4.3!

Initial and Final Values of Laplace Transforms

1!

Definition of the Laplace Transform

!

1.1!

Mathematical Transforms

!

1.2!

Fourier Transform

5! Laplace Transforms of Discontinuous and Periodic ! Functions

!

1.3!

Laplace Transform

!

5.1!

Laplace Transform of Unit Step Functions

!

1.4!

Existence Requirements

!

5.2!

t-Shifting Property

!

1.5!

Laplace Transforms of Simple Functions

!

5.3!

Laplace Transform of Rectangular Pulses

!

5.4!

Laplace Transform of Unit Impulse Functions

!

5.5!

Laplace Transform of Periodic Functions

2!

Properties of the Laplace Transform

!

2.1!

Linearity Property

!

2.2!

s-Shifting Property

6!

Inverse Laplace Transforms

!

2.3!

Scaling Property

!

6.1!

Integral Transform Pairs

!

6.2!

The Inverse Laplace Transform

!

6.3!

Inverse Laplace Transforms from a Transform Table

3! Laplace Transforms of Derivatives and Integrals !

3.1!

Laplace Transforms of Derivatives of Continuous

!

!

Functions

7! Convolution

!

3.2!

Laplace Transforms of Derivatives of Piecewise

!

7.1!

The Convolution Integral

!

!

Continuous Functions

!

7.2!

Laplace Transform of the Convolution Integral

!

3.3!

Laplace Transforms of Integrals

!

7.3!

Inverse Laplace Transform of the Convolution

!

!

Integral v

8!

Laplace Transforms of Series Expansions

!

11.4! Series

!

8.1!

Laplace Transform of Power Series

!

11.5! Power Series

!

8.2!

Inverse of a Laplace Transform Series

!

11.6! Taylor Series

!

11.7! Laurent Series

9! !

Solution of Linear Ordinary Differential Equations having Constant Coefficients

!

11.8! Singularities

!

11.9! Residues

!

9.1!

Ordinary Differential Equations

!

11.10!Residue Integration

!

9.2!

Solving Linear ODEs using Laplace Transforms

!

9.3!

Solving Linear ODEs having Discontinuous Forcing

12! Inverse Laplace Transforms from Residue Integration

!

!

Functions

!

12.1! Inverse Laplace Transform

!

9.4!

Solving Simultaneous ODEs using Laplace

!

12.2! Inverse Laplace Transform of a Unit Step Function

!

!

Transforms

!

12.3! Inverse Laplace Transform of the Unit Impulse

!

!

10! Solution of Partial Differential Equations

!

12.4! Inverse Laplace Transform of a Function of

!

10.1! Partial Differential Equations

!

!

!

10.2! Solving PDEs using Laplace Transforms

!

12.5! Residue Integration of the Inverse Laplace Transform

!

10.3! Obtaining Laplace Transforms using Partial

!

12.6! Pole Position Determines System Stability

!

!

Differentiation

11! Complex Variables !

11.1! Contours in the Complex Plain

!

11.2! Derivatives of Complex Functions

!

11.3! Important Theorems for Determining Residues

Function Exponential Order

13! Solution of Linear Ordinary Differential Equations ! having Variable Coefficients !

13.1! Differential equations of Second Degree

!

13.2! Differential equations of Higher Degree

vi

14! Evaluating Integrals Equations with the Laplace !

Transform

!

14.1! Integral Equations

!

14.2! Laplace Transform of Integral Equations

Appendix A!

The Greek Alphabet

Appendix B!

Integration by Parts

Appendix C!

Partial Fractions

!

C.1! General Procedure

!

C.2! Calculating Partial Fractions

!

C.3! Heaviside Cover-up Method

!

C.4! Heaviside Expansion Formula

Appendix D!

Summary of Propositions

Appendix E!

Laplace Transform Tables

Appendix F!

Properties of Laplace Transforms

Appendix G!

Power Series

Appendix H!

Special Functions

References

vii

Chapter 1 Definition of the Laplace Transform

L

{ f (t )} = lim

T →∞

∫

T

0−

f ( t ) e− s t dt

6

!

In this chapter we will define the Laplace transform and

transformed function (or image function) is designated as

discuss requirements for its existence. We will then present the

output of the system.

Laplace transforms of some simple functions. We will begin by

!

reviewing the general nature of mathematical transforms.

integral transforms can be written as:

Mathematical transformations of a function f (t) that are

F (s) =

∫

b

f ( t ) K ( s, t ) dt !

1.1! MATHEMATICAL TRANSFORMS

!

!

where the function F ( s ) is the transform of the function f (t) ,

In mathematics a transformation is a mapping of one set

of numbers into another set of numbers. This mapping is performed by some procedure that is invertible, but which may not be injective (one-to-one). The mapping operation is known as a transform. !

If a set of numbers can be represented by some function,

then a mathematical transform that operates on this function to produce another function is a particular type of mathematical operator. The transformed function will have an independent variable different from that of the original function. A new functional relation is then established with a new independent variable. Therefore the transformed function can be considered to be in a transform space that is different from the space of the original function. !

An operator is known in engineering as a system. The

original function is designated as input for the system, and the

a

(1.1-1)

and where K ( s, t ) is called the kernel for the transformation. The original function f (t) is a function of a real variable t and exists in t-space (space containing the t dimension), and the transformed function F ( s ) is a function of a variable s and exists in s-space. The kernel K ( s, t ) is a function of both t and s and so functions in both t-space and s-space. Integral

transforms are defined by their kernels. The integral of an integral transform is evaluated on an interval ( a, b ) where a can be − ∞ . and b can be ∞ . !

The purpose of transforming a mathematical function is to

obtain one that is simpler to interpret than is the original function. For example, integral transforms can be used to convert differential and integral equations into algebraic equations (the algebraic equations are generally easier to solve). Such transformations from calculus to algebra are known as operational calculus. Once a solution in transform space is 7

obtained, the transformation is inverted to bring the solution

transform is K ( ω, t ) = e− i ω t . The Fourier transform therefore

back to the original space (see Figure 1.1-1).

represents a function f (t) as a linear combination of complex exponential functions e− i ω t . This makes the Fourier transform very useful for analyzing the frequency content of a signal f (t) . ! !

From Euler’s formula we have: e−i ω t = cos ( ω t ) − i sin ( ω t ) !

(1.2-2)

where e− i ω t is a bounded periodic function since e− i ω t = 1 . The existence of the Fourier transform integral depends, therefore, on the function f (t) . If f (t) diverges as t → ∞ , the Fourier transform of f (t) will not exist, and equation (1.2-1) has no meaning. Many functions do not have a Fourier transform. Figure 1.1-1! Integral transform solution of a problem.

!

1.2! FOURIER TRANSFORM !

Fourier transform, which is defined as: !

F (ω ) =

∫

−∞

f ( t ) e−i ω t dt !

It may be possible to keep f (t) from diverging as t → ∞ by

multiplying it by a convergence factor in the form of an

Probably the most familiar integral transform is the ∞

1.3! LAPLACE TRANSFORM exponential e− σ t , where σ is chosen to be some positive realvalued parameter that is large enough to ensure convergence of

(1.2-1)

where ω is a real-valued parameter, and where the real-valued variable t often represents time. The Fourier transform F ( ω ) of

the integral: !

∫

∞

−∞

f ( t ) e− σ t dt !

(1.3-1)

This in turn will ensure the existence of the integral:

f (t) is a complex-valued function. The kernel for the Fourier 8

!

F ( σ, ω ) =

∫

∞

−∞

f ( t ) e− σ t e−i ω t dt !

(1.3-2)

variable t , into F ( s ) , a function of a complex variable s . The function f (t) is generally real in practice, but can be complex.

where F ( σ, ω ) is a function of both σ and ω , and where f (t) is

!

Since we choose σ > 0 , the convergence factor e− σ t can

assumed to be integrable in any finite interval 0 ≤ t ≤ T . We see

cause the Laplace integral to diverge if t is negative. Therefore

and that f ( t ) e− σ t is an attenuated function. We have:

that we have the one-sided or unilateral Laplace transform:

that F ( σ, ω ) is the Fourier integral of the function f ( t ) e− σ t ,

! !

F ( σ, ω ) =

∫

∞

f (t ) e

− ( σ+i ω ) t

−∞

dt !

(1.3-3)

We will now let σ and ω be the rectangular components

s = σ + iω !

(1.3-4)

!

F (s) =

∫

−∞

f (t ) e

−st

∫

∞

∫

∞

−

f ( t ) e− σ t e−i ω t dt !

(1.3-6)

f ( t ) e− s t dt !

(1.3-7)

or !

where σ ∈! and ω ∈! . Equation (1.3-3) can then be written: ∞

!

F (s) =

0

of a complex variable s so that: !

the Laplace transform is usually restricted to values of t ≥ 0 so

F (s) =

0

−

which is referred to as simply the Laplace transform. The integral in this equation is known as the Laplace transform

dt !

(1.3-5)

which is an integral transform known as the bilateral Laplace transform. The kernel of the bilateral Laplace transform is K ( s, t ) = e− s t , where s is complex, and is not constrained to be

only along the imaginary axis as i ω is for the Fourier transform. The bilateral Laplace transform can therefore be considered to be a generalization of the Fourier transform. The bilateral Laplace transform changes f (t) , a function of a real

integral. We see that modification of the Fourier transform to be applicable to many more functions leads to the development of the Laplace transform. The unilateral Laplace transform is the subject of this book. !

All non-negative real numbers in the t-plane upon which

the Laplace transform acts form the domain of definition of the Laplace transform. Points in the t-plane that are mapped by the Laplace transform into the complex s-plane form the range of the Laplace transform. 9

Example 1.3-1

transform does not process f (t) for negative values of t . The

Show that the Laplace transform F ( s ) is a complex function.

Laplace transform operates on functions only over the interval

)

⎡⎣ 0 − , ∞ , where we are using the interval notation:

Solution:

Finite and half-open interval!

We have by definition: !

F (s) =

∞

∫

0−

f (t ) e

−st

dt =

∫

∞

0−

f (t ) e

− ( σ+i ω ) t

!

dt

!

e

F (s) =

∫

∫

∞

0

−

discontinuity of f (t) occurs at t = 0 (see Lundberg et al., 2007).

f ( t ) e− σ t ( cos ( ω t ) − i sin ( ω t )) dt

F (s) =

0

−

f ( t ) e− σ t cos ( ω t ) dt − i

∫

∞

t = 0 + are all equivalent as a limit of the Laplace transform

0

−

right (from the positive side).

f ( t ) e− σ t sin ( ω t ) dt

and so F ( s ) is clearly a complex function. !

If no discontinuity of f (t) exists at t = 0 , then t = 0 − , t = 0 , and integral, where t = 0 + means that t = 0 is approached from the

or !

as t = 0 − , where t = 0 − means that t = 0 is approached from the

transform. This becomes an important consideration when any

we can write: !

The lower limit of the Laplace transform integral is written

included within the interval of integration of the Laplace

= cos ( ω t ) − i sin ( ω t )

∞

⇒ ! a ≤ t < b ! (1.3-8)

left (from the negative side). The point t = 0 is then entirely

Using Euler’s formula: −i ω t

[ a, b ) !

Because physical processes are generally causal, the

restriction to t ≥ 0 in the Laplace transform is not severe. Any physical function f (t) is expected to have a beginning at which time it becomes of interest. We can arbitrarily take t = 0 to

!

The function f (t) serves as the input to the Laplace

transform. The output from the Laplace transform is designated

F ( s ) , and is referred to as the Laplace transform F ( s ) of f (t) . The Laplace transform operator is denoted by L . The Laplace transform of f (t) is then given by: !

L

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s ) !

(1.3-9)

correspond to this beginning time. In any case, the Laplace 10

!

The historical origin of the Laplace transform is generally

!

For the Laplace integral to converge to a limit, certain

considered to be some of the early work of Euler beginning in

continuity and bounding requirements must apply. The limit in

1737. This was followed by Laplace’s studies of probability

equation (1.4-1) will generally exist only for certain values of

theory (1812), and Bateman’s study of radioactivity (1910). The

Re s .

Laplace transform techniques are closely related to operational calculus methods developed by Oliver Heaviside (1892, 1894, 1912) to analyse transient disturbances in electrical networks. Laplace transform theory and techniques were further extended in studies by Lerch (1903), Bromwich (1916), and Doetsch (1937). A history of the Laplace transform is given by Lützen (1979) and Deakin (1981, 1982, 1992, 2018).

Example 1.4-1 Determine the Laplace transform for f (t) = 0 . Solution: We have: !

L {0} = lim

T →∞

∫

T

0

−

( 0 ) e− s t dt = 0

1.4! EXISTENCE REQUIREMENTS !

The Laplace transform F ( s ) of a function f (t) is an

improper integral that must be evaluated in terms of a limit: ! L

∞

{ f (t )} = ∫ f (t ) e 0

−

−st

dt = lim

T →∞

∫

T

0

−

f ( t ) e− s t dt = F ( s ) ! (1.4-1)

Example 1.4-2 Determine the Laplace transform for f (t) = 1 . Solution: We have:

For the Laplace transform of a function f (t) to exist, the integral in equation (1.4-1) must converge to a limit F ( s ) as

!

L {1} = lim

T →∞

T → ∞ . If no limit exists, the Laplace integral is said to diverge, and f (t) then does not have a Laplace transform. Most physical processes can be represented by functions that have a Laplace transform.

∫

T

0−

(1) e

T

−st

⎡ e− s t ⎤ ⎡ e− sT 1 ⎤ dt = lim ⎢ − ⎥ ⎥ = lim ⎢ T →∞ ⎣ −s ⎦ 0 T → ∞ ⎣ −s −s ⎦

or !

L {1} =

1 ! s

Re s > 0 11

Note that F ( s ) = 1 s is holomorphic over the entire s-plain except s = 0 , but that F ( s ) only converges for Re s > 0 .

1.4.1! !

!

A function f (t) will be continuous at a point t = t 0 if it

satisfies the three conditions given in equation (1.4-2), and if

f (t 0+ ) = f (t 0− ) . If we have f (t 0+ ) ≠ f (t 0− ) then a discontinuity

CONTINUOUS AND DISCONTINUOUS

exists at t = t 0 .

FUNCTIONS

!

Before considering the continuity requirements for the

( )

If both f t 0+

( ) ( )

f t 0+ − f t 0−

( )

and f t 0−

are finite so that the difference

is finite, then t = t 0 is a jump discontinuity. An

existence of a Laplace transform, we will briefly review the

example of a jump discontinuity is shown Figure 1.4-1. If a

definitions of continuous and discontinuous functions.

function f (t) has a jump discontinuity at a point t = t 0 , then

!

A function f (t) is defined as continuous at a point t = t 0 if

( )

the following three conditions are satisfied:

!

!

⎧ ⎪ 1. ⎪⎪ ⎨ 2. ⎪ ⎪ 3. ⎪⎩

f (t) is undefined at t = t 0 , and cannot be assumed to be the

( )

average of f t 0+ and f t 0− .

f ( t ) exists at t = t 0 lim f (t) exists

t → t0

! !

!

!

!

(1.4-2)

lim f ( t ) → f ( t 0 )

t → t0

A function f (t) can approach a point t = t 0 from either the

right or the left. We will use the following notation to describe these two directions of approach to a point t = t 0 :

( )

t → t0

( )

t → t0

!

f t 0+ means lim f ( t ) is approached from the right

!

f t 0− means lim f ( t ) is approached from the left

Figure 1.4-1! A jump discontinuity in f ( t ) at t = t 0 . 12

!

( )

( )

If either or both f t 0+ and f t 0− are infinite so that the

( ) ( )

difference f t 0+ − f t 0−

is infinite, then t = t 0 is an infinite

discontinuity. !

A function f ( t ) can be continuous at a point when the

point is approached from one direction, but not continuous when the point is approached from the other direction. The function is then considered to be discontinuous at the point. Such a discontinuity is an infinite discontinuity. A function

1.! Within each open segment f ( t ) will be continuous (see Figure 1.4-2). 2.! Finite right hand and finite left hand limits of f ( t )

will exist as f ( t ) approaches the respective end points from within the segment.

3.! At any point t 0 that is a juncture between two of the segments, f ( t ) will have a jump discontinuity at

( )

( )

that has an infinite discontinuity when approached from any

which the limits f t 0− and f t 0+ of f ( t ) at the sides of

direction is known as a pole (see Chapter 10).

the discontinuity both exist (see Figure 1.4-1).

1.4.2! !

CONTINUITY REQUIREMENTS OF THE LAPLACE TRANSFORM

For the Laplace transform F ( s ) of a function f ( t ) to exist,

4.! Any finite interval will have only a finite number of jump discontinuities, and so a finite number of segments.

f ( t ) must be defined and integrable on the interval of

integration. This in turn requires the function f ( t ) to be piecewise continuous on the interval of integration. For a function f ( t ) to be continuous at a point t = t 0 , f ( t ) must satisfy equation (1.4-2). !

For a function to be piecewise continuous on a finite

interval, it must be possible to partition the interval into a finite number of segments such that: Figure 1.4-2! A piecewise continuous function f ( t ) . 13

A function f ( t ) that is piecewise continuous on an

!

n−1

L

{ f (t )} = Tlim →∞ ∑ ∫

ti+1

f ( t ) e− s t dt !

interval will be bounded on each segment of the interval. A

!

function f ( t ) will be piecewise continuous over ⎡⎣ 0 − , ∞ if f ( t ) is piecewise continuous over ⎡⎣ 0 − , T ⎤⎦ for all T > 0 .

where ti , i = 1, 2, !, n − 1 are the points of discontinuity between

)

Example 1.4-3

!

⎧ 1 0≤t σ 0 ! (1.4-8)

since e− i ω t = 1 . !

A function f ( t ) that satisfies equation (1.4-7) is limited to

Figure 1.4-3! Examples of functions of exponential order. !

If f ( t ) is of exponential order, then

( )

f ( t ) = O eσ 0 t

as

t → ∞ . The constant σ 0 is the minimum value for which equation (1.4-7) is valid. It is always possible to replace σ 0 with a constant σ1 where σ1 > σ 0 since we then have eσ 0 t < eσ1 t for

no more than exponential growth. Such a function is said to be

t ≥ T0 .

of exponential order. If f ( t ) is of exponential order, then the

!

M eσ 0 t when T0 is chosen to be large enough (see Figure 1.4-3).

all t < T0 . Most functions arising from physical problems can be

The concept of exponential order has been developed to meet

expected to be of exponential order. All rational algebraic

the Laplace transform existence requirements for a function

functions are of exponential order. Only functions that increase

rate at which f ( t ) can increase will be limited to no more than

f (t ) .

A function f ( t ) that satisfies equation (1.4-7) for all t ≥ T0

will be of exponential order even if f ( t ) > M eσ 0 t for some or

more rapidly than the exponential factor eσ 0 t when t ≥ T0 are not of exponential order (see Example 1.4-8). 15

Proposition 1.4-1:

Example 1.4-5

If f ( t ) is a bounded function such that f ( t ) ≤ M for t ≥ T0 ,

Show that f (t) = t n where n is a positive integer is of

then f ( t ) is of exponential order.

exponential order for t ≥ T0 .

Proof:

Solution:

If σ 0 = 0 equation (1.4-7) becomes:

!

f ( t ) ≤ M e( 0 ) t = M !

!

t ≥ T0 !

From the Maclaurin series for eσ 0 t we have: (1.4-9)

∞

σ0 t

and so f ( t ) is bounded by the constant M . We see then that

!

order.

and so:

■

Note that the symbol

■

!

signifies the end of a proof in this

book.

∑

tn
0 .

1.4.4!

Solution: We have: !

=

n=0

any function f ( t ) that is bounded for t ≥ T0 is of exponential

!

e

σ 0n t n σ 0n t n > n! n!

f (t) = 2 e3t sin ( t ) = 2 e3t sin ( t ) ≤ 2 e3t

Therefore f (t) = 2 e3t sin ( t ) is of exponential order for t > 0 .

!

CONVERGENCE REQUIREMENTS OF THE LAPLACE TRANSFORM

Since the Laplace transform F ( s ) is defined in terms of an

improper integral, it is important to determine the conditions under which this integral converges. It is also important to know the region of the complex s -plane for which F ( s ) exists since F ( s ) is a function of s .

16

Proposition 1.4-2, Absolute Convergence of the Laplace Transform: If f ( t ) is a piecewise continuous function for t ≥ 0 , and if f ( t ) ≤ M eσ 0 t , then the Laplace transform:

L

!

{ f (t )} = ∫ f (t ) e− s t dt = F ( s ) !

!

!

∞

{ f (t )} = ∫ f (t ) e 0−

−st

dt =

∫

T0

0−

f (t ) e

−st

dt +

∫

∞

f ( t ) e− s t dt

(1.4-11)

∫

0

!

−

f ( t ) e− s t dt ≤

∫

0−

e− s t dt ! (1.4-13)

f ( t ) e− s t is

∫

T0

0−

f ( t ) e− s t dt ≤ K

∫

T0

0−

⎛ 1 e− σ T0 ⎞ e− σ t e− i ω t dt ≤ K ⎜ − σ ⎟⎠ ⎝σ

!

(1.4-14)

sum of integrals on segments in each of which f ( t ) e− s t is

T0

!

∞

0−

dt ≤ K

T0

The first integral on the right in equation (1.4-12) is then the

We then have: !

∫

f (t ) e

−st

bounded on the interval 0 ≤ t ≤ T0 . Therefore

!

We can write: L

dt ≤

T0

Since s = σ + i ω and e− i ω = 1 for all ω , we see that e− s t is also

(1.4-10)

0−

Proof:

!

0−

f (t ) e

−st

absolutely integrable on the interval 0 ≤ t ≤ T0 , and we have:

∞

converges absolutely for Re s > σ 0 where σ 0 ≥ 0 .

!

∫

!

T0

bounded and continuous. Therefore this integral converges absolutely over the interval ⎡⎣ 0 − , T0 ⎤⎦ . !

For t ≥ T0 we are given that f ( t ) ≤ M eσ 0 t where M is

independent of s . Therefore:

∫

T0

0

−

f ( t ) e− s t dt +

∫

∞

f ( t ) e− s t dt ! (1.4-12)

T0

Since f ( t ) is taken to be a piecewise continuous function

in the interval 0 ≤ t ≤ T0 , some finite positive number K will

!

⎧⎪ ∞ if σ < σ 0 lim M e− ( σ − σ 0 ) t = ⎨ ! t→∞ ⎪⎩ 0 if σ > σ 0

(1.4-15)

If σ > σ 0 where Re s = σ we will have:

exist that bounds the function f ( t ) over the interval 0 ≤ t ≤ T0

!

right is integrable, and we can write:

since e− i ω t = 1 for all ω . Since f ( t ) ≤ M eσ 0 t , we obtain:

so that we have f ( t ) ≤ K . Therefore the first integral on the

f ( t ) e− s t = f ( t ) e− σ t e− i ω t = f ( t ) e− σ t !

t ≥ T0 ! (1.4-16)

17

f ( t ) e− s t ≤ M e− ( σ − σ 0 ) t !

!

t ≥ T0 !

(1.4-17)

For the second integral on the right of equation (1.4-12) we then have:

∫

!

∞

f ( t ) e− s t dt ≤ M

T0

∫

∞

e− ( σ − σ 0 ) t dt !

(1.4-18)

From the comparison theorem for improper integrals, we can now conclude that the Laplace transform integral converges,

)

and converges absolutely for Re s > σ 0 over ⎡⎣ 0 − , ∞ .

Proposition 1.4-3, Existence of the Laplace Transform: If f ( t ) is a piecewise continuous function for t ≥ 0 , and if

T0

or ∞

∫

!

f (t ) e

−st

T0

M dt ≤ e− ( σ − σ 0 ) t σ0 − σ

f ( t ) ≤ M eσ 0 t where Re s > σ 0 with σ 0 ≥ 0 and t ≥ T0 , then the

∞

!

Laplace transform:

(1.4-19)

T0

∫

!

f ( t ) e− s t

T0

(1.4-20)

on [T0 , ∞ ] . From Equation (1.4-20) we have:

∫

!

f (t ) e

T0

! ! ! !

−st

M ! dt ≤ σ − σ0

∫

0−

f (t ) e

−st

∫

∞

0

−

f ( t ) e− s t dt = F ( s ) !

(1.4-23)

Proof: !

Re s > σ 0 !

(1.4-21)

⎛ 1 e− σ T0 ⎞ M ! dt ≤ K ⎜ − + σ ⎟⎠ σ − σ ⎝σ

Re s > σ 0

0

(1.4-22)

Follows from Proposition 1.4-2 since L −

)

{ f (t )}

absolutely over ⎡⎣ 0 , ∞ where Re s > σ 0 with σ 0 ≥ 0 . !

From equations (1.4-14), (1.4-21), and (1.4-12) we have: ∞

{ f (t )} =

exists.

M dt ≤ e− ( σ − σ 0 ) T0 ! σ − σ0

Therefore if σ > σ 0 this improper integral converges absolutely ∞

L

!

We then have Re s > σ 0 : ∞

■

converges ■

Propositions 1.4-2 and 1.4-3 prove that if a function f ( t )

satisfies the two conditions: 1.!

f ( t ) is piecewise continuous.

2.!

f ( t ) is of exponential order.

then these conditions are sufficient to ensure that its Laplace transform exists. These conditions are, however, not necessary for the existence of a Laplace transform. 18

!

Certain functions that are not piecewise continuous or

are not of exponential order can still have Laplace transforms. For example, f ( t ) = 1

t is not piecewise continuous on the

)

interval ⎡⎣ 0 − , ∞ since f ( t ) → ∞ as t → 0 + , which is a discontinuity that is not finite. Nevertheless, f ( t ) = 1 t does

!

however, as discussed in Chapter 6. Proposition 1.4-5, Riemann Lebesque Lemma:

From Propositions 1.4-1 and 1.4-2 we see that many of the

(1.4-25)

Re s→ ∞

uniformly. Proof:

functions, and exponential functions. Proposition 1.4-4, Uniqueness of the Laplace Transform:

{ f (t )} = F ( s ) exists, then f (t ) has

only the one Laplace transform F ( s ) .

! !

From equation (1.4-22) we have:

∫

∞

f (t ) e

0−

−st

⎛ 1 e− σ T0 ⎞ M ! dt ≤ K ⎜ − + ⎟ σ ⎠ σ−σ ⎝σ

Re s > σ 0

0

! !

Proof:

(1.4-26)

As Re s = σ → ∞ equation (1.4-26) becomes:

From the definition of the Laplace transform given in !

equation (1.4-1) we have: !

lim F ( s ) → 0 !

!

among these functions are polynomials, sine and cosine

!

)

function of exponential order over ⎡⎣ 0 − , ∞ , then:

most common functions do have Laplace transforms. Included

If the Laplace transform L

{ f (t )} = F ( s ) where f (t ) is a piecewise continuous

If L

have a Laplace transform. !

The inverse Laplace transform is not necessarily unique,

L

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s ) !

(1.4-24)

This is a one-to-one relation between f ( t ) and F ( s ) and so f ( t ) has the unique Laplace transform F ( s ) .

lim

Re s→ ∞

∫

∞

∫

∞

0

−

f ( t ) e− s t dt → 0 !

(1.4-27)

f ( t ) e− s t dt = 0 !

(1.4-28)

Therefore: !

■

lim

Re s→ ∞

0−

or 19

lim F ( s ) → 0 !

!

(1.4-29)

Re s→ ∞

where equation (1.4-29) applies uniformly over Re s > σ 0 (see Proposition 1.4-7). !

■

If equation (1.4-29) does not apply for a Laplace transform

⎡ σM ⎤ lim s F ( s ) = lim ⎢ K 1− e− σ T0 + ! σ→ ∞ σ→ ∞ σ − σ 0 ⎥⎦ ⎣

(

!

true, however. It is possible for a function g ( s ) → 0 as s → ∞

⎡ ⎢ M lim s F ( s ) = lim ⎢ K 1− e− σ T0 + σ σ→ ∞ σ→ ∞ 1− 0 ⎢ σ ⎣

(

!

)

lim s F ( s ) = K + M !

!

Proposition 1.4-6:

and so s F ( s ) is bounded as Re s → ∞ .

{ f (t )} = F ( s ) where f (t ) is a piecewise continuous

)

function of exponential order over ⎡⎣ 0 , ∞ , then s F ( s ) is bounded as Re s → ∞ . −

Proof: !

∫

∞

0

−

⎛ 1 e− σ T0 ⎞ M ! f ( t ) e− s t dt ≤ K ⎜ − + σ ⎟⎠ σ − σ 0 ⎝σ

! ! Since Re s = σ , we can write:

(1.4-32)

(1.4-33)

σ→ ∞

■

Example 1.4-6 Determine the Laplace transform for f (t) = eat where a is a real constant. Solution:

From equation (1.4-22) we have:

! F (s) =

⎤ ⎥ ⎥! ⎥ ⎦

Therefore:

without g ( s ) being the Laplace transform of any function.

If L

(1.4-31)

or

F ( s ) of a function f ( t ) , then f ( t ) cannot be a piecewise continuous function of exponential order. The converse is not

)

The function f (t) = eat is continuous in − ∞ < t < ∞ . We have: Re s > σ 0

(1.4-30)

!

{ }

L eat = lim

T →∞

∫

T

0−

e− s t eat dt =

1 ⎡ lim e− ( s−a ) T − 1⎤⎥ ⎢ T a − s ⎣ →∞ ⎦

If Re ( s − a ) > 0 the function f (t) = eat is of exponential order and the limit exists. We then have:

20

!

{ }

L eat =

1 ! s−a

Re s > a

We see that the Laplace transform changes a transcendental function e

at

into a simpler algebraic function 1 ( s − a ) . Note

that if a = 0 we have: !

{ }

L eat =

1.4.5! !

REGION OF CONVERGENCE OF THE LAPLACE TRANSFORM

The region of convergence of the Laplace transform is

defined to be the the set of values of s for which the Laplace transform exists. If f ( t ) is a piecewise continuous function of

1 s

exponential order, the Laplace transform integral will converge absolutely in the region Re s > σ 0 as given in Proposition1.4-2, and so will converge to the right of the vertical line Re s = σ 0 in

as given in Example 1.4-2.

the s-plane (see Figure 1.4-4). Example 1.4-7 Determine the Laplace transform for f (t) = e− at where a is a real constant. Solution: The function f (t) = e− at is continuous in − ∞ < t < ∞ and of exponential order. We have: !

{ } = lim ∫

L e

− at

T →∞

T

0−

e− s t e−at dt =

−1 ⎡ lim e− ( s + a ) T − 1⎤⎥ ⎢ s + a ⎣ T →∞ ⎦

Therefore: !

{ }

L e− at =

1 s+a

Figure 1.4-4! Example of a region of convergence and a region of absolute convergence of the Laplace transform for a function of exponential order. 21

!

The right half-plane Re s > σ 0 is known as the region of

absolute convergence, and is a function of σ 0 . The term right half-plane is used to mean a half-plane that is bounded on the left side but which extends infinitely on the right side. The constant

Re s > σ 0

is

called

the

abscissa

of

Proof: ! !

absolute

We have from equation (1.4-20):

∫

∞

f ( t ) e− σ t dt ≤

T0

M e− ( σ − σ 0 ) T0 ! σ0 − σ

Re s > σ 0 ! (1.4-35)

convergence, and the line Re s = σ 0 is not included within the

for large enough T0 . Let σ ≥ σ1 > σ 0 where σ1 is some given

region of absolute convergence.

value of σ . We then have:

!

If the region of convergence of the Laplace transform of

f ( t ) is to the right of a vertical line Re s = σc where σc < σ 0 ,

!

∫

∞

f ( t ) e− σ t dt ≤

T0

then σc is called the abscissa of convergence or the abscissa of

M M ≤ = MU ! σ − σ 0 σ1 − σ 0

(1.4-36)

simple convergence. The region of convergence Re s > σ c can

where MU > 0 is an upper bound for the expression in equation

be larger than the region of absolute convergence Re s > σ 0 .

(1.4-36) for all Re s ≥ σ1 > σ 0 . Since:

Depending upon the function, the abscissa of convergence can be positive, zero, or negative.

!

Proposition 1.4-7, Uniform Convergence of the Laplace Transform:

we have:

f (t ) ≤ M e

!

L

!

∫

∞

0−

f ( t ) e− s t dt = F ( s ) !

f ( t ) e− σ t dt ≤

∫

∞

f ( t ) e− σ t dt !

(1.4-37)

T0

f ( t ) e− σ t dt ≤ MU !

(1.4-38)

T0

for t ≥ T0 , then the Laplace transform:

{ f (t )} =

∫

∞

T0

If f ( t ) is a piecewise continuous function having σ0 t

∫

∞

(1.4-34)

where MU is independent of σ . We see that the integral in equation (1.4-38) converges absolutely. Therefore from the

converges uniformly with respect to s in the open half-plane

Weierstrass M-test for integrals, we also see that the Laplace

Re s > σ 0 .

transform of f ( t ) converges uniformly with respect to s .

■

22

Proposition 1.4-8:

Example 1.4-8

If f ( t ) is a piecewise continuous function having

Show that the Laplace transform does not converge for the

f ( t ) ≤ M eσ 0 t for t ≥ T0 , then the Laplace transform:

L

!

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s ) !

2

continuous function f (t) = e t . (1.4-39)

Solution: 2

The function f (t) = e t is not of exponential order since it

converges absolutely and uniformly with respect to s in the

increases more rapidly than eat when t ≥ T0 . We have:

open half-plane Re s > σ 0 .

!

Proof: !

{ }= ∫

L e

t2

∞

0

Follows from Propositions 1.4-2 and 1.4-7.

■

−

t2 −st

e e

dt =

∫

∞

0

−

et

2

−s t

dt

Adding and subtracting s 2 4 , the exponent of this integral can be rewritten as:

Proposition 1.4-9: If the Laplace transform of a function f ( t ) with f ( t ) ≤ M eσ 0 t

!

converges absolutely with respect to s for Re s ≥ σ1 > σ 0 , then it

{ }= e

L e

t2

( )

− s2 4

∫

∞

0−

e (t−s 2 ) dt 2

converges uniformly and absolutely with respect to s for

For any given value of s we then have as t → ∞ :

Re s = σ1 where σ1 ≥ σ 0 .

!

Proof:

{ }→ ∞

L et

2

Therefore the Laplace transform does not converge for the

!

Follows from Proposition 1.4-8.

!

As noted previously, the Laplace transform is an improper

2

■

continuous function f (t) = e t , and so the Laplace transform 2 does not exist for f (t) = e t .

integral that must be evaluated in terms of a limit. We will

!

generally do this implicitly for the rest of the book without

not sufficient to ensure that the Laplace transform of the

explicitly showing the limiting process.

function exists.

From Example 1.4-8 we see that continuity of a function is

23

∞

⎡ t e− s t ⎤ 1 L {t } = ⎢ − ⎥ + s ⎦ t = 0− s ⎣

1.5! LAPLACE TRANSFORMS OF SIMPLE FUNCTIONS

!

!

and so:

Using the defintion of the Laplace transform given as

equation 1.3-9, the Laplace transforms of many functions can be easily determined. Example 1.5-1 Determine the Laplace transform for the unit ramp function

f (t) = t .

∫

∞

0

−

t e− s t dt

!

u = t!

du = dt !

dv = e− s t dt v=−

Using: !

∫ u dv = u v − ∫ v du

we have:

1 ! s2

Re s > 0

Determine the Laplace transform for the ramp function

⎧ t 0 ≤ t ≤1 f (t ) = ⎨ t >1 ⎩ 1

Solution:

We can integrate by parts (see Appendix B). We let: !

t = 0−

=

e− s t dt

Example 1.5-2

!

We have:

L {t } =

∞

0

−

defined by

Solution:

!

!

1 L {t } = − 2 e− s t s

∫

∞

e

−st

The ramp function f (t) is shown in Figure 1.5-1. Its Laplace transform is: !

L

∞

{ f (t )} = ∫ f (t ) e 0

s

−st

−

dt =

∫

1

0

−

te

−st

dt +

∫

∞

e− s t dt

1

Integrating by parts, we let: !

u =t!

dv = e− s t dt

!

du = dt !

v=−

e− s t s 24

we then have: 1

!

⎡ t e− s t ⎤ 1 L { f ( t )} = ⎢ − + ⎥ s ⎦ t = 0− s ⎣

∫

∞

1

0−

e

−st

⎡ e− s t ⎤ dt + ⎢ − ⎥ ⎣ s ⎦ t =1

Determine the Laplace transform for f (t) = t 2 . Solution:

or 1

!

Example 1.5-3

⎡ t e− s t ⎤ e− s t L { f ( t )} = ⎢ − − ⎥ s ⎦ t = 0− s 2 ⎣

1

∞

⎡ e− s t ⎤ + ⎢− ⎥ s ⎦ t =1 t = 0− ⎣

We have: !

{ }

L t2 =

0

and so: !

1− e− s ! L { f ( t )} = s2

Re s > 0

∫

∞ −

t 2 e− s t dt

Integrating by parts, we let: !

u = t2 !

dv = e− s t dt

!

du = 2t dt !

v=−

e− s t s

We then have: !

∞

{ }

2 ⎡ 1 ⎤ + L t 2 = ⎢ − e− s t t 2 ⎥ ⎣ s ⎦ t = 0− s

∫

∞

0

−

t e− s t dt

or !

{ }

L t

2

∞

⎡ t 2 e− s t ⎤ 2 = ⎢− + F (t ) ⎥ s ⎦ t = 0− s ⎣

Using L’Hôpital’s rule and Example 1.5-1, we have: ! Figure 1.5-1! Ramp function for Example 1.5-2.

{ }

L t2 =

2 2 1 = 3! 2 s s s

Re s > 0 25

!

Example 1.5-4

{ }

L tn =

Determine the Laplace transform for f (t) = t n .

or for k = n :

Solution:

!

We have: !

{ }= ∫

L t

n

∞

0−

n −st

t e

{ }

L tn =

!

{ }

L t

dt ! ∞

⎡ t n e− s t ⎤ n = ⎢− + ⎥ s ⎦ t = 0− s ⎣

∫

∞

0−

t

n−1 − s t

e

!

dt

!

{ }

L tn =

{ }

!

{ }

L t

∞

⎡ t n e− s t ⎤ n = ⎢− + L t n−1 ⎥ s ⎦ t = 0− s ⎣

{ }

!

{ }

L t

Re s > 0

n! ! s n+1

Re s > 0

∞ −

t α e− s t dt !

(1.5-1)

becomes: !

{ }

n = L t n−1 s

n ⎛ n − 1⎞ = ⎜ ⎟ L s⎝ s ⎠

{ }= ∫

L t

α

Letting u = s t so that t = u s and dt = du s , equation (1.5-1)

{ }

L tα =

1 s

α+1

∫

∞

0

−

u α e−u du !

(1.5-2)

Using the definition of the gamma function:

from which we obtain: n

n! n! n! 1 L t 0 = n L {1} = n ! n s s s s

0

This gives us a recursive relation: n

{ }

Re s > 0

We will now consider the Laplace transform of t α where

! L t

{ }

L t n−k !

α is any real number, and where α > −1 . We have:

or n

s

k

Therefore:

Integrating by parts, we have: n

n ( n − 1) ( n − 2 )!( n − k + 1)

{t }! n−2

Proceeding in this way, we get:

Re s > 0

!

Γ ( α + 1) =

∫

∞

0−

u α e− u du !

α > −1 !

(1.5-3)

we can write: 26

! ! !

L

{t } = α

Γ ( α + 1) s α+1

!

α > −1 !

(1.5-4)

If α = n , where n is a nonnegative integer, then:

Γ ( n + 1) = n! !

(1.5-5)

and we have: !

{ }

L tn =

n! s n +1

!

Re s > 0 !

(1.5-6)

as determined in Example 1.5-4.

27

Chapter 2 Properties of the Laplace Transform

L

{

}

f ( t ) ebt = F ( s − b )

28

!

Certain properties of the Laplace transform make it

possible to construct the transforms of many different functions using Laplace transforms already obtained for other less complicated functions.

L {c1 f ( t ) + c2 g ( t )} = lim

!

T →∞

∫

T

∫

T

0

⎡⎣ c1 f ( t ) + c2 g ( t ) ⎤⎦ e− s t dt ! (2.1-3)

−

or

2.1! LINEARITY PROPERTY !

Since this is an improper integral, we have:

L {c1 f ( t ) + c2 g ( t )} = lim

!

The Laplace transform operator L is a linear operator.

This means that the Laplace transform of any linear

T →∞

0

⎡⎣ c1 f ( t ) e− s t + c2 g ( t ) e− s t ⎤⎦ dt −

! !

(2.1-4)

combination of functions equals the same linear combination of

Since integration is a linear operation, we have:

their Laplace transforms. The linearity of the Laplace transform the Laplace transform is defined by the following proposition.

⎡ ! L {c1 f ( t ) + c2 g ( t )} = lim ⎢ c1 T →∞ ⎣ ! !

Proposition 2.1-1, Linearity Property of Laplace Transforms:

or

is a direct result of the linearity of integration. The linearity of

The Laplace transform is a linear operator. If f ( t ) and g ( t ) are real-valued functions having Laplace transforms that exist, then: L {c1 f ( t ) + c2 g ( t )} = c1 L

!

{ f (t )} + c2 L {g (t )}!

(2.1-1)

where c1 and c2 are arbitrary constants.

! L {c1 f ( t ) + c2 g ( t )} = c1 lim

T →∞

From the definition of the Laplace transform we can write:

Therefore:

!

L {c1 f ( t ) + c2 g ( t )} =

0

⎡⎣ c1 f ( t ) + c2 g ( t ) ⎤⎦ e− s t dt !

(2.1-2)

⎤ g ( t ) e− s t dt ⎥ 0− ⎦

∫

(2.1-5)

∫

T

0−

f (t ) e

−st

dt + c2 lim

T →∞

∫

T

0−

g ( t ) e− s t dt

and so:

!

−

dt + c2

T

(2.1-6)

Proof:

∫

0−

f (t ) e

−st

! !

! L {c1 f ( t ) + c2 g ( t )} = c1

∞

∫

T

∫

∞

0

−

f ( t ) e− s t dt + c2

L {c1 f ( t ) + c2 g ( t )} = c1 L

!

∫

∞

0

−

g ( t ) e− s t dt ! (2.1-7)

{ f (t )} + c2 L {g (t )} !

(2.1-8)

■

29

on the right converges absolutely for σ > σ g . Therefore we see

Proposition 2.1-2, Convergence of a Sum of Laplace Transforms:

that the Laplace transform

L {c1 f ( t ) + c2 g ( t )}

(

converges

)

absolutely for σ > σ m where σ m = max σ f , σ g . ■

If f ( t ) and g ( t ) are piecewise continuous functions of

{ f (t )} converges for σ > σ f and {g (t )} converges for σ > σ g , then L {c1 f (t ) + c2 g (t )}

exponential order, and if L L

(

Example 2.1-1

)

Determine the Laplace transform for f (t) = c where c is a real

converges absolutely for σ > σ m where σ m = max σ f , σ g .

constant.

Proof: !

From Proposition 2.1-1 we have:

!

L {c1 f ( t ) + c2 g ( t )} = c1 L

Solution:

{ f (t )} + c2 L {g (t )} !

We have from Proposition 2.1-1 and Example 1.4-2:

(2.1-9)

!

where c1 and c2 are arbitrary constants. We can write: !

∫

∞

0

−

⎡⎣ c1 f ( t ) + c2 g ( t ) ⎤⎦ e− s t dt ≤

∫

∞

0

−

⎡⎣ c1 f ( t ) + c2 g ( t ) ⎤⎦ e− σ t dt

! !

(2.1-10)

or ! !

∫

0−

ce

−st

dt = c

∫

∞

0−

e− s t dt = c L {1} =

c s

From Proposition 2.1-1 and Example 1.5-4 we see that the

Laplace transform of any finite power series of t n will exist. For an infinite series, however, a linear combination of Laplace transformed terms of the series will not generally equal the

∫

∞

0−

⎡⎣ c1 f ( t ) + c2 g ( t ) ⎤⎦ e− s t dt ≤ c1

∫

∞

0−

+ c2

Laplace transform of the series sum (see Chapter 8).

f ( t ) e− σ t dt

∫

∞

0

!

!

L {c} =

∞

−

Example 2.1-2

g (t ) e

− σt

dt ! (2.1-11)

From Proposition 1.4-2 we know that first integral on the

right converges absolutely for σ > σ f , and the second integral

Determine the Laplace transform for f (t) =

N

∑a t

n

n

.

n=0

Solution: We have from Proposition 2.1-1 and Example 1.5-4: 30

⎧ N ⎫ a a 2! a2 3! a2 N ! aN ⎪ n⎪ !L ⎨ an t ⎬ = 0 + 21 + 3 + 4 +!+ N +1 ! s s s ⎪⎩ n = 0 ⎪⎭ s s

∑

!

Re s > 0

!

L {cosh ( at )} =

!

1.4-7 that it is possible to obtain the Laplace transforms for the hyperbolic cosine and sine functions using eat and e− at .

⎧ eat − e−at ⎫ 1 1 at −at L {sinh ( at )} = L ⎨ ⎬= L e − L e 2 2 2 ⎩ ⎭

{ }

L {sinh ( at )} =

Determine the Laplace transforms for cosh ( at ) and sinh ( at ) where a is a real constant.

ea t + e−a t ! cosh ( at ) = 2

Solution:

eat − e−at sinh ( at ) = 2

From Example 1.4-6 we have: !

From Examples 1.4-6 and 1.4-7 we have:

1 ! s−a

{ }

L e−at =

1 ! s+a

Re s > a

{ }

L ei t =

1 s+i s 1 = 2 = 2 +i 2 s − i s +1 s +1 s +1

From Euler’s formula ei t = cos ( t ) + i sin ( t ) and Proposition 2.1-1 we can then write:

With Proposition 2.1-1 we then have: !

1 ⎤ a ⎡ 1 ⎢⎣ s − a − s + a ⎥⎦ = s 2 − a 2

Determine the Laplace transforms for cos ( t ) and sin ( t ) .

We will use:

{ }

1 2

Example 2.1-4

Solution:

L eat =

{ }

and so: !

Example 2.1-3

!

1 ⎤ s ⎡ 1 ⎢⎣ s − a + s + a ⎥⎦ = s 2 − a 2

We also have:

We also see from Proposition 2.1-1 and Examples 1.4-6 and

!

1 2

⎧ eat + e−at ⎫ 1 1 at −at L {cosh ( at )} = L ⎨ ⎬= L e + L e 2 2 ⎩ ⎭ 2

{ }

{ }

!

L {cos ( t )} =

s s +1 2

!

L {sin ( t )} =

1 2

s +1

!

Re s > 1

or 31

2 Determine the Laplace transform for cos ( t ) .

convolution of the two functions (see Chapter 7).

is equal, however, to the

Solution:

Example 2.1-6

We have:

Determine the Laplace transforms for the damped cosine and

sine functions: e−bt cos ( t ) and e−bt sin ( t ) where b > 0 is a real

1 cos ( t ) = (1+ cos ( 2t )) 2 2

constant.

With Proposition 2.1-1 we can write: !

{

}

L cos 2 ( t ) =

Solution:

1 1 L {1} + L {cos ( 2t )} 2 2

From Euler’s formula ei t = cost + i sint we have: ! e

From Example 2.1-4 we then have: !

{

}

L cos 2 ( t ) =

11 1 s s2 + 2 + = 2 s 2 s2 + 4 s s2 + 4

(

)

−bt

e− (b−i ) t + e− (b+i ) t cos ( t ) = ! 2

functions into algebraic functions. Laplace transforms of

!

→∞.

The Laplace transform of the product of two functions

}

1 ! s + (b − i )

L e− (b+i ) t =

{

}

1 s + (b + i )

{

}

1 ! ( s + b) − i

L e− (b+i ) t =

{

}

1 ( s + b) + i

or

tangent, cotangent, secant, and cosecant do not exist because for these functions f ( t ) e

{

L e− (b−i ) t =

The Laplace transform therefore changes sine and cosine

− st

e− (b−i ) t − e− (b+i ) t sin ( t ) = 2i

From Example 1.4-7 we can write: !

!

{ f (t )} L {g (t )}

The product

!

!

L

Example 2.1-5

L e− (b−i ) t =

f ( t ) and g ( t ) is not equal to the product of their Laplace

From Proposition 2.1-1 we then have:

transforms L

! L e−b t cos ( t ) =

!

L

{ f (t )} and

{ f (t ) g (t )} ≠

L

L { g ( t )} :

{ f (t )} L {g (t )} !

(2.1-12)

{

}

⎤ 1⎡ 1 1 s+b + ⎢ ⎥= 2 ⎣ ( s + b ) − i ( s + b ) + i ⎦ ( s + b )2 + 1 32

{

}

! L e−b t sin ( t ) =

1 ⎡ 1 1 − ⎢ 2i ⎣ ( s + b ) − i ( s + b ) + i

⎤ 1 ⎥= 2 ⎦ ( s + b) + 1

{

}

L sinh 2 ( t ) =

!

1⎡ 1 1 2⎤ 2 + − = 4 ⎢⎣ s − 2 s + 2 s ⎥⎦ s s 2 − 4

(

)

Example 2.1-7

2.2! S-SHIFTING PROPERTY

Determine the Laplace transform for 1+ 2t − 3sint .

!

Solution:

that translates the Laplace transform by a constant. Shifting the

Using Proposition 2.1-1 we have: !

L {1+ 2t − 3sin ( t )} = L {1} + 2 L {t } − 3 L {sint }

and so from Examples 1.4-2, 1.5-1. and 2.1-4: !

1 2 3 L {1+ 2t − 3sin ( t )} = + 2 − 2 s s s +1

There is a Laplace transform property known as shifting

Laplace transform of a function is equivalent to the exponential scaling of the function as we will now show. Proposition 2.2-1, First Shift Theorem (s-Shifting Property): If L

L

!

Example 2.1-8

Proof:

Determine the Laplace transform for sinh 2 ( t ) .

!

Solution:

! L 2

⎡ e t − e− t ⎤ e2t e− 2t 1 sinh ( t ) = ⎢ + − ⎥ = 4 4 2 ⎣ 2 ⎦

bt

(2.2-1)

From the definition of the Laplace transform we can write:

{ f (t ) e } = ∫ bt

∞

⎡⎣ f ( t ) ebt ⎤⎦ e− s t dt = −

∫

∞

0−

f ( t ) e− ( s− b ) t dt ! (2.2-2)

Let u = s − b :

2

From Examples 1.4-6 and 1.4-7 and Proposition 2.1-1 we have:

{ f (t ) e } = F ( s − b ) !

0

We have: !

{ f (t )} = F ( s ) , then:

!

L

{ f (t ) e } = ∫ bt

∞

0

−

f ( t ) e− u t dt = F ( u ) !

(2.2-3)

Therefore: 33

L

!

{ f (t ) e } = F ( s − b ) ! −bt

(2.2-4)

■

Example 2.2-1 Using the s-shift property determine the Laplace transform

Proposition 2.2-2: If L

for ebt where b is a real constant.

{ f (t )} = F ( s ) is absolutely convergent for Re s > σ 0 , then

{ f (t ) e } is absolutely convergent for Re s > σ

L

bt

0

+ Reb .

Proof: !

Solution: From Example 1.5-3 we have: !

From the definition of the Laplace transform we can write:

!L

{

}

f ( t ) ebt =

∫

∞

0

⎡ f ( t ) ebt ⎤⎦ e− s t dt = − ⎣

∫

∞

0−

f ( t ) e− ( s− b ) t dt ! (2.2-5)

L

! Since L

{ f (t ) e } = ∫

∞

0−

f ( t ) e− ( σ − Re b ) t dt !

{ f (t )} = F ( s ) is absolutely convergent for

(2.2-6) Re s = σ > σ 0

!

L

! then L

{ f (t )} = ∫ f (t ) e 0−

−σt

L

{(1) e } = L {e } = s −1 b bt

bt

Example 2.2-2 Determine the Laplace transform for t 2e− 5t . Solution:

where: ∞

1 s

Applying the s-shift property we then obtain:

where s = σ + i ω . Since e− i ω = 1 and e− i Im b = 1 we have: bt

L {1} =

From Example 1.5-3 we have:

dt !

{ f (t ) e } is absolutely convergent for Re s > σ

(2.2-7)

bt

0

+ Reb .

!

{ }

L t2 =

2 s3

Applying the s-shift property we then obtain:

■

!

{

}

L t 2e− 5t =

2

( s + 5 )3 34

Solution: Example 2.2-3

From Example 1.5-4 we have:

Determine the Laplace transforms for the damped cosine and sine functions: e

−bt

cos ( t ) and e

−bt

sin ( t ) where b > 0 is a real

!

{ }

L tn =

n! s n+1

Re s > 0

!

constant.

Applying the s-shift property we then obtain:

Solution:

!

{

}

L t n eb t =

From Example 2.1-4 we have: !

L {cos ( t )} =

s s2 + 1

L {sin ( t )} =

!

1 s2 + 1

!

Re s > 1

Applying the s-shift property we then obtain:

{

}

! L e−b t cos ( t ) =

{

! L e

−b t

}

sin ( t ) =

s+b

( s + b)

2

( s − b )n+1

Re s > b

!

Example 2.2-5 Determine the Laplace transform for sinh ( 3t ) sin ( 2t ) . Solution:

+1

We have:

1

!

( s + b )2 + 1

⎧ e3t − e− 3t ⎫ L {sinh ( 3t ) sin ( 2t )} = L ⎨ sin ( 2t ) ⎬ 2 ⎩ ⎭

or

Compare this with Example 2.1-6.

! Example 2.2-4

L {sinh ( 3t ) sin ( 2t )} =

1 L 2

{e

3t

sin ( 2t ) − e− 3t sin ( 2t )

}

Applying the s-shift property we then obtain: n bt

Determine the Laplace transform for t e positive integer.

n!

where n is a

!

L {sinh ( 3t ) sin ( 2t )} =

⎤ 1⎡ 2 2 − ⎢ ⎥ 2 ⎢⎣ ( s − 3)2 + 4 ( s + 3)2 + 4 ⎥⎦ 35

and so: L {sinh ( 3t ) sin ( 2t )} =

!

1

−

1

( s − 3)2 + 4 ( s + 3)2 + 4

scaling that scales the Laplace transform by a constant.

{ f (bt )} = b1 F ⎛⎜⎝ bs ⎞⎟⎠ !

(2.3-1)

From the definition of the Laplace transform we can write: ∞

{ f (bt )} = ∫ f (bt ) e− s t dt ! 0−

τ = bt !

{ f (bt )} = b1 F ⎛⎜⎝ bs ⎞⎟⎠ !

F (b s ) =

b>0!

(2.3-5)

1 ⎧ ⎛ t ⎞⎫ L ⎨ f ⎜ ⎟ ⎬! b ⎩ ⎝ b⎠ ⎭

(2.3-6)

F (b s ) =

!

∫

∞

0−

f ( τ ) e− b s τ dt !

Re s >

σ ! b

(2.3-7)

Changing variables: !

dτ ! dt = b

From the definition of the Laplace transform we can write:

(2.3-2)

Changing variables: !

(2.3-4)

where b > 0 .

!

L

f ( τ ) e− ( s b ) τ dτ !

Proof:

Proof:

!

0

−

{ f (t )} = F ( s ) , then:

!

where b > 0 .

!

∫

∞

Proposition 2.3-2, Scaling Property (t Scaling): If L

{ f (t )} = F ( s ) , then: L

1 b

■

Proposition 2.3-1, Scaling Property (s Scaling):

!

L

!

The Laplace transform operation has a property known as

If L

{ f (bt )} =

and so:

2.3! SCALING PROPERTY !

L

!

(2.3-3)

t = bτ !

dτ =

dt ! b

(2.3-8)

we have:

we have: 36

F (b s ) =

!

1 b

∫

∞

⎛t⎞ f ⎜ ⎟ e− s t dt ! 0− ⎝ b ⎠

(2.3-9)

Determine the Laplace transform for both eb t cos ( at ) and

and so:

1 ⎧ ⎛ t ⎞⎫ F (b s ) = L ⎨ f ⎜ ⎟ ⎬ ! b ⎩ ⎝ b⎠ ⎭

!

Example 2.3-2

b > 0!

(2.3-10)

eb t sin ( at ) where a and b are real constants.

Solution:

■

From Example 2.3-1 and Proposition 2.2-1: Example 2.3-1 Determine the Laplace transform for cos ( at ) and sin ( at ) where a is a real constant.

!

{

}

{

}

L ebt cos ( at ) =

!

L ebt sin ( at ) =

From Example 2.1-4 we have: L {cos ( t )} =

s s2 + 1

!

L {sin ( t )} =

1 s2 + 1

Applying Proposition 2.3-1 we then obtain: !

!

s a 1 s L {cos ( at )} = = 2 2 a ⎛ s⎞ s + a2 ⎜⎝ a ⎟⎠ + 1

L {sin ( at )} =

( s − b )2 + a 2

and

Solution:

!

s−b

1 1 a = 2 2 a ⎛ s⎞ s + a2 + 1 ⎜⎝ a ⎟⎠

!

Re s > 1

a

( s − b )2 + a 2

Example 2.3-3 Determine the Laplace transform for cos 2 t . Solution: Using trigonometric formula: !

cos 2 t =

1 ⎡1+ cos ( 2t ) ⎤⎦ 2⎣

and Proposition 2.1-1 we have:

37

!

{

}

L cos 2 t =

1 1 L {1} + L {cos ( 2t )} 2 2

Therefore from Examples 1.4-2 and 2.3-1: !

{

}

L cos 2 t =

11 1 s s2 + 2 + = 2 s 2 s2 + 4 s s2 + 4

(

)

38

Chapter 3 Laplace Transforms of Derivatives and Integrals

L

( )

− f t = s F s − f 0 ′ ( ) { ( )}

39

!

The primary application of the Laplace transform is in the

Since f ( t ) is a continuous function of exponential order over

solution of differential equations. The use of Laplace transforms

[ 0, ∞ ) and f ′ (t ) is a piecewise-smooth function over [ 0, ∞ ) , we

in solving ordinary differential equations is the subject of

can integrate by parts. Letting:

Chapter 9, and their use in solving partial differential equations

!

u = e− s t !

dv = f ′ ( t ) dt !

(3.1-3)

we will now determine the Laplace transforms of derivatives

!

du = − s e− s t dt !

v = f (t ) !

(3.1-4)

and of integrals of functions.

we have:

is the subject of Chapter 10. In preparation for these chapters

3.1! LAPLACE TRANSFORMS OF DERIVATIVES OF CONTINUOUS FUNCTIONS

and f ′ ( t ) is a piecewise-smooth function over [ 0, ∞ ) , then

{ f ′ (t )} exists for all Re s > σ 0 , and we have: L

{ f ′ (t )} = s F ( s ) − f ( 0 − ) !

where L

!

t = 0−

+s

∫

0

−

f ( t ) e− s t dt !

(3.1-5)

(3.1-1)

{ f (t )} = F ( s ) .

lim f ( t ) e− s t − f ( 0 − ) + s F ( s ) ! { f ′ (t )} = t→ ∞

(3.1-6)

Since f ( t ) is of exponential order, we have from equation (1.4-8): !

f ( t ) e− s t = f ( t ) e− σ t e− i ω t ≤ M e− ( σ − σ 0 ) t !

σ > σ 0 ! (3.1-7)

and so: !

Proof: !

L

!

If f ( t ) is a continuous function of exponential order over [ 0, ∞ )

!

{ f ′ (t )} = f (t ) e

∞

and so:

Proposition 3.1-1, Laplace Transforms of Derivatives:

L

L

!

−st ∞

lim f ( t ) e− s t → 0 !

(3.1-8)

t→∞

and we have:

From the definition of the Laplace transform we can write:

L

{ f ′ (t )} = ∫

∞

0

−

f ′ ( t ) e− s t dt !

(3.1-2)

!

L

Therefore L

{ f ′ (t )} = s F ( s ) − f ( 0 − ) !

{ f ′ (t )} exists if Re s > σ 0 .

(3.1-9) ■

40

Note that f ′ ( t ) need not be of exponential order for this

!

Example 3.1-2

proposition. A function f ( t ) can be of exponential order and

Determine the Laplace transform for f (t) = sin 2 t .

yet its derivative f ′ ( t ) not be of exponential order.

Differentiation in the t -plain is seen to be equivalent to

Solution:

multiplication in the s -plain. Differential operations become

We have:

algebraic operations.

!

!

f ′ ( t ) = 2 sint cost = sin 2t

Example 3.1-1

and so from equation (3.1-1):

Determine the Laplace transform for f ′(t) where f (t) = sin ( t ) .

!

Solution:

we can write:

We have from Example 2.1-4:

!

!

L

L

!

L

!

L

as expected.

{

}

L sin 2 t =

L {sin 2t } s

=

1 2 s s2 + 4

If f ( t ) and f ′ ( t ) are both continuous functions of exponential

{ f ′ (t )} = s s 21+ 1 − 0 = s 2 s+ 1 { f ′ (t )} =

} ( )

Proposition 3.1-2, Laplace Transforms of Second Derivatives:

)

order over ⎡⎣ 0 − , ∞ , and if f ′′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then L { f ′′ ( t )} exists for all Re s > σ 0 , and we

)

From Example 2.1-4 we see: !

{

L {sin 2t } = s L sin 2 t − f 0 −

Since f 0 − = 0 , we have from Example 2.3-1:

{ f ′ (t )} = s F ( s ) − f ( 0 )

where f (0) = 0 . We then have:

{ f ′ (t )} = s F ( s ) − f ( 0 − )

( )

{ f (t )} = L {sin (t )} = s 21+ 1 = F ( s )

From Proposition 3.1-1 we can write: !

L

L {cos ( t )}

have: !

L

{ f ′′ (t )} = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − ) !

(3.1-10) 41

where L

{ f (t )} = F ( s ) .

We can use Proposition 3.1-1 to write: !

Proof: !

From the definition of the Laplace transform we can write:

L

!

{ f ′′ (t )} = ∫

∞

0

−

f ′′ ( t ) e− s t dt !

(3.1-11)

L

{ f ′′ (t )} = s ⎡⎣ s F ( s ) − f ( 0− )⎤⎦ − f ′ ( 0− ) !

(3.1-17)

L

{ f ′′ (t )} = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − ) !

(3.1-18)

or !

Since f ( t ) and f ′ ( t ) are continuous functions of exponential

Therefore L

over [ 0, ∞ ) , we can integrate by parts. Letting:

!

order over [ 0, ∞ ) , and f ′′ ( t ) is a piecewise-smooth function

{ f ′′ (t )} exists if Re s > σ 0 .

■

Note that f ′′ ( t ) need not be of exponential order.

!

u = e− s t !

dv = f ′′ ( t ) dt !

(3.1-12)

Example 3.1-3

!

du = − s e− s t dt !

v = f ′ (t )!

(3.1-13)

Determine the Laplace transform for f ′′(t) where

f (t) = sin ( t ) .

we have:

L

!

{ f ′′ (t )} = f ′ (t ) e

−st ∞

+s

t = 0−

∞

∫

0−

f ′ (t ) e

−st

Solution:

dt ! (3.1-14)

and so: ! L

!

( )

{ f ′′ (t )} = lim f ′ (t ) e− s t − f ′ 0− + s t→ ∞

∫

∞

0

−

f ′ ( t ) e− s t dt ! (3.1-15)

L

{ f ′′ (t )} = − f ′ ( 0 ) + s ∫ −

∞

0−

f ′ (t ) e

L

{ f (t )} = L {sin (t )} = s 21+ 1 = F ( s )

and !

Since f ′ ( t ) is of exponential order, we obtain: !

From Example 2.1-4 we have:

f ′ ( t ) = cos ( t )

From Proposition 3.1-2 we can write: −st

dt !

(3.1-16)

!

L

{ f ′′ (t )} = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − ) 42

where f (0) = 0 and f ′(0) = 1. Therefore:

L

!

1 1 L { f ′′ ( t )} = s 2 − s (0) − 1 = − 2 s +1 s +1 2

!

L

{ f ′′ (t )} =

L

!

{ − sin (t )}

Proposition 3.1-3, Laplace Transforms of Higher Order Derivatives:

Re s > σ 0 , and we have:

!

L

!

!

{ f ( ) (t )} = s F ( s ) − s f ( 0 ) − s n

where L

n

n−1

−

n−2

( )

n−1−k

( )

f (k ) 0− !

(3.1-20)

k=0

Note that f ( n ) ( t ) need not be of exponential order.

If f ( t ) is a piecewise continuous function of exponential order

)

exponential order over ⎡⎣ 0 − , ∞ , and if f ( n ) ( t ) is a piecewisesmooth function over ⎡⎣ 0 − , ∞ , then L f ( n ) ( t ) exists for all

{

∑s

Proposition 3.2-1:

If f ( t ) , f ′ ( t ) , f ′′ ( t ) ,!, f ( n−1) ( t ) , are continuous functions of

)

n−1

3.2! LAPLACE TRANSFORMS OF DERIVATIVES OF PIECEWISE CONTINUOUS FUNCTIONS

as expected.

)

(t )} = s F ( s ) − n

■

or !

{f

(n)

over ⎡⎣ 0 − , ∞ with a single finite discontinuity at t = t 0 , and if f ′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then:

}

( )

L

!

f ′ 0 − − !− f ( n−1) 0 −

{ f (t )} = F ( s ) .

{ f ′ (t )} = s F ( s ) − f ( 0− ) − ⎡⎣ f (t0+ ) − f (t0− )⎤⎦ e− s t

where L

(3.1-19)

)

0

! (3.2-1)

{ f (t )} = F ( s ) and Re s > σ 0 .

Proof: !

For a function f ( t ) having a single jump discontinuity at

( )

( )

Proof:

t = t 0 similar to that shown in Figure 1.4-1, f t 0+ and f t 0− are

!

both finite and exist. We can write:

Follows by mathematical induction from Propositions

3.1-1 and 3.1-2. We then have:

!

L

{ f ′ (t )} = ∫

t 0−

0

−

f ′ ( t ) e− s t dt +

∫

∞

t 0+

f ′ ( t ) e− s t dt !

(3.2-2) 43

Integrating by parts, and letting: −st

!

u=e

!

du = − s e− s t dt !

!

Proposition 3.2-2:

dv = f ( t ) dt !

(3.2-3)

If f ( t ) is a piecewise continuous function of exponential order

v = f (t ) !

(3.2-4)

over ⎡⎣ 0 − , ∞ with finite discontinuities at t = t 0 , t1, t 2 , !, t n , and if f ′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then:

)

we have: L

! !

{ f ′ (t )} = f (t ) e− s t

t 0− t=0

−

+s

∫

t 0−

0

−

f ( t ) e− s t dt + f ( t ) e− s t +s

!

∫

∞

f ( t ) e− s t dt !

t 0+

∞

n

t = t 0+

!

L

!

{ f ′ (t )} = f (t0− ) e− s t

0

( )

− f 0− + s −f

!

( )e t 0+

∫

0

− s t0

−

+s

∫

t 0+

!

f ( t ) e− s t dt ! (3.2-6)

t = t 0− and t = t 0+ , we can combine the integrals: L

!

{ f ′ (t )} = f ( ) e

− s t0

( ) − f ( )e

−f 0

−

t 0+

− s t0

+s

∫

∞

0−

f (t ) e

!

−st

L

! ■

{ f ′ (t )} = s F ( s ) − f ( 0 ) − ⎡⎣ f ( ) − f ( )⎤⎦ e t 0−

■.

LAPLACE TRANSFORMS OF INTEGRALS

Proposition 3.3-1: the integral of f ( t ) is of exponential order. Proof: !

− s t0

Follows from Proposition 3.2-1.

3.3!

dt

or

t 0+

{ f (t )} = F ( s ) and Re s > σ 0 .

If f ( t ) is a piecewise-smooth function of exponential order, then

(3.2-7) −

(3.2-9)

Proof:

f ( t ) e− s t dt ∞

k

k=0

where L

Since the function f ( t ) does not add to the integrals between

t 0−

{ f ′ (t )} = s F ( s ) − f ( 0 ) − ∑ ⎡⎣ f (t k+ ) − f (t k− )⎤⎦ e− s t !

L

−

(3.2-5)

or t 0−

)

! (3.2-8)

Since f ( t ) is a piecewise-smooth function, it is integrable

)

on ⎡⎣ 0 − , ∞ . We now let: 44

g (t ) =

!

∫

t

and so:

f ( τ ) dτ !

(3.3-1)

0

)

where g ( t ) will then be continuous over ⎡⎣ 0 − , ∞ except at points at which f ( t ) has a discontinuity. We also have:

d g′ ( t ) = dt

!

∫

t

f ( τ ) dτ = f ( t ) !

(3.3-2)

Since f ( t ) is piecewise-smooth, over the finite interval

⎡⎣ 0 , T0 ⎤⎦ we know that f ( t ) will be bounded by some positive constant K so that: −

! ! !

f (t ) ≤ K !

0 ≤ t ≤ T0 !

f ( t ) ≤ M eσ 0 t !

t > T0 !

g (t ) ≤

e σ 0 t dτ !

(3.3-7)

T0

(

)

M σ0 t e − eσ 0 T0 ! σ0

(3.3-8)

g ( t ) ≤ ( K − M ) T0 +

!

M σ0 t e ! σ0

(3.3-9)

Selecting M > K : g (t ) ≤

M σ0 t e ! σ0

(3.3-10)

Therefore g ( t ) is of exponential order for Re s > σ 0 and t > T0 . ■

∫

T0

f ( τ ) dτ +

0

∫

t

f ( τ ) dτ !

(3.3-5)

Proposition 3.3-2: If f ( t ) is a piecewise-smooth function of exponential order, then

T0

the Laplace tranform of the integral of f ( t ) exists.

or !

∫

t

Since eσ 0 T0 > σ 0 T0 approximating eσ 0 T0 with σ 0 T0 :

We can write: !

g ( t ) ≤ K T0 +

!

! (3.3-4)

dτ + M

or

(3.3-3)

Since f ( t ) is of exponential order, we also have:

∫

T0

0

0

Therefore g′ ( t ) is piecewise continuous. !

g (t ) ≤ K

!

g (t ) ≤

∫

T0

0

f ( τ ) dτ +

∫

t

T0

f ( τ ) dτ !

(3.3-6)

Proof: !

Follows from Propositions 3.3-1 and 1.4-3.

■

45

If f ( t ) is a piecewise-smooth function of exponential order, and if L

{ f (t )} = F ( s ) , then: ⎧ L ⎨ ⎩

!

∫

0−

or

⎧ Determine L ⎨ ⎩ Solution:

)

−

over ⎡⎣ 0 , ∞ . We now let:

g (t ) =

∫

t

f ( τ ) dτ !

!

)

where g ( t ) will then be continuous over ⎡⎣ 0 , ∞ except at points at which f ( t ) has a discontinuity. From Propositions −

!

for Re s > σ 0 and t > T0 ., and that g ( t ) has a Laplace transform

{ g (t )} = G ( s ) .

Since g ( t ) is of exponential order, using Proposition 3.1-1

L

{ f (t )} =

L

{ g′ (t )} = s G ( s ) − g ( 0 + ) !

( )

∫

L {cos ( τ )} = F ( s ) =

s s +1 2

⎧ L ⎨ ⎩

⎫ F (s) 1 s 1 cos ( τ ) dτ ⎬ = = = 2 = L {sin ( t )} 2 s s s + 1 s + 1 0 ⎭

∫

t

If f ( t ) is a piecewise-smooth function of exponential order, and if L

(3.3-13)

where from equation (3.3-12) we see g 0 + = 0 . Therefore we have:

⎫ cos ( τ ) dτ ⎬ . 0 ⎭ t

Proposition 3.3-4, Laplace Transform of Double Integrals:

we can write: !

(3.3-15)

From Proposition 3.3-3 we obtain:

3.3-1 and 3.3-2 we also know that g ( t ) is of exponential order

!

Re s > σ 0 !

We have from Example 2.1-4:

(3.3-12)

0

L

0

⎫ F (s) f ( τ ) dτ ⎬ = ! s ⎭

Example 3.3-1

Since f ( t ) is a piecewise-smooth function, it is integrable

!

∫

t

(3.3-14)

■

(3.3-11)

Proof: !

⎧ L ⎨ ⎩

!

⎫ F (s) f ( τ ) dτ ⎬ = ! s ⎭

t

{ f (t )} = F ( s ) = s G ( s ) = s L { g (t )}!

L

!

Proposition 3.3-3, Laplace Transform of Integrals:

!

{ f (t )} = F ( s ) , then: ⎧ L ⎨ ⎩

t

τ

0−

0−

∫∫

⎫ F (s) f ( γ ) dγ dτ ⎬ = 2 ! s ⎭

(3.3-16) 46

!

Since f ( t ) is a piecewise-smooth function, it is integrable

)

over ⎡⎣ 0 − , ∞ . We now let:

g (t ) =

!

t

τ

0−

0−

∫∫

f ( γ ) dγ dτ !

g′ ( t ) =

∫

0

f ( γ ) dγ !

(3.3-18)

⎧ Determine L ⎨ ⎩

!

we can write:

(3.3-23)

t

⎫ sin ( 2 γ ) dγ dτ ⎬ . 0 ⎭

∫∫ 0−

τ

{ g′′ (t )} = s 2G ( s ) − s g ( 0 − ) − g′ ( 0 − ) !

L

where L

{ g (t )} = G ( s ) and g ( 0 − ) = g′ ( 0 − ) = 0 . Therefore we

(3.3-20)

have:

{ f (t )} = s 2G ( s ) !

L {sin ( 2 τ )} = F ( s ) =

2 s +4 2

From Proposition 3.3-4 we obtain: !

!

!

0−

Example 3.3-2

(3.3-19)

Since g ( t ) is of exponential order, using Proposition 3.1-2

{ g′′ (t )} = L

0−

∫∫

⎫ F (s) f ( γ ) dγ dτ ⎬ = 2 ! s ⎭

From Example 2.3-1 we have:

g′′ ( t ) = f ( t ) !

L

τ

Solution:

Therefore g′ ( t ) and g′′ ( t ) are piecewise continuous. !

t

■

and !

⎧ L ⎨ ⎩

(3.3-17)

)

t

(3.3-22)

and so: !

where g ( t ) will then be continuous over ⎡⎣ 0 − , ∞ except at points at which f ( t ) has a discontinuity. We also have: !

F ( s ) = s 2G ( s ) !

!

Proof:

⎧ L ⎨ ⎩

t

⎫ F (s) 1 2 2 sin ( 2 γ ) dγ dτ ⎬ = 2 = 2 2 = 4 s s s + 4 s + 4s 2 0 ⎭

∫∫ 0−

τ

Proposition 3.3-5, Laplace Transform of Multiple Integrals: If f ( t ) is a piecewise-smooth function of exponential order, and

(3.3-21)

if L

{ f (t )} = F ( s ) , then:

or 47

! !

⎧ L ⎨ ⎩ !

t

τ1

τ2

0−

0−

∫∫ ∫ ∫ 0−

!

τn

0−

⎫ F (s) f ( γ n ) dγ n dγ n −1 ! dγ 1 dτ ⎬ = n ! s ⎭ (3.3-24)

Proof: !

Follows using the same procedures as in Propositions 3.3-3

and 3.3-4.

■

48

Chapter 4 Derivatives and Integrals of Laplace Transforms

L {t f ( t )} = − F ′ ( s )

49

!

In this chapter we will determine the derivatives and

integrals of Laplace transforms of functions. We will show that derivatives of all orders of the Laplace transform of a function

F′ ( s) =

!

f ( t ) will exist if f ( t ) is piecewise-smooth and of exponential

or

order with Re s > σ 0 .

!

4.1! DERIVATIVES OF LAPLACE TRANSFORMS

and so: !

Proposition 4.1-1: if L

!

L { t f ( t )} = −

d L ds

{ f (t )} = − F ′ ( s ) !

(4.1-1)

where F ′ ( s ) exists for Re s > σ 0 .

!

Proof:

!

From the definition of the Laplace transform, we have:

d F′ ( s) = L ds

{ f (t )} = dsd

∫

∞

0

!

−

f (t ) e

−st

dt !

(4.1-2)

Since f ( t ) is piecewise-smooth function of exponential order, the Laplace transform F ( s ) converges uniformly for Re s > σ 0

(see Proposition 1.4-7). Therefore the order of differentiation

∞

0−

∫

∞

0−

f ( t ) e− s t dt =

∫

∞

f (t )

0−

∂ −st e dt ! ∂s

(4.1-3)

( − t f (t )) e− s t dt !

(4.1-4)

F ′ ( s ) = L {− t f ( t )} !

Therefore:

!

!

F′ ( s) =

If f ( t ) is a piecewise-smooth function of exponential order, and

{ f (t )} = F ( s ) , then we have:

∫

d ds

L { t f ( t )} = − F ′ ( s ) = −

(4.1-5)

{ f (t )} !

d L ds

(4.1-6)

We can verify this equation using the definition: d L ds

{ f (t )} = F ′ ( s ) = Δslim →0

F ( s + Δs ) − F ( s ) Δs

!

(4.1-7)

We can now write: ! !

⎛ e − t ( s +Δs ) − e−s t ⎞ F ′ ( s ) − L {− t f ( t )} = lim ⎜ ⎟ f ( t ) dt Δs 0 − Δs → 0 ⎝ ⎠

∫

∞

−

∫

∞

0−

( − t f (t )) e− s t dt !

(4.1-8)

or

and integration in equation (4.1-2) can be interchanged. 50

!

F ′ ( s ) − L {− t f ( t )} =

!

⎡ ⎛ e− t Δs − 1 ⎞ ⎤ + t ⎥ f ( t ) e− s t dt ⎢ Δslim ⎜ ⎟ ⎣ → 0 ⎝ Δs ⎠ ⎦

∞

∫

0

−

!

(4.1-9)

or using L’Hôpital’s rule: !

F ′ ( s ) − L {− t f ( t )} =

∫

∞

0

−

!

d L ds

Determine the Laplace transform for t 2 using Proposition 4.1-1. Solution:

( − t + t ) f (t ) e− s t dt = 0 !

(4.1-10)

We have from Example 1.5-1: !

Therefore: !

Example 4.1-1

{ f (t )} = F ′ ( s ) = L { − t f (t )} !

(4.1-11)

Since f ( t ) is a piecewise-smooth function of exponential

order, we know from Proposition 1.4-3 that L

{ f (t )} exists for

L {t} =

1 = F (s) s2

Let f ( t ) = t . From Proposition 4.1-1 we can write: !

{ }

L { t f ( t )} = L t 2 = − F ′ ( s ) = −

d 1 2 = 3 2 ds s s

which agrees with the results of Example 1.5-3.

Re s > σ 0 . Since we also have:

!

t < ejt !

j>0!

(4.1-12)

Determine the Laplace transform for t e−at .

we can write: !

t f ( t ) ≤ Me ( j + σ 0 ) t !

Example 4.1-2

t ≥ T0 !

(4.1-13)

and so if f ( t ) is a piecewise-smooth function of exponential

order, then t f ( t ) will also be a piecewise-smooth function of exponential order. Therefore L {t f ( t )} must also exist for Re s > σ 0 , and so F ′ ( s ) must exist for all Re s > σ 0 .

Solution: −at Let f ( t ) = e . We have from Example 1.4-7:

!

{ }

L e−at =

1 = F (s) s+a

From Proposition 4.1-1 we can write:

■

!

L

{ t e } = − F′ ( s ) = − dsd s +1 a = ( s +1a ) −at

2

51

Example 4.1-3 Determine the Laplace transform for t sin ( at ) .

∫

!

Let f ( t ) = sin ( at ) . We have from Example 2.3-1

L { sin ( at )} =

d a 2as 2 2 = ds s + a s2 + a2

(

)

t m f ( t ) e− s t dt = ( −1)

m

dm F (s) ! ds m

(4.1-16)

∫

∞

0

−

t m f ( t ) e− s t dt !

(4.1-17)

(see Proposition 1.4-7). The order of differentiation and integration in equation (4.1-17) can then be interchanged.

{ f (t )} = F ( s ) , then F ( s ) exists for Re s > σ 0 , and is

!

d m+1 m m+1 F ( s ) = ( −1) ds

∫

∞

0−

t m f (t )

∂ −st e dt ! ∂s

(4.1-18)

or

given by: n

}

f ( t ) = ( −1) F ( n ) ( s ) ! n

(4.1-14)

where n is a positive integer. Proof: !

(4.1-15)

the Laplace transform F ( s ) converges uniformly for Re s > σ 0

(n)

{t

m

Since f ( t ) is piecewise-smooth function of exponential order, 2

If f ( t ) is a piecewise-smooth function of exponential order, and

L

}

f ( t ) = ( −1) F ( m ) ( s ) !

d m+1 m d m+1 F ( s ) = ( −1) ds ds

!

Proposition 4.1-2:

!

∞

0−

From Proposition 4.1-1 we can write:

L

m

We now can write:

a = F (s) 2 s + a2

! L { t f ( t )} = L { t sin ( at )} = − F ′ ( s ) = −

{t

or

Solution:

!

L

!

We can prove this proposition by mathematical induction.

We will assume this theorem is true for n = m . We then have:

d m+1 m+1 m+1 F ( s ) = ( −1) ds

!

∫

∞

0

−

t m+1 f ( t ) e− s t dt !

(4.1-19)

and so for n = m + 1 : L

!

{t

n

}

f ( t ) = ( −1) F ( n ) ( s ) ! n

(4.1-20)

and so by mathematical induction the theorem is true for all

n > 0.

■

52

!

From Proposition 4.1-2 we see that if a function has the

form t n f ( t ) where f ( t ) is a piecewise-smooth function of

Example 4.1-5

exponential order, then the Laplace transform L

Determine the Laplace transform for ( t − 3) e5 t .

obtained by calculating the nth derivative of multiplied by ( −1) . n

{t

n

}

f (t )

L

is

{ f (t )}

2

Solution: We are given:

Example 4.1-4

!

Determine the Laplace transform for t 2 sin ( at ) .

Let f ( t ) = e5t . We have from Example 1.4-6:

Solution: Let f ( t ) = sin ( at ) . We have from Example 2.3-1: !

L {sin ( at )} =

a s +a 2

2

(t − 3)2 e5 t = t 2 e5 t − 6t e5 t + 9 e5 t

!

{ }

L e5t =

From Proposition 4.1-2 we can write:

= F (s)

!

L

From Proposition 4.1-2 we can write: !

{

}

L t sin ( at ) 2

d⎛ a ⎞ 2as = − ⎜⎝ 2 ⎟ ds s + a 2 ⎠ s2 + a2

(

and so: !

(

)

{(t − 3) e } = (−1) F′′ ( s ) − 6 (−1) F′ ( s ) + 9 F ( s ) 2

5t

2

We have:

d2 ⎛ a ⎞ = ( −1) F ′′ ( s ) = 2 ⎜ 2 ⎟ ds ⎝ s + a 2 ⎠ 2

!

We have: !

1 = F (s) s−5

d 5t −1 ! F′ ( s) = e = ds ( s − 5 )2

( )

F ′′ ( s ) =

( )

d 2 5t 2 = 2 e ds ( s − 5 )3

and so: 2

!

)

L

{(t − 3) e } = ( s −25) 2

5t

3

−

6

( s − 5)

2

+

9 s−5

2 2 d2 ⎛ a ⎞ 2 a 3s − a = = L t 2 sin ( at ) ⎟ 2⎜ 2 2 3 ds ⎝ s + a ⎠ s2 + a2

(

)

{

} 53

Since derivatives of F ( s ) exist for all orders for Re s > σ 0 ,

! Proposition 4.1-3, Holomorphic Laplace Transform:

from the definition of the Laplace transform we have:

If f ( t ) is a piecewise-smooth function of exponential order, and L

{ f (t )} = F ( s ) , then F ( s ) is holomorphic in its region of

absolute convergence Re s > σ 0 . If f ( t ) is a piecewise-smooth function of exponential

order, we see from Propositions 4.1-1 and 4.1-2 that all higher order derivatives of F ( s ) will exist for the entire region of

or

{ f (t )} is holomorphic in its region of absolute convergence

Proof:

(4.1-22)

0−

e− s t t g ( t ) dt !

(4.1-23)

d ds

∫

∞

∫

∞

e− s t g ( t ) dt !

(4.1-24)

0−

{ f ′′ (t )} !

(4.1-25)

d ⎡ 2 s F ( s ) − s f 0 − − f ′ 0 − ⎤⎦ ! ds ⎣

( )

(4.1-26)

dF ( s ) − 2 s F ( s ) + f 0− ! ds

(4.1-27)

0−

e− s t f ′′ ( t ) dt = −

d L ds

Using Proposition 3.1-2 we obtain:

If f ( t ) is a piecewise-smooth function of exponential order, and

{ f (t )} = F ( s ) , then for all Re s > σ 0 : L {t f ′′ ( t )} = −s 2

L {t f ′′ ( t )} = −

!

■

Proposition 4.1-4:

!

∞

d L {t g ( t )} = − ds

absolute convergence Re s > σ 0 . We can then conclude that

region, F ( s ) is defined and differentiable in its entire region of

L

e− s t t f ′′ ( t ) dt !

−

From Proposition 4.1-1 we then have: !

Re s > σ 0 (see Section 11.2.4).

0

∫

L {t g ( t )} =

!

absolute convergence Re s > σ 0 . Since F ′ ( s ) then exists in this

L

∫

∞

Letting g ( t ) = f ′′ ( t ) we can write:

Proof: !

L {t f ′′ ( t )} =

!

dF ( s ) − 2 s F ( s ) + f 0− ! ds

( )

L {t f ′′ ( t )} = −

!

( )

Therefore: (4.1-21)

L {t f ′′ ( t )} = −s 2

!

( )

■

54

4.2! INTEGRALS OF LAPLACE TRANSFORMS

∫

!

∞

F ( u ) du =

s

Proposition 4.2-1: If f ( t ) is a piecewise-smooth function of exponential order, and ⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists, then: t →0 ⎝ t ⎠ ⎧ f (t ) ⎫ L ⎨ ⎬= ⎩ t ⎭

!

where L

∫

∞

(4.2-1)

∫

{ f (t )} = F ( s ) .

∫

∞

u =s

e−u t du dt !

∫

∞

∫

∞

∞

⎡ 1 ⎤ F ( u ) du = f ( t ) ⎢ − e−u t ⎥ dt ! ⎣ t ⎦ u=s 0+

∫

∞

F ( u ) du =

0+

s

Since f ( t ) is a piecewise-smooth function, f ( t ) t will also

f (t ) − s t e dt ! t

⎧ f (t ) ⎫ L ⎨ ⎬= ⎩ t ⎭

!

∫

∞

F ( u ) du !

exponential order as f ( t ) . Finally, since lim+ ( f ( t ) t ) exists, the

Determine the Laplace transform for

Laplace transform of f ( t ) t will also exist over ⎡⎣ 0 , ∞ .

Solution:

t →0

!

+

)

From the definition of the Laplace transform we can write:

∫

s

F ( u ) du =

∫ ∫ u=s

∞

t=0

f ( t ) e−u t dt du !

+

(4.2-2)

Since the integrand is uniformly convergent (see Proposition 1.4-7), we can change the order of integration:

(4.2-6)

■

Example 4.2-1

∞

(4.2-5)

s

order and f ( t ) t ≤ f ( t ) , we see that f ( t ) t will be of the same

∞

(4.2-4)

and we have:

be a piecewise-smooth function. Since f ( t ) is of exponential

!

(4.2-3)

Therefore: !

Proof: !

∞

s

s

t=0

+

f (t )

and so: !

F ( u ) du !

∫

∞

sin ( at ) t

.

⎛ f (t ) ⎞ Let f ( t ) = sin ( at ) . First we check to see if lim ⎜ ⎟ exists. + t →0 ⎝ t ⎠ We have using L’Hôpital’s rule:

!

⎛ sin ( at ) ⎞ ⎛ f (t ) ⎞ lim+ ⎜ = lim+ ⎜ = lim a cos ( at ) = a ⎟ t →0 ⎝ t ⎠ t →0 ⎝ t ⎟⎠ t → 0+ 55

and so the limit exists. Then from Proposition 4.2-1 we have: !

⎧ sin ( at ) ⎫ L ⎨ ⎬= ⎩ t ⎭

∫

∞

F ( u ) du =

s

∫

∞

s

!

a du 2 u + a2

⎧ sin ( at ) ⎫ L ⎨ ⎬= t ⎩ ⎭

∫

∞

s

!

∞

1 2

⎛ u⎞ ⎜⎝ ⎟⎠ + 1 a

du ⎡ −1 u ⎤ = tan a ⎢⎣ a ⎥⎦ u = s

!

⎧ sin ( at ) ⎫ π −1 ⎛ s ⎞ −1 ⎛ a ⎞ L ⎨ ⎬ = − tan ⎜⎝ ⎟⎠ = tan ⎜⎝ ⎟⎠ a s ⎩ t ⎭ 2

! Example 4.2-2 e a t − eb t . t

!

}

− ebt du

⎧ eat − ebt ⎫ L ⎨ ⎬= t ⎩ ⎭

s

1 ⎤ ⎡ 1 − ⎢⎣ u − a u − b ⎥⎦ du

⎧ eat − ebt ⎫ ⎛ u − a⎞ L ⎨ ⎬ = ln ⎜⎝ ⎟ t u − b⎠ ⎩ ⎭

⎛ a⎞ 1− ⎛ s−a⎞ ⎜ u⎟ = lim ln ⎜ − ln ⎜ ⎟ ⎝ s − b ⎟⎠ u→∞ b s 1− ⎜⎝ u ⎟⎠

∞

⎛e −e ⎞ ⎛ ae − be ⎞ ⎛ f (t ) ⎞ lim+ ⎜ = lim+ ⎜ = lim+ ⎜ ⎟ ⎟ ⎟⎠ = a − b t →0 ⎝ t ⎠ t →0 ⎝ t 1 ⎠ t →0 ⎝ at

⎧ eat − ebt ⎫ ⎛ s−b⎞ L ⎨ ⎬ = ln ⎜⎝ s − a ⎟⎠ t ⎩ ⎭

⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists where a is a constant, then: t →0 ⎝ t + a ⎠

⎛ f (t ) ⎞ Let f ( t ) = e − e . First we check to see if lim+ ⎜ ⎟ exists. t →0 ⎝ t ⎠ We have using L’Hôpital’s rule: bt

s

at

If f ( t ) is a piecewise-smooth function of exponential order, and

bt

at

∫

{e

−1

Proposition 4.2-2:

Solution:

at

s

L

or

See also Example 6.2-4.

Determine the Laplace transform for

∫

∞

F ( u ) du =

∞

and so:

and so: !

∫

∞

or

or !

⎧ eat − ebt ⎫ L ⎨ ⎬= t ⎩ ⎭

bt

and so the limit exists. Then from Proposition 4.2-1 we have:

!

⎧ f (t ) ⎫ s a L ⎨ ⎬=e ⎩t + a⎭

where L

∫

∞

e−u a F ( u ) du !

(4.2-7)

s

{ f (t )} = F ( s ) . 56

■

Proof: !

From the definition of the Laplace transform we can write:

!

es a

∫

∞

e−u a F ( u ) du = es a

s

∞

∫∫

∞

0−

s

e−u a f ( t ) e−u t dt du !

Proposition 4.2-3:

If f ( t ) is a piecewise-smooth function of exponential order, and

(4.2-8)

⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists, then: t →0 ⎝ t ⎠

Since the integrand is uniformly convergent (see Proposition 1.4-7), we can change the order of integration:

es a

!

∫

∞

e−u a F ( u ) du = es a

s

∫

∞

f (t )

0−

∫

∞

⎧ f (t ) ⎫ L ⎨ ⎬= ⎩ t ⎭

!

e−u (t + a ) du dt !

(4.2-9)

s

where L

and so: ! e

sa

∫

∞

e

−u a

F ( u ) du = e

∫

s

!

es a

∫

∞

e− u a F ( u ) du =

s

∫

∞

f (t )

−

t+a

0

e− s t dt !

Follows from Proposition 4.2-1 when s → 0 .

■

⎛ f (t ) ⎞ if lim+ ⎜ 2 ⎟ exists, then: t →0 ⎝ t ⎠

(4.2-11)

f ( t ) ( t + a ) will also be a piecewise-smooth function of order.

Therefore

the

Laplace

transform

f ( t ) ( t + a ) will exist, and we have: !

{ f (t )} = F ( s ) .

If f ( t ) is a piecewise-smooth function of exponential order, and

Since f ( t ) is a piecewise-smooth function of exponential order, exponential

(4.2-13)

Proposition 4.2-4:

Therefore: !

F ( u ) du !

0

Proof:

∞

∞

1 − u (t + a ) ⎤ ⎡ f (t ) ⎢ − e ⎥⎦ dt ! (4.2-10) ⎣ t+a 0− u=s

sa

∫

∞

⎧ f (t ) ⎫ s a L ⎨ ⎬=e ⎩t + a⎭

∫

∞

s

e−u a F ( u ) du !

⎧ f (t ) ⎫ L ⎨ 2 ⎬= ⎩ t ⎭

!

of

where L (4.2-12)

∞

∫∫ s

∞

u1

F ( u 2 ) du 2 du1 !

(4.2-14)

{ f (t )} = F ( s ) .

Proof: 57

Since f ( t ) is a piecewise-smooth function, f ( t ) t

!

also be a piecewise-smooth function. Since exponential order and f ( t ) t

2

2

will

f ( t ) is of

≤ f ( t ) , we see that f ( t ) t will

(

) exists, the Laplace transform of exist over ⎡⎣ 0 , ∞ ) . lim f ( t ) t

! !

!

2

f (t ) t

2

t

g (t ) t

=

t

2

!

(4.2-15)

!

(4.2-16)

From the Proposition 4.2-1 we can write: !

⎧ f (t ) ⎫ ⎧ g (t ) ⎫ L ⎨ 2 ⎬= L ⎨ ⎬= ⎩ t ⎭ ⎩ t ⎭

⎧ f (t ) ⎫ ! L ⎨ n ⎬= ⎩ t ⎭

∞

∫

s

G ( u1 ) du1 !

(4.2-17)

⎧ f (t ) ⎫ ⎧ g (t ) ⎫ L ⎨ 2 ⎬= L ⎨ ⎬= ⎩ t ⎭ ⎩ t ⎭

Using Proposition 4.2-1 again:

!

∫

s

⎧ f (t ) ⎫ L ⎨ ⎬ du1 ! ⎩ t ⎭

(4.2-18)

∞

∞

∞

u1

u2

un−1

∫ ∫ ∫ !∫ s

F ( un ) dun dun−1 !du1 !

(4.2-20)

{ f (t )} = F ( s ) .

Follows from Propositions 4.2-1 and 4.2-4 by mathematical ■

4.3! INITIAL AND FINAL VALUES OF LAPLACE TRANSFORMS !

∞

∞

Proof: induction.

where L { g ( t )} = G ( s ) . Therefore: !

(4.2-19)

⎛ f (t ) ⎞ if lim+ ⎜ n ⎟ exists, then: t →0 ⎝ t ⎠

where L f (t )

F ( u 2 ) du 2 du1 !

u1

If f ( t ) is a piecewise-smooth function of exponential order, and

We then have: !

s

Proposition 4.2-5:

will also

Let g (t ) =

∫∫

∞

■

+

f (t )

∞

2

be of the same exponential order as f ( t ) . Finally, since t → 0+

⎧ f (t ) ⎫ L ⎨ 2 ⎬= ⎩ t ⎭

The following two theorems provide information about a

function f ( t ) from its Laplace transform even when the actual

function f ( t ) is undetermined. The initial value theorem shows that the limit of the Laplace transform of a function f ( t )

as s → ∞ is related to the limits of f ( t ) as t → 0 + . The Final 58

value theorem shows that the limit of the Laplace transform of

( )

s F (s) − f 0

−

=

∫

0+

f ′ ( t ) e− s t dt +

a function f ( t ) as s → 0 is related to the limit of f ( t ) as t → ∞ .

!

!

Taking the limit as s → ∞ :

The initial value theorem provides information regarding

the short-term performance of f ( t ) , while the final value theorem

provides

information

regarding

the

long-term

performance of f ( t ) . If the function f ( t ) is known, the limiting

!

(

( ))

lim s F ( s ) − f 0

s→∞

0

−

−

∫

∞

0

⎡ = lim ⎢ s→∞ ⎢⎣

0+

∫

0−

+

f ′ ( t ) e− s t dt !

∫

f ′ ( t ) e− s t dt +

∞

0+

⎤ f ′ ( t ) e− s t dt ⎥ ⎥⎦

values of a Laplace transform can provide a check on the

!

correctness of the Laplace transform.

Since f 0 − and f 0 + are not functions of s , we have:

Proposition 4.3-1, Initial Value Theorem:

{ f (t )} = F ( s ) where f (t ) is a continuous function of exponential order over [ 0, ∞ ) and f ′ ( t ) is a piecewise-smooth function of exponential order over [ 0, ∞ ) , then: If L

( )!

lim ( s F ( s )) = f 0

!

s→∞

+

(4.3-1)

!

From the definition of the Laplace transform and from

Proposition 3.1-1 we have: !

!

L

( )

( ) = f ( 0 ) − f ( 0 ) + lim ∫

lim ( s F ( s )) − f 0

s→∞

−

+

−

s→∞

∞

0+

f ′ ( t ) e− s t dt

!

(4.3-5) L

{ f ′ (t )}

converges uniformly for

Re s > σ 0 (see Proposition 1.4-7), and so we can interchange the

integration and limiting process.

lim

s→∞

∫

∞

0+

f ′ (t ) e

−st

dt =

∫

∞

lim f ′ ( t ) e− s t dt !

0+ s → ∞

(4.3-6)

Since s is not a function of t , we can let s → ∞ prior to integrating:

{ f ′ (t )} = s F ( s ) − f ( 0− ) = ∫

We can write:

(4.3-4)

The Laplace transform

!

Proof:

!

( )

!

(4.3-3)

∞

0−

f ′ ( t ) e− s t dt !

(4.3-2)

!

∫

∞

lim f ′ ( t ) e− s t dt = 0 !

0 s→∞ +

(4.3-7)

and so equation (4.3-5) becomes:

59

!

( ) ( ) ( )

lim ( s F ( s )) − f 0 − = f 0 + − f 0 − !

!

s→∞

(4.3-8)

Proof:

If f ( t ) is discontinuous at the origin, we will have

( ) ( )

f 0 + ≠ f 0 − , and so equation (4.3-8) becomes:

( )

s→∞

(4.3-9) +

−

and so:

( )

s→∞

(4.3-10) !

( ) = f (0 ) !

lim ( s F ( s )) = f 0

!

s→∞

−

+

(4.3-11)

( ) − f ′(0 ) = ∫

s F (s) − s f 0 2

−

−

0+

{ f (t )} = F ( s ) where f (t ) and f ′ (t ) are continuous functions of exponential order over [ 0, ∞ ) , and f ′′ ( t ) is a piecewise-smooth function of exponential order on [ 0, ∞ ) , then: If L

(

( )) = f ′ ( 0 ) !

lim s F ( s ) − s f 0

+

−

f ′′ ( t ) e− s t dt +

∫

∞

0

+

f ′′ ( t ) e− s t dt !

Taking the limit as s → ∞ :

(4.3-12)

(

( )) =

( )

lim s 2 F ( s ) − s f 0 − − f ′ 0 −

s→∞

⎡ lim ⎢ s→∞ ⎢⎣

!

Proposition 4.3-2, Initial Value Theorem for Derivative:

s→∞

f ′′ ( t ) e− s t dt

(4.3-14)

Note also that certain functions such as sin at and cos at

do not have a unique limit, but oscillate with t .

!

0

−

(4.3-13)

!

−

∫

! !

■

2

( )

∞

!

0

or

!

!

( )

{ f ′′ (t )} = s 2 F ( s ) − s f 0− − f ′ 0− =

We can write:

lim ( s F ( s )) − f 0 − = 0 !

!

L

!

( ) = f (0 ) ,

If f ( t ) is continuous at the origin, we have f 0

!

From the definition of the Laplace transform and

Proposition 3.1-2 we can write:

lim ( s F ( s )) = f 0 + !

!

!

∫

0+

0−

f ′′ ( t ) e− s t dt +

∫

∞

0+

⎤ f ′′ ( t ) e− s t dt ⎥ ! ⎥⎦

(4.3-15)

( )

( ) lim ( s F ( s ) − s f ( 0 )) − f ′ ( 0 ) = f ′ ( 0 ) − f ′ ( 0 )

Since f ′ 0 − and f ′ 0 + are not functions of s , we have: ! !

2

−

−

+

−

s→∞

+ lim

s→∞

∫

∞

0+

f ′′ ( t ) e− s t dt !

(4.3-16)

60

The Laplace transform L

{ f ′ (t )}

converges uniformly for

Re s > σ 0 (see Proposition 1.4-7), and so we can interchange the

(

( )) = f ′ ( 0 ) = f ′ ( 0 ) !

lim s 2 F ( s ) − s f 0 −

!

s→∞

−

+

(4.3-22)

■

integration and limiting process. !

lim

s→∞

∫

∞

0+

f ′′ ( t ) e

−st

dt =

∫

∞

lim f ′′ ( t ) e

0+ s → ∞

Proposition 4.3-3, Final Value Theorem: −st

dt !

{ f (t )} = F ( s ) where f (t ) is a continuous function of exponential order over [ 0, ∞ ) and f ′ ( t ) is a piecewise-smooth function of exponential order on [ 0, ∞ ) , then:

(4.3-17)

If L

Since s is not a function of t , we can let s → ∞ prior to integrating:

∫

!

∞

lim f ′′ ( t ) e

−st

0+ s → ∞

dt = 0 !

(

s→∞

−

+

−

!

( )

( )

f ′ 0 + ≠ f ′ 0 − , and so equation (4.3-19) becomes:

(

( )) = f ′ ( 0 ) !

lim s 2 F ( s ) − s f 0 −

!

s→∞

!

If

or

is

continuous

( ) = f ′ ( 0 ) , and so:

f′ 0

!

f (t )

+

+

at

the

we

have

−

(

( ))

( )

lim s 2 F ( s ) − s f 0 − − f ′ 0 − = 0 !

s→∞

!

L

{ f ′ (t )} = s F ( s ) − f ( 0 ) = ∫ −

∞

0

−

f ′ ( t ) e− s t dt !

(4.3-24)

We can write: (4.3-20)

origin,

We have from Proposition 3.1-1:

(4.3-19)

If f ( t ) is discontinuous at the origin, we will have

!

(4.3-23)

t→∞

Proof:

( )) − f ′ ( 0 ) = f ′ ( 0 ) − f ′ ( 0 ) !

lim s F ( s ) − s f 0

!

−

s→0

(4.3-18)

and so equation (4.3-16) becomes: 2

lim ( s F ( s )) = lim f ( t ) !

!

!

lim

s→0

∫

∞

0

−

(

( )) !

f ′ ( t ) e− s t dt = lim s F ( s ) − f 0 −

The Laplace transform L

s→0

{ f (t )} = F ( s )

(4.3-25)

converges uniformly

for Re s > σ 0 (see Proposition 1.4-7), and so we interchange the (4.3-21)

integration and limiting process: !

∫

∞

(

( )) !

lim f ′ ( t ) e− s t dt = lim s F ( s ) − f 0 −

0− s → 0

s→0

(4.3-26)

61

Since s is not a function of t , we can let s → 0 prior to

!

L

integrating. Equation (4.3-26) then becomes:

∫

!

∞

0−

(

( ))

f ′ ( t ) dt = lim s F ( s ) − f 0 − ! s→0

(4.3-27)

( )

∫

!

0

−

( )

f ′ ( t ) dt = lim ( s F ( s )) − f 0 − ! s→0

(4.3-28)

( )

( )

lim f ( t ) − f 0 − = lim ( s F ( s )) − f 0 − !

t→∞

s→0

(4.3-29)

( )

f 0 + = lim ( s F ( s )) = lim s→∞

s =1 s→∞ s + 2

From Proposition 4.3-3 we have the final value when t → ∞ : !

lim f ( t ) = lim ( s F ( s )) = lim

t→∞

s→0

s =0 s→0 s + 2

These values agree with the function f ( t ) = e− 2t for t = 0 and

t→∞.

and so: !

the initial value when t = 0 : !

or !

− 2t

From Proposition 4.3-1 and using L’Hôpital’s rule, we have

Since f 0 − is not a function of s , we have: ∞

{ e } = s +1 2

lim ( s F ( s )) = lim f ( t ) !

s→0

(4.3-30)

t→∞

assuming that lim f ( t ) exists. t→∞

■

Example 4.3-1 Determine the initial and final values of the function

f ( t ) = e− 2t . Solution: From Example 1.4-7 the Laplace transform of f ( t ) = e− 2t is:

62

Chapter 5 Laplace Transforms of Discontinuous and Periodic Functions

{

}

L U (t − t0 )

e− s t0 = s

63

!

In this chapter we will consider the Laplace transforms of

discontinuous and periodic functions. The discontinuous

where t 0 is a constant (see Figure 5.1-1). The discontinuity or jump from 0 to 1 for U ( t − t 0 ) occurs at t = t 0 (see Figure 5.1-1).

functions we will consider involve either unit step functions or unit impulse functions. Discontinuous and periodic functions often provide the forcing functions for differential equations.

5.1! LAPLACE TRANSFORM OF UNIT STEP FUNCTIONS !

The unit step function (also called the Heaviside step

function) is a discontinuous function that has the shape of a stair step having unit value, jumping from 0 to 1 at the origin (see Figure 5.1-1).

5.1.1! DEFINITION OF UNIT STEP FUNCTIONS !

The unit step function U ( t ) is a function of t , and is

defined as: ! ! !

⎧ 0 t 0 . Therefore the region of absolute convergence for the unit step function is the entire right half-plane σ > 0 . !

t 0 ≥ 0 ! (5.1-5)

filtering property of the unit step function. The filtering action of the unit step function is illustrated in Figure 5.1.2.

Both the unit step function and the shifted unit step

function are clearly of exponential order since we have M = 1 and σ 0 = 0 : U ( t ) ≤ M eσ 0 t ≤ 1 !

!

(5.1-3)

Multiplication of a unit step function by a constant c yields a step function of amplitude c !

By forming the product of a function f ( t ) with the shifted

unit step function U ( t − t 0 ) , the product will be zero for all

t < t 0 , but will equal the function f ( t ) for all t ≥ t 0 . We then

have: !

⎧⎪ 0 t < t0 ! f (t ) U (t − t0 ) = ⎨ ⎪⎩ f ( t ) t ≥ t 0

We also have:

t0 ≥ 0 !

(5.1-4) Figure 5.1-2! Examples of filtering action of the unit step function on the unit ramp function f ( t ) = t . 65

Example 5.1-2

Example 5.1-4

Express the function f ( t ) defined by:

Express the f ( t ) shown in Figure 5.1-3 in terms of unit step

!

⎧ t ⎪ f (t ) = ⎨ t 2 ⎪ t3 ⎩

functions:

0≤t 0 !

(6.2-10)

We can conclude then that

f (t ) ≡ g (t ) !

if F ( s ) = G ( s ) for some Re s > σ 0 . Therefore over any set of points for which f ( t ) is a continuous function of exponential

We will let:

h (t ) =

∫

t

0

[1] ⎡⎣ f ( τ ) − g ( τ )⎤⎦ dτ !

(6.2-5)

If f ( t ) and g ( t ) are continuous functions of exponential order over some Re s > σ 0 , then from equation (6.2-5) we see that over this set of points: !

{0} = 0 !

−1

and so:

!

Proof: !

h (t ) = L

h′ ( t ) = f ( t ) − g ( t ) !

order, F ( s ) will have a unique inverse Laplace transform. !

■

Integrals only operate on continuous sequences of points,

and not on single isolated points. Therefore if f ( t ) = g ( t ) over a set of points except for isolated points of finite discontinuity in

f ( t ) or g ( t ) at which f ( t ) ≠ g ( t ) , then we will still have

(6.2-6)

F ( s ) = G ( s ) . In such cases, however, the inverse Laplace

86

transform will not be unique because the differences in f ( t )

where k is a finite constant. We see then that f ( t ) ≠ g ( t ) .

transform. Information is essentially lost in the process of the

Solution:

direct transformation. It is then possible for f ( t ) and g ( t ) to

We have:

and g ( t ) at points of discontinuity are invisible to the Laplace

differ by a null function n ( t ) :

h′ ( t ) = f ( t ) − g ( t ) = n ( t ) ≠ 0 !

!

∫

∞

∫

∞

0

(6.2-11)

f ( t ) e− s t dt =

−

∫

∞

0

h ( t ) e− s t dt

−

and

where:

h (t ) =

!

F (s) =

!

t

∫ n (τ) dτ = 0 !

t > 0!

! (6.2-12)

0

so that L {h ( t )} = 0 and we have F ( s ) = G ( s ) when f ( t ) ≠ g ( t ) . The discontinuous unit step function U ( t ) and the number 1,

0

Example 6.2-1

! G ( s ) = lim

ε1→ 0

transform where: !

f (t ) = h (t ) !

!

⎧ h (t ) 0− ≤ t < T ⎪⎪ g ( t ) = ⎨ k ≠ h (T ) t =T ⎪ t >T ⎪⎩ h ( t )

g ( t ) e− s t dt

∫

T − ε1

0−

h (t ) e

−st

dt +

∫

T

T

ke

−st

dt + lim

ε2 → 0

∫

∞

h ( t ) e− s t dt

T + ε2

or !

Show that the functions f ( t ) and g ( t ) have the same Laplace

−

where:

for example, both have the same Laplace transform 1 s . A null function cannot be a continuous function.

G (s) =

G (s) =

∫

∞

0−

h ( t ) e− s t dt

Therefore F ( s ) = G ( s ) although f ( t ) ≠ g ( t ) .

t ≥ 0− Example 6.2-2 Show that if n ( t ) is a null function, then so is: !

L {n ( t )} = N ( s ) 87

Solution:

Solution:

We have by definition:

From Proposition 4.1-1 we have:

∫

!

t

n ( τ ) dτ = 0 !

N (s) =

τ=0

∫

∞

0

−

!

n ( t ) e− s t dt

Therefore:

Let − st

!

dv = n ( τ ) dτ

!

u=e

!

du = − s e− s t dt !

!

v=

∫

t

!

We then can write:

⎡ n ( t ) e− s t dt = − ⎢ e− s t 0− ⎣

∫

!

∞

⎤ n ( τ ) dτ ⎥ + τ=0 ⎦ t=0

∫

t

!

N (s) =

∫

0

−

{

L {t f ( t )} = L eb t − ea t

∫

s e− s t

t =0

∫

t

n ( τ ) dτ dt

!

τ=0

or !

f (t ) =

n ( t ) e− s t dt = 0

(

1 bt at e −e t

Example 6.2-4

Example 6.2-3

!

Determine the inverse Laplace transform f ( t ) for: ⎛ s − a⎞ F ( s ) = log ⎜ ⎝ s − b ⎟⎠

}

t f ( t ) = ebt − eat

Therefore if n ( t ) is a null function, then so is L {n ( t )} = N ( s ) .

!

1 1 − s−b s−a

We then have: ∞

and so: ∞

L {t f ( t )} =

and so:

n ( τ ) dτ

0

∞

L {t f ( t )} = − F ′ ( s )

)

Determine the inverse Laplace transform f ( t ) for: ⎛ 1⎞ F ( s ) = tan −1 ⎜ ⎟ ⎝ s⎠

Solution: From Proposition 4.1-1 we have: 88

L {t f ( t )} = − F ′ ( s )

!

1.! Generally the simplest method of inverting a Laplace

Therefore:

transform involves using an existing table of Laplace

L {t f ( t )} = −

!

−

transforms. Such a table provides quick and easy assess

1 s2

⎛ 1⎞ 1+ ⎜ ⎟ ⎝ s⎠

2

or L {t f ( t )} = −

!

−1 s2 + 1

and so: L {t f ( t )} = − L {− sin ( t )}

!

We then have: !

t f ( t ) = sin ( t )

limited to only the more common (standard) functions. This method is presented in this chapter. 2.! The most general method of inverting a Laplace transform is by residue integration using the inversion formula given in equation (6.2-4). This method requires some knowledge of complex variable theory, including knowledge of contour integration. This method is presented in Chapter 12. Complex variable theory is first reviewed in Chapter 11. 3.! A third method of inverting a Laplace transform is to use

or f (t ) =

!

6.2.2!

sin ( t ) t

METHODS OF INVERTING A LAPLACE TRANSFORM

!

to both the direct and inverse Laplace transforms, but is

There are a number of methods available to invert a

one or more of the properties of Laplace transforms listed below in Section 6.2.3 in conjunction with a Laplace transform table. 4.! A fourth method of inverting a Laplace transform is to use the convolution integral. This method is presented in Chapter 7.

Laplace transform. Some of these are: 89

5.! A fifth method of inverting a Laplace transform is to expand the transform into a series, and to invert the series term-by-term. This method is presented in Chapter 8.

numerical integration. This method will not be discussed in this book.

6.2.3!

PROPERTIES OF THE INVERSE LAPLACE TRANSFORM

Inverse Laplace transforms have a linearity property

transforms.

and g ( t ) are continuous functions. Therefore the inverse L

!

{F ( s )} = f (t ) !

{G ( s )} = g (t ) !

(6.2-14)

{ f (t )} + c2 L {g (t )} !

(6.2-15)

−1

L

L {c1 f ( t ) + c2 g ( t )} = c1 L

!

−1

where c1 and c2 are arbitrary constants. Therefore: L {c1 f ( t ) + c2 g ( t )} = c1 F ( s ) + c2 G ( s ) !

!

c1 f ( t ) + c2 g ( t ) = L

!

If f ( t ) and g ( t ) are continuous functions having Laplace

which is:

transforms L

!

{ f (t )} = F ( s ) and

L { g ( t )} = G ( s ) ,

respectively, then the inverse Laplace transform of

(6.2-16)

−1

{c1 F ( s ) + c2 G ( s )} !

(6.2-17)

{c1 F ( s ) + c2 G ( s )} = c1 L −1{F ( s )} + c2 L −1{G ( s )}

−1

!

(6.2-18)

■

{c1 F ( s ) + c2 G ( s )} = c1 L −1{F ( s )} + c2 L −1{G ( s )}

−1

! where c1 and c2 are arbitrary constants.

L

! !

c1 F ( s ) + c2 G ( s ) is a linear operator:

!

L { g ( t )} = G ( s ) where f ( t )

and so:

Proposition 6.2-2, Linearity Property of Inverse Laplace Transforms:

L

{ f (t )} = F ( s ) and

From Proposition 2.1-1 we have:

which follows from the linearity property of Laplace

!

We have: L

!

Laplace transforms exist (see Proposition 12.4-1):

6.! A sixth method of inverting a Laplace transform is to use

!

Proof:

(6.2-13)

!

The following are some important properties of inverse

Laplace transforms that can be used to extend the utility of existing transform tables: 90

1.! From Proposition 6.2-2 we have the linearity property

L

!

−1 ⎧

d ⎫ ⎨ F ( s )⎬ = L ⎩ ds ⎭

for inverse Laplace transforms: L

!

{c1 F ( s ) + c2 G ( s )} = c1 L −1{F ( s )} + c2 L −1{G ( s )}

−1

! !

for inverse Laplace transforms: !

L

−1

{F ( s − a )} = ea t f (t ) !

Re ( s − a ) > 0 ! (6.2-20)

3.! From Proposition 2.3-1 we have the scaling property for inverse Laplace transforms: !

1 L b

−1 ⎧

⎛ s⎞⎫ ⎨ F ⎜⎝ ⎟⎠ ⎬ = f ( bt ) ! ⎩ b ⎭

!

b>0!

(6.2-21)

L

(s) ⎫ ⎨ ⎬= ⎩ s ⎭

−1 ⎧ F

∫

0−

−1

L

L

(s) ⎫ ⎨ 2 ⎬= ⎩ s ⎭

−1 ⎧ F

t

∫∫ 0

−

τ

{F ( τ )} dτ !

0

−

L

−1

{F ( γ )} dγ dτ !

5.! From Proposition 4.1-1 we have:

L

!

6.3! !

−1

{e

− s t0

}

F ( s ) = f (t − t0 ) U (t − t0 ) !

(6.2-25)

INVERSE LAPLACE TRANSFORMS FROM A TRANSFORM TABLE Many simple proper rational fractions can be directly

table. Extensive tables of Laplace transforms for the most common functions have been tabulated and are readily available in the literature and on the internet. A table of some common Laplace transforms is given in Appendix E, and a

(6.2-22)

table of Laplace transform properties is given in Appendix F. !

A Laplace transform F ( s ) of a function f ( t ) generally

takes the form of a rational fraction unless the transform

4.! From Proposition 3.3-4 we have: !

property for inverse Laplace transforms:

associated with an inverse Laplace transform using a transform

4.! From Proposition 3.3-3 we have: t

(6.2-24)

6.! From Proposition 5.2-1 we have the t-plain shifting

(6.2-19)

2.! From Proposition 2.2-1 we have the s-shifting property

{F ′ ( s )} = − t f (t ) !

−1

includes an impulse function. A rational fraction is defined to (6.2-23)

be an algebraic fraction in which the numerator is a polynomial

P ( s ) of degree m , and the denominator Q ( s ) is a polynomial of degree n : 91

! !

am s m + am−1 s m−1 +!+ a1 s + a0 ( ) ! F ( s) = = bn s n + bn−1 s n−1 +!+ b1 s + b0 Q ( s) P s

(6.3-1)

The first step in obtaining an inverse Laplace transform

using a transform table is to ensure that the degree of the

If m ≥ n , we see then that the function f ( t ) will always include unit impulse functions. !

From the fundamental theorem of algebra we know that

any polynomial having only real coefficients can be factored

numerator of the transform is less than the degree of the

into a product of linear factors ( a s + b ) or quadratic factors of

denominator. This will generally be true since from Proposition

the form a s 2 + b s + c

that are irreducible (have no other real

1.4-5 we know that we must have:

factors). After factoring the denominator of equation (6.3-1) into

!

lim F ( s ) = 0 !

Re s > k !

(6.3-2)

n

(

)

!

If m ≥ n , we can then divide the numerator P ( s ) by the

denominator Q ( s ) to obtain a proper rational fraction having n > m , together with a polynomial in s :

!

()

F s = cm s m + cm−1 s m−1 +!+ c1 s + c0 +

() ! Q ( s) P1 s

(6.3-3)

m m−1 ! f ( t ) = cmδ ( ) ( t ) + cm−1 δ ( ) ( t ) +!+ c1 δ ′ ( t ) + c0 δ ( t )

() ()

⎧ P s ⎫⎪ ⎨ ⎬ ! (6.3-4) ⎪⎩ Q s ⎪⎭

−1 ⎪ 1

P (s)

c ( s − z1 ) ( s − z2 ) ( s − z 3 )!( s − zn )

!

(6.3-5)

where c is a constant multiplier and z n are the n roots of Q ( s ) . The roots can be real, imaginary, or complex, and are called the poles or the F ( s ) singularities (see Chapter 11). An imaginary a root. Along with c the roots z n completely specify Q ( s ) .

6.3.1! !

the unit impulse function and its derivatives:

+L

F (s) =

!

or complex root only occurs when its complex conjugate is also

The polynomial in s is the Laplace transform of multiples of

!

n

first-order and quadratic factors, we obtain:

Re s→ ∞

if f ( t ) is a piecewise continuous function of exponential order over ⎡⎣ 0 − , ∞ .

)

PARTIAL FRACTION EXPANSION

When using transform tables to obtain the inverse Laplace

transform when more than one pole is present, the factored Laplace transform F ( s ) in equation (6.3-5) is first changed into a sum of simpler fractions known as partial fractions. Each denominator of a partial fraction will be a single irreducible 92

first-order or quadratic factor of Q ( s ) . Only proper rational

!

fractions can be expanded as partial fractions, each of which is

f (t ) =

t4 t4 = 4! 24

also a proper rational fraction. The number of partial fractions will equal the degree of the denominator polynomial. This

Example 6.3-2

process is called partial fraction expansion or partial fraction.

Determine the inverse Laplace transform:

Many partial fractions are common functions which can be directly associated with an inverse Laplace transform using a

!

f (t ) = L

⎧ 1 ⎨ 2 ⎩⎪ s s + 4

−1 ⎪

transform table. The inverse Laplace transform of the original function is then constructed from the standard functions in the table. A review of partial fractions is given in Appendix C.

6.3.2!

INVERSE LAPLACE TRANSFORMS FOR CONTINUOUS FUNCTIONS USING LAPLACE TRANSFORM TABLES

(

)

⎫⎪ ⎬ ⎭⎪

Solution: From the Laplace transform table in Appendix E we have: !

L

⎧ 1 ⎨ 2 ⎩⎪ s s + 4

−1 ⎪

(

)

⎫⎪ 1 ⎬ = ⎡⎣1− cos ( 2t ) ⎤⎦ 4 ⎭⎪

Example 6.3-1

Example 6.3-3

Determine the inverse Laplace transform:

Determine the inverse Laplace transform:

!

f (t ) = L

−1 ⎧

1⎫ ⎨ 5⎬ ⎩s ⎭

!

f (t ) = L

−1 ⎧

1 ⎫ ⎨ ⎬ ⎩ s ( s − 5) ⎭

Solution:

Solution:

From the Laplace transform table in Appendix E we have:

Using equation (6.2-22) we have:

93

!

f (t ) = L

−1 ⎧

1 ⎫ ⎨ ⎬= ⎩ s ( s − 5) ⎭

∫

t

0−

L

−1 ⎧

1 ⎫ ⎨ ⎬ dτ ⎩τ − 5 ⎭

From the Laplace transform table in Appendix E we have: !

f (t ) =

∫

t

0

−

e5 τ dτ

!

!

e5t − 1 = 5 τ=0

s+9 ⎫ ⎨ ⎬ ⎩ s ( s + 3) ⎭

Expanding as partial fractions we have: !

t

−1⎧

Solution:

Therefore: e5 τ f (t ) = 5

f (t ) = L

s+9 3 2 = − s ( s + 3) s s + 3

Therefore using Proposition 6.2-2: !

f (t ) = L

−1⎧

s+9 ⎫ ⎨ ⎬= 3 L ⎩ s ( s + 3) ⎭

Another way to solve this problem is to let: !

F (s) =

1 s ( s − 5)

−1⎧ 1 ⎫

⎨ ⎬− 2 L ⎩s ⎭

−1⎧

1 ⎫ ⎨ ⎬ ⎩s + 3⎭

From the Laplace transform table in Appendix E we have: !

(

)

f ( t ) = 3 (1) − 2 e− 3 t = 3 − 2 e− 3 t

Expanding as partial fractions we have: !

F (s) =

1 11 1 1 =− + s ( s − 5) 5 s 5 s−5

From the Laplace transform table in Appendix E we have: !

e5 t − 1 f (t ) = 5

Example 6.3-5 Determine the inverse Laplace transform: !

f (t ) = L

−1⎧

3 ⎫ ⎨ 2 ⎬ ⎩s + 5⎭

Solution: Example 6.3-4

We can write:

Determine the inverse Laplace transform:

!

3 3 5 = 2 s +5 5 s +5 2

94

From the Laplace transform table in Appendix E we have: !

f (t ) =

3 5

sin

( 5 t)

L

{1} −

−1

L

−1⎧

1 ⎫ ⎨ ⎬+ 9 L ⎩ s + 1⎭

−1⎧ 1 ⎫

⎨ ⎬− 9 L ⎩s ⎭

−1⎧

1 ⎫ ⎨ ⎬ ⎩ s + 1⎭

f ( t ) = δ ( t ) − e− t + 9 − 9 e− t = δ ( t ) + 9 − 10 e− t

When the numerator P ( s ) and the denominator Q ( s ) are of

Example 6.3-6

the same degree, the function f ( t ) will always include a unit

Determine the inverse Laplace transform:

f (t ) = L

s2 + 9 ⎫ ⎨ ⎬= L ⎩ s ( s + 1) ⎭

From the Laplace transform table in Appendix E we have: !

!

−1⎧

impulse function.

−1⎧

s2 + 9 ⎫ ⎨ ⎬ ⎩ s ( s + 1) ⎭

Example 6.3-7 Solution:

Determine the inverse Laplace transform:

The polynomials in the numerator P ( s ) = s + 9 and the 2

f (t ) = L

denominator Q ( s ) = s + s have the same degree, and so F ( s )

!

Q(s) :

Solution:

2

−1⎧

is not a proper rational fraction. We must first divide P ( s ) by

!

s2 + 9 s−9 = 1− s ( s + 1) s ( s + 1)

Expanding as partial fractions we have: !

s +9 1 9 9 = 1− + − s ( s + 1) s +1 s s +1 2

Therefore using Proposition 6.2-2:

⎫ 2 ⎨ 2 ⎬ ⎩ s − 6 s + 13 ⎭

The denominator is an irreducible quadratic, and so the next step is to complete the square (take half the coefficient.of the middle term and square it): !

2 s 2 − 6 s + 13

=

(s

2 2

)

− 6s + 9 + 4

=

2

( s − 3)2 + 4

From the Laplace transform table in Appendix E and Proposition 2.2-1 we have: 95

!

f (t ) = L

⎧ ⎫⎪ 3t 2 ⎨ ⎬ = e sin ( 2t ) 2 ⎪⎩ ( s − 3) + 4 ⎪⎭

−1⎪

Example 6.3-9 Determine the inverse Laplace transform: !

Example 6.3-8 Determine the inverse Laplace transform: !

step is to complete the square: !

step is to complete the square: s+4 s+4 = 2 s + 2 s + 2 ( s + 1)2 + 1

⎧ ( s + 1) ⎫⎪ ⎨ ⎬+ L 2 ⎪⎩ ( s + 1) + 1 ⎪⎭

−1⎪

⎧ ⎫⎪ 3 ⎨ ⎬ 2 ⎪⎩ ( s + 1) + 1 ⎪⎭

!

−1⎪

Proposition 2.2-1 we have: !

f ( t ) = e cos ( t ) + 3e sin( t ) −t

s 2

3 1⎞ ⎛ ⎜⎝ s + ⎟⎠ + 4 2

1⎞ 1 ⎫ ⎧⎛ ⎪ ⎜⎝ s + 2 ⎟⎠ − 2 ⎪ ⎪ ⎪ f ( t ) = L −1⎨ ⎬ 2 3⎪ 1⎞ ⎪⎛ + ⎪⎩ ⎜⎝ s + 2 ⎟⎠ 4 ⎪⎭

or

From the Laplace transform table in Appendix E and −t

s2 + s + 1

=

numerator of F ( s ) is also expressed in terms of s + 1 2 :

numerator of F ( s ) is also expressed in terms of s + 1 : ⎧ ( s + 1) + 3 ⎫⎪ ⎨ ⎬= L 2 ⎪⎩ ( s + 1) + 1 ⎪⎭

s

We can use the first shift theorem (Proposition 2.2-1) if the

We can use the first shift theorem (Proposition 2.2-1) if the

−1⎪

⎫ s ⎨ 2 ⎬ ⎩ s + s + 1⎭

The denominator is an irreducible quadratic, and so the next

The denominator is an irreducible quadratic, and so the next

! f (t ) = L

−1⎧

Solution:

⎧ s+4 ⎫ f ( t ) = L −1⎨ 2 ⎬ ⎩s +2s + 2⎭

Solution:

!

f (t ) = L

!

⎧ ⎫ 1 s+ ⎪ ⎪ 1 ⎪ 2 −1⎪ f (t ) = L ⎨ L ⎬− 2 3 3⎪ 1 ⎞ ⎪⎛ + ⎪⎩ ⎜⎝ s + 2 s ⎟⎠ 4 ⎪⎭

⎧ ⎫ 3 ⎪ ⎪ ⎪ 2 −1⎪ ⎨ ⎬ 2 3⎪ 1 ⎞ ⎪⎛ + ⎪⎩ ⎜⎝ s + 2 s ⎟⎠ 4 ⎪⎭ 96

From the Laplace transform table in Appendix E and Proposition 2.2-1 we have: !

⎛ 3 ⎞ 1 −t 2 ⎛ 3 ⎞ f ( t ) = e − t 2 cos ⎜ t − e sin ⎜ t ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ 3

Example 6.3-10

t

∫∫ 0−

τ

0

sin ( 2 γ ) dγ dτ = L

−1 ⎧

2 ⎫ ⎨ 4 2⎬ ⎩ s + 4s ⎭

Expanding the left side as partial fractions we have: 2 2 1 1 = = 2− 2 s + 4s 2s 2 s2 + 4 s2 s2 + 4

(

4

)

(

)

∫

t

0−

(1− cos ( 2 τ )) dτ

and so: τ

t

τ 1 t 1 sin ( 2 γ ) dγ dτ = − sin ( 2 τ ) = − sin ( 2t ) 2 4 2 4 0 0

t

∫∫

!

0−

t

!

∫∫

6.3.3!

0−

τ

0

sin ( 2 γ ) dγ dτ = L

−1 ⎧

2 ⎫ ⎨ 4 2⎬ ⎩ s + 4s ⎭

INVERSE LAPLACE TRANSFORMS FOR DISCONTINUOUS FUNCTIONS USING TRANSFORM TABLES

2 ⎫ t 1 = − sin ( 2t ) ⎨ 4 2⎬ ⎩ s + 4s ⎭ 2 4

Determine the inverse Laplace transform:

−1 ⎧

L

0

1 2

Example 6.3-11

From the Laplace transform table in Appendix E we have: !

0−

sin ( 2 γ ) dγ dτ =

is valid.

Solution:

!

∫∫

τ

Therefore the equation:

Verify the following result from Example 3.3-2: !

t

!

!

f (t ) = L

−1⎧ 2 e ⎨ 2

⎫ ⎬ ⎩ s +1 ⎭ −πs

Evaluating the integral, we obtain: t

! or

∫∫ 0−

τ

1 sin ( 2 γ ) dγ dτ = 2 0

∫

t

0−

( − cos ( 2 γ )) 0 dτ τ

Solution: From the Laplace transform table in Appendix E and Proposition 5.2-1 we have: 97

!

f (t ) = L

−1⎧ 2 e ⎨ 2

⎫ ⎬ = 2 sin ( t − π ) U ( t − π ) = − 2 sint U ( t − π ) ⎩ s +1 ⎭ −πs

Determine the inverse Laplace transform:

or !

Example 6.3-13

⎧⎪ 0 t 0 . The product f ( t ) ∗ g ( t ) is also

definition given in equation (7.2-1) are: !

∫

t

! !

f ( t − v ) g ( v ) ( −1) dv =

f ( t ) ∗ ⎡⎣ c1 g ( t ) + c2 h ( t ) ⎤⎦ = c1 f ( t ) ∗ g ( t ) + c2 f ( t ) ∗ h ( t ) !

(7.1-4)

!

⎡⎣ f ( t ) ∗ g ( t ) ⎤⎦ ∗ h ( t ) = f ( t ) ∗ ⎡⎣ g ( t ) ∗ h ( t ) ⎤⎦ !

(7.1-5)

!

f ( t ) ∗ c1 g ( t ) = c1 f ( t ) ∗g ( t ) = c1 ( f ( t ) ∗g ( t )) !

(7.1-6)

h ( t ) are piecewise-smooth functions.

7.1.2!

∫

t

CONVERGENCE OF THE CONVOLUTION INTEGRAL

exponential order, then their convolution:

Convolution is a commutative operation as can be seen by

f (t ) ∗ g (t ) =

Other properties of convolutions that follow from the

If f ( t ) and g ( t ) are piecewise continuous functions of

letting v = t − τ so that dv = − dτ and equation (7.2-1) becomes: !

(7.1-3)

Proposition 7.1-1, Convergence of the Convolution:

PROPERTIES OF THE CONVOLUTION INTEGRAL

0

∫ g (τ ) f (t −τ ) dτ = g (t ) ∗ f (t ) ! 0

known as the generalized product.

!

t

where c1 and c2 are arbitrary constants, and f ( t ) , g ( t ) , and

0

7.1.1!

f (t ) ∗ g (t ) =

! !

THE CONVOLUTION INTEGRAL

t

We then have:

f ( t −τ ) g (τ ) dτ !

0

(7.1-2)

!

f (t ) ∗ g (t ) =

∫

t

f ( τ ) g ( t − τ ) dτ !

(7.1-7)

0

is also a continuous function of exponential order, and so converges absolutely. 102

If σ f < σg , let a = σ g − σ f and M = M f Mg . We then have:

Proof: Since f ( t ) and g ( t ) are piecewise continuous functions

!

over any finite interval 0 ≤ t < τ , the function f ( t ) ∗ g ( t ) will also be piecewise continuous over 0 ≤ t < τ . Since f ( t ) and g ( t ) are functions of exponential order, we

!

also have: !

f ( t ) ≤ Mf e

σf t

!

g ( t ) ≤ Mg e

σg t

where σ f

∫

f ( τ ) g ( t − τ ) dτ ≤

0

!

∫

t

Mf e

σf τ

Mg e

σg t

e

( σ f −σg ) t − 1 σ f − σg

! !

σg ( t−τ )

dτ !

(7.1-10)

σ f ≠ σg !

! !

(7.1-11)

If σ f > σg , let a = σ f − σg and M = M f Mg . We then have: !

∫

0

−e a

σf t

≤ Me

σg t

!

∫

t

∫

t

∫

t

f ( τ ) g ( t − τ ) dτ ≤

t

∫Me

σg τ

f

Mg e

σ g ( t−τ )

dτ !

(7.1-14)

0

!

f ( τ ) g ( t − τ ) dτ ≤ M e

σg t

0

t

∫ dτ !

(7.1-15)

0

!

f ( τ ) g ( t − τ ) dτ ≤ M

f ( τ ) g ( t − τ ) dτ ≤ M t e

σg t

!

(7.1-16)

0

0

t

σg t

(7.1-13)

0

0

f ( τ ) g ( t − τ ) dτ ≤ M f Mg e

≤M

e

If σ f = σg , let M = M f Mg . We then have:

!

∫

σg t

and so:

or t

−e −a

!!

(7.1-9)

and σ g are non-negative real constants, and

t

σf t

or

T0 = max (T1, T2 ) . Therefore:

!

e

0

(7.1-8)

t ≥ T2 ≤ T0 !

!

∫

f ( τ ) g ( t − τ ) dτ ≤ M

!

t ≥ T1 ≤ T0 !

!

!

t

e

σf t

−e a

σg t

≤ Me

σf t

!

(7.1-12)

From equations (7.2-12), (7.2-13), and (7.2-16) we see that if

f ( t ) and g ( t ) are both piecewise continuous functions of exponential order, then their convolution: !

f (t ) ∗ g (t ) =

∫

t

f ( τ ) g ( t − τ ) dτ !

(7.1-17)

0

is also a piecewise continuous function of exponential order. From Proposition 1.4-2 we then know that their convolution converges absolutely.

■

103

t

f (t ) ∗ g (t ) = e

!

Proposition 7.1-2:

− bt

If f ( t ) and g ( t ) are piecewise continuous functions of Therefore:

exponential order, then their convolution:

f (t ) ∗ g (t ) =

!

∫

t

⎡ e− ( a−b ) τ ⎤ ⎢ ⎥ ⎣ b − a ⎦ τ= 0

f ( τ ) g ( t − τ ) dτ !

f (t ) ∗ g (t ) = e

!

(7.1-18)

0

−bt

⎡ e− ( a−b) t − 1⎤ e− a t − e− b t ⎢ ⎥= b−a ⎣ b−a ⎦

has a Laplace transform.

7.2!

Proof: !

Follows from Propositions 7.1-1 and 1.4-3.

LAPLACE TRANSFORM OF THE CONVOLUTION INTEGRAL

■

Proposition 7.2-1, Laplace Convolution Theorem: Example 7.1-1 Determine the convolution f ( t ) ∗ g ( t ) where f ( t ) = e

g (t ) = e

− bt

− at

If f ( t ) and g ( t ) are piecewise continuous functions of and

exponential order with Laplace transforms of L

.

and L

Solution:

L

!

{ g (t )} = G ( s ) , respectively, then: { f (t ) ∗ g (t )} = F ( s ) G ( s ) !

{ f (t )} = F ( s ) (7.2-1)

We have from equation (7.1-1): !

f (t ) ∗ g (t ) =

∫

t

f ( τ ) g ( t − τ ) dτ =

0

0

or !

∫

t

e− a τ e− b (t−τ ) dτ

Proof: ! !

f ( t ) ∗ g ( t ) = e− bt

t

∫e 0

− ( a−b ) τ

dτ

From the definition of the Laplace transform we have: L

{ f (t ) ∗ g (t )} = ∫

∞

t=0

−

f ( t ) ∗ g ( t ) e− s t d t !

(7.2-2)

Using equation (7.1-1) we obtain:

and so: 104

!

L

⎡ e− s t ⎢ t = 0− ⎣

{ f (t ) ∗ g (t )} = ∫

⎤ f ( τ ) g ( t − τ ) dτ ⎥ d t ! (7.2-3) τ=0 ⎦

∞

∫

t

We can rewrite equation (7.2-3) as: !

L

{ f (t ) ∗ g (t )} =

⎡ ⎢ t = 0− ⎣

∫

∞

⎤ e− s t f ( τ ) g ( t − τ ) dτ ⎥ d t ! τ=0 ⎦ t

∫

(7.2-4)

where the region of integration includes 0 ≤ τ ≤ t (vertical bar in shaded region of Figure 7.2-1), and 0 − ≤ t ≤ ∞ . !

Since the integrals are uniformly convergent (see

Proposition 1.4-7), we can change the order of integration: L

!

⎡ ⎢ τ=0 ⎣ ∞

{ f (t ) ∗ g (t )} = ∫

⎤ e− s t f ( τ ) g ( t − τ ) d t ⎥ dτ ! (7.2-5) t=τ ⎦

∫

∞

where the region of integration includes τ ≤ t ≤ ∞ (horizontal bar in shaded region of Figure 7.2-1) and 0 ≤ τ ≤ ∞ . We will now let v = t − τ where we hold τ fixed. We then

!

have dv = dt , and equation (7.2-5) becomes: L

!

{ f (t ) ∗ g (t )} = ∫

∞

τ=0

∫

∞

v= 0

e− s ( v +τ ) f (τ ) g ( v ) dv dτ !

(7.2-6)

!

and so: !

L

{ f (t ) ∗ g (t )} =

Figure 7.2-1! Integration regions for L

{ f (t ) ∗ g (t )} .

From the definition of the Laplace transform equation

(7.2-7) becomes: ⎡ f (τ ) e− s τ ⎢ τ=0 ⎣

∫

∞

⎤ g ( v ) e− s v dv ⎥ dτ ! (7.2-7) v= 0 ⎦

∫

∞

! L

{ f (t ) ∗ g (t )} = G ( s ) ∫

∞

f (τ ) e− sτ dτ !

τ=0

Re s > σ 0 !

(7.2-8)

105

or

! L

!

{ f (t ) ∗ g (t )} = F ( s ) G ( s ) !

Re s > σ 0 !

(7.2-9)

and so:

!

{ f (t )} L {g (t )} !

L

Re s > σ 0 ! (7.2-10)

■

{ f (t ) ∗δ (t )} =

L

{ f (t )} L {δ (t )} !

(7.2-13)

!

L

{ f (t ) ∗δ (t )} =

L

{ f (t )} (1) !

(7.2-14)

f ( τ ) δ ( t − τ ) dτ = f ( t ) !

(7.2-15)

and so:

Proposition 7.2-2:

F1 ( s ) , F2 ( s ) , !, Fn ( s ) , respectively, then: L

{ f1 (t ) ∗ f2 (t ) ∗!∗ fn (t )} = F1 ( s ) F2 ( s ) ! Fn ( s ) !

■

! (7.2-11)

Proof:

The Laplace transform of the convolution of two functions

f ( t ) ∗ g ( t ) can give the impression that the Laplace transform is multiplicative. In general, however, this is not true: L

!

Follows from Proposition 7.2-1.

If f ( t ) = 1 and g ( t ) = 1 show that the convolution

f (t ) ∗ g (t ) ≠ f (t ) g (t ) .

function δ ( t ) is f ( t ) :

∫

t

0

Proof:

(7.2-16)

Example 7.2-1

The convolution of any function f ( t ) with the unit impulse

f ( t ) ∗δ ( t ) =

{ f (t ) g (t )} ≠ F ( s ) G ( s ) !

■

Proposition 7.2-3:

!

∫

t

0

functions of exponential order with Laplace transforms of

!

f ( t ) ∗δ ( t ) =

!

If f1 ( t ) , f2 ( t ) , !, fn ( t ) are piecewise continuous

!

L

or

{ f (t ) ∗ g (t )} =

L

!

We have from equation (7.2-10):

f ( τ ) δ ( t − τ ) dτ = f ( t ) !

Solution: (7.2-12)

We have from equation (7.2-10): !

L

{ f (t ) ∗ g (t )} = L { f (t )} L {g (t )} = 1s 1s = s12 106

Therefore: !

Solution:

{ f (t ) ∗ g (t )} ≠ L { f (t ) g (t )} = L {1} = 1s

L

We have: !

Example 7.2-2 If g ( t ) = 1 show that the convolution f ( t ) ∗ g ( t ) ≠ f ( t ) g ( t ) .

We have from equation (7.2-10): !

∫

{

}

⎧ L ⎨ ⎩

⎫ τ 2 e3 (t−τ ) dτ ⎬ = L t 2 L e3t 0 ⎭

∫

t

{ } { }

Therefore:

{ f (t ) ∗ g (t )} = L { f (t )} L {g (t )} = L { f (t )} 1s

L

⎫ τ 2 e3 (t−τ ) dτ ⎬ = L t 2 ∗ e3t 0 ⎭ t

and so: !

Solution:

⎧ L ⎨ ⎩

!

⎧ L ⎨ ⎩

⎫ 2 1 τ 2 e3 (t−τ ) dτ ⎬ = 3 0 ⎭ s s−3

∫

t

and so: !

{ f (t ) ∗ g (t )} =

L

F (s) s

=

L

{ f (t )} s

Therefore: !

f (t ) ∗ g (t ) ≠ f (t ) g (t ) = f (t )

Determine the Laplace transform of: !

∫τ 0

2

e

3 ( t−τ )

Determine the Laplace transform of: !

∫

t

0

e− 3 (t−τ ) cos ( 4 τ ) dτ

Solution:

Example 7.2-3

t

Example 7.2-4

dτ

We have: !

⎧ L ⎨ ⎩

⎫ e− 3 (t−τ ) cos ( 4 τ ) dτ ⎬ = L e− 3t ∗ cos ( 4 t ) 0 ⎭

∫

t

{

}

and so: 107

!

⎧ L ⎨ ⎩

t

∫

e

− 3 ( t−τ )

0

⎫ cos ( 4 τ ) dτ ⎬ = L e− 3t ⎭

{ } L {cos ( 4 t )}

⎧ L { f ( t ) ∗ g ( t )} = L ⎨ ⎩

∫

!

t

0

⎫ F (s) f ( τ ) dτ ⎬ = s ⎭

Therefore: !

⎧ L ⎨ ⎩

7.3!

⎫ 1 s e− 3 (t−τ ) cos ( 4 τ ) dτ ⎬ = 2 0 ⎭ s + 3 s + 16 t

∫

!

∫

t

0−

⎫ F (s) f ( τ ) dτ ⎬ = s ⎭

∫

t

f ( τ ) g ( t − τ ) dτ =

0

∫

t

f ( τ ) dτ

1 s ( s − 2)

.

We will take:

0

{ f (t ) ∗ g (t )} = L { f (t )} L {g (t )} = L { f (t )} 1s =

Therefore:

Example 7.3-1

Solution:

!

From equation (7.2-10) we have: L

(7.3-1)

Determine the inverse Laplace transform for

If g ( t ) = 1 we have from equation (7.1-1):

!

{F ( s ) G ( s )} = f (t ) ∗ g (t ) !

Laplace transforms of some functions.

Solution:

f (t ) ∗ g (t ) =

−1

This equation can be very useful in obtaining the inverse

using the Laplace convolution theorem (Proposition 7.2-1).

!

L

!

Derive Proposition 3.3-3:

⎧ L ⎨ ⎩

The inverse Laplace transform of the convolution of two

functions f ( t ) ∗ g ( t ) can be obtained from equation (7.2-1):

Example 7.2-5

!

INVERSE LAPLACE TRANSFORM OF THE CONVOLUTION INTEGRAL

F (s) s

1 s ( s − 2)

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

where: !

F (s) =

1 ! s

G (s) =

1

( s − 2) 108

We then have: !

f (t ) = L

−1 ⎧ 1 ⎫

g (t ) = L

⎨ ⎬ = 1! ⎩s ⎭

!

−1 ⎧

1 ⎫ 2t ⎨ ⎬=e ⎩s − 2⎭

L

!

−1 ⎧

1 ⎫ 2t ⎨ ⎬ = f ( t ) ∗ g ( t ) = (1) ∗ e ⎩ s ( s − 2) ⎭

!

L

!

⎫ ⎨ ⎬= ⎩ s ( s − 2) ⎭ 1

t

∫ (1) e

2 ( t−τ )

0

1 dτ = − e2 (t−τ ) 2

−1 ⎧

1 ⎫ 1 2t ⎨ ⎬ = e −1 ⎩ s ( s − 2) ⎭ 2

(

)

Example 7.3-2

Solution:

!

( s + 1) ( s − 2 )

1 ⎫ −t ⎨ ⎬=e ! ⎩ s + 1⎭

g (t ) = L

−1 ⎧

1 ⎫ 2t ⎨ ⎬=e ⎩s − 2⎭

L

−1 ⎧

⎫ 1 −t 2t ⎨ ⎬ = f (t ) ∗ g (t ) = e ∗ e ⎩ ( s + 1) ( s − 2 ) ⎭

L

−1 ⎧

⎫ ⎨ ⎬= ⎩ ( s + 1) ( s − 2 ) ⎭ 1

∫

t

e−τ e2 (t−τ ) dτ = e2 t

0

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

t

∫e

− 3τ

dτ

0

1

( s + 1) ( s − 2 )

L

−1 ⎧

⎫ e2 t −3τ = − e ⎨ ⎬ 3 s + 1 s − 2 ( ) ( ) ⎩ ⎭ 1

t

= τ=0

(

1 2t e − e− t 3

)

. Example 7.3-3 Determine the inverse Laplace transform for

We will take: 1

−1 ⎧

or !

Determine the inverse Laplace transform for

1 s−2

Therefore:

τ=0

! L

f (t ) = L

t

or !

G (s) =

and so:

Therefore: −1 ⎧

1 ! s +1

We then have:

and so: !

F (s) =

Solution:

(

1

s s2 + 9

)

.

We will take:

where: 109

!

1

(

s s2 + 9

)

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

Example 7.3-4 Determine the inverse Laplace transform for

where: !

F (s) =

1 ! s

G (s) =

(s

1 2

+9

)

We then have: !

f (t ) = L

−1 ⎧ 1 ⎫

g (t ) = L

⎨ ⎬ = 1! ⎩s ⎭

−1 ⎧

1 ⎫ 1 ⎨ 2 ⎬ = sin ( 3t ) ⎩s + 9⎭ 3

L

⎧ 1 ⎨ 2 ⎩⎪ s s + 9

−1 ⎪

(

)

⎫⎪ 1 ⎬ = f ( t ) ∗ g ( t ) = (1) ∗ sin ( 3t ) 3 ⎭⎪

!

L

⎧ 1 ⎨ 2 ⎪⎩ s s + 9

−1 ⎪

(

)

⎫⎪ 1 ⎬= ⎪⎭ 3

!

0

L

⎧ 1 ⎨ 2 ⎩⎪ s s + 9

−1 ⎪

( s + a) ( s + b)

(

)

t ⎫⎪ 1 1 = cos 3 t − τ = (1− cos 3t ) ( ) ⎬ 9 τ=0 9 ⎭⎪

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

where: !

1 ! s+a

F (s) =

G (s) =

1 s+b

We then have: −1 ⎧

1 ⎫ − at ⎨ ⎬=e ! ⎩s + a⎭

g (t ) = L

−1 ⎧

1 ⎫ −bt ⎨ ⎬=e ⎩s + b⎭

and so: !

or !

1

t

∫ (1) sin ⎡⎣ 3 (t − τ)⎤⎦ dτ

.

We will take:

! f (t ) = L

Therefore:

( s + a) ( s + b)

Solution:

and so: !

1

L

−1 ⎧

⎫ 1 − at −bt ⎨ ⎬ = f (t ) ∗ g (t ) = e ∗ e ⎩ ( s + a) ( s + b) ⎭

Therefore: !

L

−1 ⎧

⎫ ⎨ ⎬= ⎩ ( s + a) ( s + b) ⎭ 1

∫

t

e− a τ e− b (t−τ ) dτ

0

or 110

!

L

−1 ⎧

⎫ −bt ⎨ ⎬=e s + a s + b ( ) ( ) ⎩ ⎭ 1

t

∫e

(b−a ) τ

0

e− b t (b−a ) τ dτ = e b−a

t

L

⎧ 1 ⎨ 2 ⎪⎩ ( s − 2 )

−1 ⎪

L

τ=0

−1 ⎧

⎫ e− b t (b−a ) t e− a t − e− b t ⎡ ⎤ − 1⎦ = ⎨ ⎬= ⎣e b−a ⎩ ( s + a) ( s + b) ⎭ b − a 1

!

⎧ 1 ⎨ 2 ⎪⎩ ( s − 2 )

−1 ⎪

L

Determine the inverse Laplace transform for

1

( s − 2)

2

.

Solution:

We will take:

We will take:

( s − 2)

2

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

where: !

F (s) =

1 ! s−2

⎧ 1 ⎫ 2t ! f ( t ) = L −1 ⎨ ⎬=e ⎩s − 2⎭

and so:

2τ

e e

2( t−τ )

dτ = e

0

2t

∫

t

dτ = t e2t

0

!

(s

s 2

)

+1

= L

2

(s

s 2

)

+1

2

.

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

where:

G (s) =

1 s−2

We then have: !

∫

t

Determine the inverse Laplace transform for

Solution:

1

⎫⎪ ⎬= ⎪⎭

Example 7.3-6

Example 7.3-5

!

⎫⎪ 2t 2t ⎬ = f (t ) ∗ g (t ) = e ∗ e ⎪⎭

Therefore:

and so: !

!

⎧ 1 ⎫ 2t g ( t ) = L −1 ⎨ ⎬=e ⎩s − 2⎭

!

F (s) =

s ! s +1 2

G (s) =

1 s +1 2

We then have: !

f (t ) = L

−1 ⎧

s ⎫ ⎨ 2 ⎬ = cos ( t ) ! ⎩ s + 1⎭

g (t ) = L

−1 ⎧

1 ⎫ ⎨ 2 ⎬ = sin ( t ) ⎩ s + 1⎭

and so: 111

!

⎧ s ⎪ L −1 ⎨ ⎪⎩ s 2 + 1

(

)

2

⎫ ⎪ ⎬ = f ( t ) ∗ g ( t ) = cos ( t ) ∗sin ( t ) ⎪⎭

L

(

)

2

⎫ 1 ⎪ 1 ⎬ = t sin ( t ) + ⎡⎣ cos ( t ) − cos ( t ) ⎤⎦ 4 ⎪⎭ 2

Therefore:

Therefore: !

!

⎧ s ⎨ ⎪⎩ s 2 + 1

−1 ⎪

⎧ s ⎪ L −1 ⎨ ⎪⎩ s 2 + 1

(

)

2

⎫ ⎪ ⎬= ⎪⎭

t

∫ cos(τ) sin (t − τ) dτ

!

L

⎧ s ⎨ ⎪⎩ s 2 + 1

−1 ⎪

0

(

)

2

⎫ ⎪ 1 ⎬ = t sin ( t ) ⎪⎭ 2

Using the trigonometric identity: !

2sin ( θ ) cos ( φ ) = sin ( θ + φ ) + sin ( θ − φ )

Determine the inverse Laplace transform for

where !

τ = φ!

t−τ=θ

and!

L

(

)

2

⎫ ⎪ 1 ⎬= ⎪⎭ 2

∫

t

0

⎡⎣sin ( t ) + sin ( t − 2 τ ) ⎤⎦ dτ

or

.

!

1 s s +1

= L

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

where:

and so: !

s s +1

We will take:

⎧ s ⎨ ⎪⎩ s 2 + 1

−1 ⎪

1

Solution:

we have: !

Example 7.3-7

⎧ s −1 ⎪ L ⎨ ⎪⎩ s 2 + 1

(

)

2

⎫ ⎪ 1 ⎬ = τ sin ( t ) ⎪⎭ 2

t

+ τ=0

t 1 cos ( t − 2 τ ) τ = 0 4

!

F (s) =

1 ! s

G (s) =

1 s +1

We then have: !

f (t ) = L

−1 ⎧ 1 ⎫

⎨ ⎬ = 1! ⎩s ⎭

g (t ) = L

−t ⎫ e ⎨ ⎬= ⎩ s +1 ⎭ πt

−1 ⎧

1

112

and so: !

L

−1 ⎧

⎫ e− t ⎨ ⎬ = f (t ) ∗ g (t ) = πt ⎩s s +1 ⎭ 1

Therefore: !

L

−1 ⎧

⎫ 1 ⎨ ⎬= π ⎩s s +1 ⎭ 1

t

e− τ

0

τ

∫

dτ = erf

( t)

113

Chapter 8 ∞

Laplace Transforms of Series Expansions

L

{ f (t )} = ∑ n=0

an n! s n+1

114

!

In this chapter we will determine the Laplace transform of

a converging power series. We will also consider the inversion of Laplace transforms that are converging power series. !

While representing a function or its Laplace transform in

the form of a power series does not provide the same physical insight as does a closed form, for complicated problems a representation as a power series may be the easiest to obtain. In such cases it is important to determine if term-by-term

Representing summation by an integral, we can write equation (8.1-2) in the form:

G ( x) =

!

0

A power series representing a function f ( t ) will have the

form: f (t ) =

( ) = A(s) = ∫

!

G e

∑a t !

n = 0, 1, 2, !!

n

n

(8.1-1)

!

−

a ( t ) e −s t dt !

(8.1-4)

A power series that consists of terms having positive

powers of t as in equation (8.1-1) can converge very slowly. Therefore it can be advantageous to convert the series into one having negative powers of t . This can be accomplished if the series having the form:

n

! { f (t )} = ∑ an L {t } = ∑ an sn! n+1

(8.1-5)

n=0

It is not always possible to obtain the Laplace transform of

an infinite series by summing term-by-term the Laplace transforms of the terms in the series. If a function f ( t ) can be

∞

n

!

∞

n

n=0

by considering the power series:

∑a x !

L

!

improper integral analogous to a power series. We can see this

n=0

∞

∞

The Laplace transform can be considered to be an

G ( x) =

−s

series can be Laplace transformed term-by-term to obtain a

∞

where an are constant coefficients.

!

(8.1-3)

equivalent to a power series:

n=0

!

a ( t ) x t dt !

0

8.1! LAPLACE TRANSFORM OF POWER SERIES

!

−

Replacing x with e− s , we have the Laplace transform as an

transformation of the series is possible.

!

∫

∞

n = 0, 1, 2, !!

(8.1-2)

expanded in a power series, however, it is possible to obtain L

{ f (t )}

by Laplace transforming the series term-by-term and 115

summing, provided the coefficients of the series follow certain constraints.

!

L

{ }

⎧ ⎪ = L ⎨ f (t ) − ⎪⎩

∑

{ f (t )} − ∑ an L { t }

⎧ ⎪ ≤ L ⎨ f (t ) − ⎪⎩

⎫ an t ⎬ ! (8.1-10) ⎪⎭ n=0

{ f (t )} −

N

∑

an L t

n

n=0

Proposition 8.1-1, Laplace Transform of a Power Series:

)

If a function f ( t ) is continuous over ⎡⎣ 0 − , ∞ and is of exponential order, and if f ( t ) can be represented by a

f (t ) =

!

!

L

n=0

n = 0, 1, 2, !!

n

n

(8.1-6)

N

!

where

M >0!

and Re s > σ 0 , then L

(8.1-7)

{ f (t )} can be represented by the

L

! { f (t )} = ∑ an sn! n+1

L

{ f (t )} − ∑ an L { t } n

n=0

(8.1-8)

Proof: A power series converges absolutely and uniformly within

its circle of convergence. Therefore the series can be integrated term-by-term as each term is Laplace transformed. We can

⎧ ⎪ ≤ L ⎨ ⎪⎩

⎫ ⎪ an t ⎬ ! ⎪⎭ n = N +1 ∞

∑

n

(8.1-11)

(8.1-7) we then have: !

n=0

!

{ f (t )} − ∑ an L { t }

N

∞

∑

n⎪

From the constraints on the coefficients given in equation

Laurent series: !

L

n

n=0

σ 0n ! an ≤ M n!

N

and so:

n=0

!

n

∞

∑a t !

(8.1-9)

or N

converging power series:

⎫ ⎪ an t n ⎬ ! ⎪⎭ n=0 N

⎧ ⎪ ≤ML ⎨ ⎪⎩

∞

∑

n = N +1

( σ 0 t )n n!

⎫ ⎪ ⎬! ⎪⎭

(8.1-12)

Since we also have: ∞

!

e

σ0 t

=

∑ n= 0

( σ 0 t )n n!

!

(8.1-13)

we can write:

write: 116

N

!

L

{ f (t )} − ∑ an L { t } n

n=0

N ⎧ ( σ 0 t )n ⎪ σ0 t ≤ M L ⎨e − n! ⎪⎩ n= 0

∑

⎫ ⎪ ⎬! ⎪⎭

! !

Therefore we have: ∞

(8.1-14)

Taking the Laplace transform: N

!

L

{ f (t )} − ∑ an L { t } n

n=0

N ⎛ σ 0n 1 ≤M⎜ − n+1 ⎜⎝ s − σ 0 n = 0 s

∑

⎞ ⎟ ! (8.1-15) ⎟⎠

∞

L

!

lim

N →∞

L

{ f (t )} − ∑ an L { t } n

n=0

!

⎛ 1 1 ≤ M lim ⎜ − N→ ∞ ⎜ s − σ s 0 ⎝

σ 0n ⎞ ⎟ sn ⎟ ⎠ n= 0 N

∑

(8.1-16)

(8.1-20)

n=0

∞

L

{ f (t )} − ∑ an L { t n } n=0

!

From equation (8.1-20) we see that the neighborhood of

t = 0 for f ( t ) corresponds to the neighborhood of s = ∞ for the Laplace transform of f ( t ) . Example 8.1-1

and so using the geometric series: ⎛ ⎞ 1 1 ⎟ ⎜ 1 ! (8.1-17) ≤M⎜ − σ0 ⎟ s − σ0 s 1− ⎜⎝ ⎟ s ⎠

Determine the Laplace transform f ( t ) = et using its Taylor series: !

or ∞

L

{ f (t )} − ∑ an L { t n } n=0

!

! { f (t )} = ∑ an sn! n+1

■

!

!

(8.1-19)

or

N

!

{ f (t )} = ∑ an L {t n } ! n=0

or !

L

!

⎛ 1 1 ⎞ ≤M⎜ − = 0! ⎝ s − σ 0 s − σ 0 ⎟⎠ (8.1-18)

t t2 t3 tn e = 1+ + + +! = 1! 2! 3! n! t

∞

∑ n=0

tn n!

Solution: We have: !

an =

1 1 σ 0 , then L

n=0

L

!

{ }

1 1 1 1 L et = + 2 + 3 +! n = s s s s

∞

∑s

1 n+1

n=0

1 = s

∞

{ f (t )} can be represented by the series:

∑ n=0

Γ n + ν +1 { f (t )} = ∑ an (s n + ν +1 ) !

(8.1-23)

n=0

1 sn

Proof: !

From equation (1.5-4) we have:

Therefore we can write: !

(8.1-22)

∞

or !

M >0!

{ }

L tα =

!

{ }

1 1 1 L et = = s 1− 1 s − 1 s

Γ ( α + 1) ! s α+1

α > −1 !

(8.1-24)

Letting α = n + ν we have from Proposition 8.1-1 and equation (8.1-24):

!

Proposition 8.1-1 is a special case of Watson’s lemma that

applies also for cases where the exponent of t is not an integer.

∞

L

!

Γ n + ν +1 { f (t )} = ∑ an (s n + ν +1 ) !

ν > −1 !

(8.1-25)

n=0

Proposition 8.1-2, Watson’s Lemma:

■

)

If a function f ( t ) is continuous over ⎡⎣ 0 , ∞ and is of −

exponential order, and if f ( t ) can be represented by a

Proposition 8.1-3, Convergence of the Laplace Transform of a Power Series:

converging power series: !

f (t ) =

If f ( t ) can be represented by a converging Taylor series:

∞

∑a t n

n+ν

!

ν > −1 !

(8.1-21)

!

f (t ) =

n=0

where

∞

∑a t ! n

n

n = 0, 1, 2, ! !

(8.1-26)

n=0

and if: 118

M σ 0n ! an ≤ n!

!

(8.1-27)

where M > 0 , then the series L

{ f (t )} will be uniformly

convergent in Re s > σ 0 .

du = f ′ ( t ) dt !

!

L

!

Follows from Propositions 8.1-1 and 1.4-7 since the power

series is uniformly convergent.

■

{ f (t )} = −

e− s t f ( t )

∞

s

f (0) 1 L { f ( t )} = + s s

Proposition 8.1-4:

)

−

order over ⎡⎣ 0 , ∞ , then: ∞

L

!

{ f (t )} = ∑

f (n) ( 0 ) s

n=0

n+1

!

(8.1-28)

t=0

!

From the definition of the Laplace transform we have:

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s ) !

(8.1-29)

u = f (t ) !

0

−

f ′ ( t ) e− s t dt !

(8.1-32)

0

−

f ′ ( t ) e− s t dt !

(8.1-33)

Integrating by parts again, we let: !

u = f ′ (t ) !

dv = e− s t dt !

(8.1-34)

!

du = f ′′ ( t ) dt !

e− s t ! v=− s

(8.1-35)

f ( 0 ) ⎡ e− s t f ′ ( t ) ⎤ 1 ! L { f ( t )} = −⎢ ⎥ + 2 2 s ⎣ s ⎦ t=0 s

dv = e− s t dt !

(8.1-30)

∫

∞

0−

f ′′ ( t ) e− s t dt ! (8.1-36)

or !

Integrating by parts, we let: !

∫

∞

∞

L

∫

∞

so that:

Proof: !

1 s

+

or !

If f ( t ) , f ′ ( t ) , f ′′ ( t ) ,! are continuous functions of exponential

(8.1-31)

so that:

Proof: !

e− s t ! s

v=−

L

{ f (t )} =

f (0) s

+

f ′ (0) s2

1 + 2 s

∫

∞

0

−

f ′′ ( t ) e− s t dt !

(8.1-37)

From repeated integration by parts, we then obtain: 119

L

!

f 0 f′ 0 f ′′ 0 f ′′′ 0 { f (t )} = (s ) + s(2 ) + s(3 ) + s 4( ) +!!

that converges for Re s > σ0 with lim F ( s ) = 0 , then F ( s ) can

(8.1-38)

s→∞

be inverted term-by-term to obtain:

or ∞

L

!

{ f (t )} = ∑ n=0

f (n) ( 0 ) s

n+1

f (t ) =

! !

∞

∑ n! t ! an

n

(8.2-2)

n=0

(8.1-39)

where f ( t ) converges absolutely and uniformly in its circle of

■

convergence.

8.2! INVERSE OF A LAPLACE TRANSFORM SERIES

Proof:

!

Representing f ( t ) by a power series, we have:

If a Laplace transform does not have a standard form

!

By definition F ( s )

found in a table, it may be useful to expand the transform into a series and then invert the series term-by-term. If the transform

!

f (t ) =

is an analytic function of 1 s .

∞

∑b t ! n

(8.2-3)

n

n=0

can be expanded as a converging Laurent series of powers of 1 s , then the inverted transform will have the form of a series

where f ( t ) will converge absolutely and uniformly in its circle

of powers of t . For small values of t , this procedure can

of convergence. From Proposition 8.1-1 we have:

provide a solution of adequate accuracy. ! Proposition 8.2-1, Inverse of a Laplace Transform Series: If the Laplace transform F ( s ) of a function f ( t ) can be represented by a series: ∞

!

L

{ f (t )} = F ( s ) = ∑ an s n+1 ! n=0

1

(8.2-1)

⎧ ∞ ⎫ ⎪ n⎪ L { f ( t )} = L ⎨ bn t ⎬ = ⎪⎩ n = 0 ⎪⎭

∑

∞

∑ bs n! ! n n+1

(8.2-4)

n=0

We will let bn = an n! so that: !

⎧ ∞ a ⎫ ⎪ n n⎪ L { f ( t )} = L ⎨ t ⎬= n! ⎪⎩ n = 0 ⎪⎭

∑

∞

∑a s1 n

n+1

!

(8.2-5)

n=0

120

Therefore we see that:

⎧ ∞ 1 −1⎪ L ⎨ an n+1 ⎪⎩ n = 0 s

∑

!

⎫ ⎪ ⎬= ⎪⎭

∞

∑ n=0

Example 8.2-1

an n t = f (t ) ! n!

(8.2-6)

Determine the inverse Laplace transform f ( t ) of: !

■

F (s) =

If the Laplace transform F ( s ) of a function f ( t ) can be

Solution:

represented by a series:

We have:

L

!

{ f (t )} = F ( s ) = ∑ an s n +1ν +1 !

ν > −1 !

(8.2-7)

n=0

s→∞

be inverted term-by-term to obtain: f (t ) =

!

∑ n=0

!

F (s) =

∞

∑ s1

n+1

1 1 1 1 1 1 = + 2 + 3 +! = = s s s 1− 1 s − 1 s s

From the transform tables in Appendices E and F we then have:

1 an t n + ν +1 ! Γ ( n + ν + 1)

(8.2-8)

!

f ( t ) = et

where f ( t ) converges absolutely and uniformly in its circle of

From Proposition 8.2-1 with an = 1 we also have:

convergence.

!

Follows from Propositions 8.1-2 and 8.2-1.

f (t ) =

∞

∑ n=0

Proof: !

n+1

n=0

that converges for Re s > σ0 with lim F ( s ) = 0 , then F ( s ) can ∞

∑ s1 n=0

Proposition 8.2-2, Inverse of a Laplace Transform Series:

∞

∞

1 n t = et n!

■

Example 8.2-2 Determine the inverse Laplace transform f ( t ) of: 121

!

F (s) =

1 2 s +1

!

f (t ) = t −

t3

( 3!)

2

+

t5

( 5!)

2

−

t7

( 7!)

2

+!

Solution: We have: !

F (s) =

1 s2 1 1 1 1 = = 2 − 4 + 6 − 8 +! 2 s + 1 1+ 1 s s s s s2

From the transform table in Appendix E we then have: !

f (t ) = t −

t3 t5 t7 + − +! = sin ( t ) 3! 5! 7!

Example 8.2-3 From its series expansion determine the inverse Laplace transform f ( t ) of: !

F (s) =

1 1 sin s s

Solution: The series expansion of F ( s ) is: !

F (s) =

1 1 1 1 − + − +! 2 4 6 s 3! s 5! s 7! s 8

From the transform table in Appendix E we then have: 122

Chapter 9 Solution of Linear Ordinary Differential Equations having Constant Coefficients

L ⎡⎣ a f ′′ ( t ) + b f ′ ( t ) + c f ( t ) ⎤⎦ = 0

123

!

In this chapter we will consider the application of Laplace

where f ( t ) is the forcing function or driving function. The

transform techniques to the solution of linear ordinary

coefficients ci ( t ) are functions of the independent variable t .

differential equations. The Laplace transform changes such

!

equations together with their initial conditions into algebraic

degree. The order of an ODE is the order of the highest

equations, thereby considerably simplifying the solution of the

derivative within the ODE. For example, equation (9.1-1) is a

differential equation.

general linear ODE of n th order. The degree of an ODE is the

9.1!

power to which the highest order derivative is raised after the

!

ORDINARY DIFFERENTIAL EQUATIONS Equations containing derivatives are called differential

equations. These equations can result whenever describing any process involving the rate of change in some variable (known as the dependent variable). If the dependent variable is a function of only one independent variable, the resulting differential equation is called an ordinary differential equation (ODE). A solution of a differential equation is any function that

Differential equations are classified according to order and

ODE has been rationalized. A linear ODE is a degree one ODE. Since the Laplace transform is a linear operator, it is compatible with any linear ODE having an independent variable with a range from zero to infinity. !

The form of a linear ODE having constant coefficients ci

is: ! cn

d n y (t ) d n−1y ( t ) dy ( t ) + c + c0 y ( t ) = f ( t ) ! n−1 n n−1 +!+ c1 dt dt dt

(9.1-2)

satisfies the differential equation. Substitution of the solution

If the forcing function f ( t ) = 0 , then the ODE is said to be

into the differential equation will then reduce the equation to

homogeneous; otherwise it is nonhomogeneous.

an identity.

!

!

The form of a general linear ODE having y ( t ) as the

dependent variable and t as the independent variable is: d y (t ) d y (t ) dy ( t ) + cn−1 ( t ) + c0 ( t ) y ( t ) = f ( t ) n n−1 +!+ c1( t ) dt dt dt ! (9.1-1)

! cn ( t ) !

n

n−1

A solution of the ODE obtained by setting f ( t ) = 0 in

equations (9.1-1) or (9.1-2) is known as a complementary or general solution. It represents the steady-state solution of the ODE. An n th order ODE will have n arbitrary constants of integration in the complementary solution. These constants are determined using the initial conditions. 124

!

The solution of an ODE that arises from the forcing

!

The general procedure for using the Laplace transform to

function is known as the particular solution. It represents the

obtain solutions of linear nonhomogeneous ODEs having

transient solution of the ODE. The particular solution is always

constant coefficients and closed form solutions consists of the

linearly independent of the complementary solution.

following:

!

The solution of any nonhomogeneous ODE is the sum of

1.! The entire nonhomogeneous ODE including the forcing

its complementary and particular solutions. When using

function is first transformed using the Laplace transform.

Laplace transforms to solve ODE, the entire nonhomogeneous ODE including its forcing function is transformed at the same

2.! The initial conditions are applied to the transformed

time. Unlike for other methods of solving ODE, this means that

equation. This produces an algebraic equation already

the steady-state and transient solutions of the ODE can be

incorporating the initial conditions.

obtained simultaneously. Nonhomogeneous ODEs are then

3.! The algebraic equation is solved for the transformed

solved without first solving the associated homogeneous ODE.

dependent variable.

The complete solution of an ODE is thereby found without the

4.! The transformed dependent variable solution is rewritten

necessity of first finding the complementary solution, and so

using partial fractions if necessary.

the need to determine the arbitrary constants associated with

5.! The inverse Laplace transform of each of the terms in the

the complementary solution is eliminated.

partial fraction expansion of the transformed solution is

9.2! !

SOLVING LINEAR ODES USING LAPLACE TRANSFORMS

determined. 6.! In this way the solution to the original nonhomogeneous

The Laplace transform is very useful for obtaining

ODE is obtained.

solutions of linear ODEs having constant coefficients. Such equations will be transformable if their driving function is

!

transformable.

9.2-1. The most direct procedure for obtaining such solutions is

Procedures for solving linear ODEs are shown in Figure

125

indicated by the dashed line, but this method can often be quite difficult. The procedure for using the Laplace transform to

a y′′ ( t ) + b y′ ( t ) + c f ( t ) = f ( t ) !

!

is then:

obtain such solutions is indicated in Figure 9.2-1 by the solid lines. It can be much simpler to execute than the more direct method.

(9.2-1)

Y (s) =

!

( a s + b ) y ( 0 ) + a y′ ( 0 ) a s2 + b s + c

+

F (s) a s2 + b s + c

!

(9.2-2)

where a , b , and c are constants. Proof: !

The Laplace transform of the ODE in equation (9.2-1) is: L {a y′′ ( t ) + b y′ ( t ) + c f ( t )} = L

!

{ f (t )} !

(9.2-3)

or a ⎡⎣ s 2 Y ( s ) − s y ( 0 ) − y′ ( 0 ) ⎤⎦ + b ⎡⎣ sY ( s ) − y ( 0 ) ⎤⎦ + cY ( s ) = F ( s )

! !

!

(9.2-4)

and so:

(a s

!

2

)

+ b s + c Y ( s ) = a s y ( 0 ) + a y′ ( 0 ) + b y ( 0 ) + F ( s ) !

Figure 9.2-1! Solution of linear ODE using Laplace transforms.

Therefore:

Proposition 9.2-1:

!

If L { y ( t )} = Y ( s ) and L

{ f (t )} = F ( s ) , where y (t ) is a continuous function of exponential order over [ 0, ∞ ) , and f ( t ) and y′ ( t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) , the Laplace transform of the second-order ODE:

Y (s) =

( a s + b ) y ( 0 ) + a y′ ( 0 ) a s2 + b s + c

+

F (s) a s2 + b s + c

!

(9.2-5)

(9.2-6)

■

!

If the initial conditions of a second-order ODE are zero, we

have: 126

!

(a s

2

)

+ b s + c Y (s) = F (s) !

(9.2-7)

and the system transfer function H ( s ) is defined to be: !

H (s) =

Y (s) 1 ! = 2 F (s) a s + b s + c

(9.2-8)

The system transfer function is then the ratio of the Laplace

!

Once the system transfer function H ( s ) is determined, the

response of a system Y ( s ) to any forcing function f ( t ) can be determined either in the s -plain: !

Y (s) = H (s) F (s) !

or in the t -plain: y (t ) = h (t ) ∗ f (t ) = L

transforms of the output function to the input forcing function.

!

When the forcing function f ( t ) is an impulse function δ ( t ) , the

as shown in Figure 9.2-2.

system response function H ( s ) is called the impulse response,

(9.2-12)

−1

{ H ( s ) F ( s )} !

(9.2-13)

and is the solution to the ODE since F ( s ) = 1 : ! !

H (s) = Y (s) ! The polynomial a s 2 + b s + c

(9.2-9) in the denominator of

equation (9.2-8) is the characteristic function for the linear homogeneous differential equation: !

a y′′ ( t ) + b y′ ( t ) + c y ( t ) = 0 !

(9.2-10)

The roots of a s 2 + b s + c depend only upon properties of the function y ( t ) , and they are the poles s1 and s2 of the impulse response: !

Figure 9.2-2! System response to a forcing function.

1 1 = ! a s + b s + c ( s − s1 ) ( s − s2 ) 2

(9.2-11)

127

Proposition 9.2-2: Let L { y ( t )} = Y ( s ) and L

Example 9.2-1

{ f (t )} = F ( s ) where y (t ) is a continuous function of exponential order over [ 0, ∞ ) , and f ( t )

!

over [ 0, ∞ ) . Given the second-order ODE:

with initial condition: y ( 0 ) = 3

and y′ ( t ) are piecewise-smooth functions of exponential order

a y′′ ( t ) + b y′ ( t ) + c f ( t ) = f ( t ) !

!

(9.2-14)

Determine the solution of the differential equation:

y′ ( t ) − 2y ( t ) = 0

Solution:

where a , b , and c are constants, then if y ( 0 ) = 0 and y′ ( 0 ) = 0

We first transform the entire ODE:

we have:

!

y (t ) =

!

∫

t

h ( t − τ ) f ( τ ) dτ !

(9.2-15)

0

H (s) =

!

Y (s) 1 ! = 2 F (s) a s + b s + c

(9.2-16)

{ y′ (t )} − 2 L { y (t )} = 0

Using Proposition 3.1-1, we have: !

where:

L

( )

⎡ s Y ( s ) − y 0 − ⎤ − 2Y ( s ) = 0 ⎣ ⎦

Applying the initial condition: !

[ s − 2] Y ( s) = 3

Proof:

and so the Laplace transform of y ( t ) is:

!

!

We have:

Y (s) = H (s) F (s) !

!

(9.2-17)

y (t ) = h (t ) ∗ f (t ) =

∫

t

h ( t − τ ) f ( τ ) dτ !

3 s−2

From the transform table in Appendix E we obtain the

From equation (7.1-2) we then have: !

Y (s) =

inverse transform: (9.2-18)

!

y ( t ) = 3e2t

0

■

128

Using the initial condition we can write: Example 9.2-2 Determine the solution of the differential equation: !

y′ ( t ) − 2y ( t ) = 0

with initial condition: y ( 2 ) = 3 Solution:

!

( )

( )

y ( 2 ) = 3 = y 0 − e2 ( 2 ) = y 0 − e 4

and so: !

( )

y 0 − = 3e− 4

Therefore: !

y ( t ) = 3e− 4 e2t = 3e− 4+2t

We first transform the entire ODE: !

L

{ y′ (t )} − 2 L { y (t )} = 0

Using Proposition 3.1-1 we have: !

( )

⎡ s Y ( s ) − y 0 − ⎤ − 2Y ( s ) = 0 ⎣ ⎦

[ s − 2 ] Y ( s ) = y ( 0− )

and so the Laplace transform of y ( t ) is: !

!

y′ ( t ) + y ( t ) = sint

Y (s) =

( )

Solution: We first transform the entire ODE: L

{ y′ (t )} + L { y (t )} =

L {sint }

y 0−

!

s−2

Using Proposition 3.1-1 we have:

From the transform table in Appendix E we obtain the inverse transform: !

Determine the solution of the differential equation: with initial condition: y ( 0 ) = 0 .

or !

Example 9.2-3

( )

y ( t ) = y 0 − e2 t

!

( )

⎡ s Y ( s ) − y 0− ⎤ + Y ( s ) = 1 ⎣ ⎦ s2 + 1

Applying the initial condition:

129

!

1 s +1

[ s + 1] Y ( s ) =

2

and so the Laplace transform of y ( t ) is: !

Y (s) =

1

( s + 1)( s 2 + 1)

Expanding as partial fractions: !

1

( s + 1)( s 2 + 1)

=

A Bs + C + 2 s +1 s +1

Y (s) =

1 1 1 s 1 1 − + 2 s + 1 2 s2 + 1 2 s2 + 1

From the transform table in Appendix E we obtain the inverse transform: !

1 1 1 y ( t ) = e− t − cost + sint 2 2 2

Example 9.2-4 Determine the solution of the differential equation: !

We first transform the entire ODE: !

y′ ( t ) + 4y ( t ) = e− 3t

with initial condition: y ( 0 ) = 1 .

L

{ y′ (t )} + 4 L { y (t )} =

{ }

L e− 3t

Using Proposition 3.1-1 we have: !

( )

⎡ s Y ( s ) − y 0 − ⎤ + 4Y ( s ) = 1 ⎣ ⎦ s+3

Applying the initial condition: !

we find: !

Solution:

1

[s + 4] Y (s) − 1 = s + 3

and so the Laplace transform of y ( t ) is: !

Y (s) =

1 1 + s + 4 ( s + 3) ( s + 4 )

Expanding as partial fractions: !

1

( s + 3)( s + 4 )

=

A B + s+3 s+4

we find: !

Y (s) =

1 1 1 1 + − = s+4 s+3 s+4 s+3

From the transform table in Appendix E we obtain the inverse transform: !

y ( t ) = e− 3t 130

Example 9.2-5

Example 9.2-6

Determine the solution of the differential equation:

Determine the solution of the differential equation:

!

y′′ ( t ) + 4y ( t ) = 0

!

y′′ ( t ) − y ( t ) = 1

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 0 .

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 0 .

Solution:

Solution:

We first transform the entire ODE:

We first transform the entire ODE:

!

L

{ y′′ (t )} + 4 L { y (t )} = 0

From Proposition 3.1-2 we have: !

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + 4Y ( s ) = 0 ⎣ ⎦

Applying the initial conditions: !

⎡⎣ s + 4 ⎤⎦ Y ( s ) − s = 0

Y (s) =

s 2 s +4

!

!

y ( t ) = cos 2t

{ y′′ (t )} − L { y (t )} =

L {1}

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ − Y ( s ) = 1 ⎣ ⎦ s

Applying the initial conditions: !

1 ⎡⎣ s 2 − 1⎤⎦ Y ( s ) = s

and so the Laplace transform of y ( t ) is: !

From the transform table in Appendix E we obtain the inverse transform:

L

From Proposition 3.1-2 we have:

2

and so the Laplace transform of y ( t ) is: !

!

Y (s) =

(

1

)

s s2 − 1

=

1 s ( s − 1) ( s + 1)

Expanding as partial fractions: !

1 A B C = + + s ( s − 1) ( s + 1) s s − 1 s + 1 131

and so the Laplace transform of y ( t ) is:

we find: !

Y (s) = −

1 1 1 + + s 2 ( s − 1) 2 ( s + 1)

!

From the transform table in Appendix E we obtain the inverse transform: !

(

1 1 = s − 4 ( s + 1) ( s + 2 ) ( s − 2 ) ( s + 1) 2

)

Expanding as partial fractions: !

1 1 y ( t ) = −1+ et + e− t 2 2

Y (s) =

1 A B C = + + ( s + 2 )( s − 2 )( s + 1) s + 2 s − 2 s + 1

we find: Example 9.2-7

!

Y (s) =

Determine the solution of the differential equation: !

y′′ ( t ) − 4y ( t ) = e−t

1 1 1 + − 4 ( s + 2 ) 12 ( s − 2 ) 3( s + 1)

From the transform table in Appendix E we obtain the

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 0 .

inverse transform: !

Solution:

y (t ) =

1 − 2t 1 2t 1 −t e + e − e 4 12 3

We first transform the entire ODE: !

L

{ y′′ (t )} − 4 L { y (t )} =

L

{e } −t

From Proposition 3.1-2 we have: !

( ) ( )

⎡ s2 Y ( s ) − s y 0− ⎣

1 − y′ 0 − ⎤⎦ − 4Y ( s ) = s +1

Applying the initial conditions: 1 ⎡⎣ s 2 − 4 ⎤⎦ Y ( s ) = ! s +1

Example 9.2-8 Determine the solution of the differential equation: !

y′′ + 4 y = cos ( 2t )

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 1 . Solution: We first transform the entire ODE: 132

!

{ y′′ (t )} + 4 L { y (t )} = L { cos ( 2t )}

L

From Proposition 3.1-2 we have: !

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + 4Y ( s ) = 2 s ⎣ ⎦ s +4

Applying the initial conditions: !

(s

2

)

+ 4 Y (s) =

s +1 s +4

!

Y (s) =

(

s2 + 4

)

!

2

+

1 s2 + 4

L

{ y′′ (t )} − 6 L { y′ (t )} + 9 L { y (t )} = 0

From Propositions 3.1-1 and 3.1-2 we have: !

( ) ( )

!

⎡⎣ s 2 Y ( s ) − 1⎤⎦ − 6 s Y ( s ) + 9Y ( s ) = 0

or !

⎡⎣ s 2 − 6 s + 9 ⎤⎦ Y ( s ) = 1

From the transform table in Appendix E we obtain the

and so the Laplace transform of y ( t ) is:

inverse transform:

!

!

t 1 y ( t ) = sin ( 2t ) + sin ( 2t ) 4 2

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ − 6 ⎡ s Y ( s ) − y 0 − ⎤ + 9Y ( s ) = 0 ⎣ ⎦ ⎣ ⎦

Applying the initial conditions:

2

and so the Laplace transform of y ( t ) is:

s

We first transform the entire ODE:

Y (s) =

1

( s − 3)2

From the transform table in Appendix E we obtain the inverse transform:

Example 9.2-9

!

y ( t ) = t e3t

Determine the solution of the differential equation: !

y′′ ( t ) − 6 y′ ( t ) + 9 y ( t ) = 0

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 1 . Solution:

Example 9.2-10 Determine the solution of the differential equation: !

y′′ ( t ) + y′ ( t ) − 6y ( t ) = 0 133

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 0

!

y (t ) =

2 − 3t 3 2t e + e 5 5

Solution: We first transform the entire ODE: !

L

Example 9.2-11

{ y′′ (t )} + L { y′ (t )} − 6 L { y (t )} = 0

Determine the solution of the differential equation:

From Propositions 3.1-1 and 3.1-2 we have: !

( ) ( )

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + ⎡ s Y ( s ) − y 0 − ⎤ − 6Y ( s ) = 0 ⎣ ⎦ ⎣ ⎦

Applying the initial conditions: !

⎡⎣ s 2 + s − 6 ⎤⎦ Y ( s ) − s − 1 = 0

and so the Laplace transform of y ( t ) is: !

Y (s) =

s +1 s +1 = s + s − 6 ( s + 3) ( s − 2 ) 2

Expanding as partial fractions: !

s +1 A B = + ( s + 3)( s − 2 ) s + 3 s − 2

Y (s) =

y′′ ( t ) + 2 y′ ( t ) + 10y ( t ) = 0

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 0 Solution: We first transform the entire ODE: !

L

{ y′′ (t )} + 2 L { y′ (t )} +10 L { y (t )} = 0

From Propositions 3.1-1 and 3.1-2 we have: !

( ) ( )

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + 2 ⎡ s Y ( s ) − y 0 − ⎤ + 10Y ( s ) = 0 ⎣ ⎦ ⎣ ⎦

Applying the initial conditions: !

⎡⎣ s 2 + 2 s + 10 ⎤⎦ Y ( s ) − s − 2 = 0

and so the Laplace transform of y ( t ) is:

we find: !

!

2 3 + 5 ( s + 3) 5 ( s − 2 )

!

Y (s) =

s+2 s + 2 s + 10 2

From the transform table in Appendix E we obtain the

The denominator is an irreducible quadratic, and so the next

inverse transform:

step is to complete the square: 134

!

Y (s) =

s+2

( s + 1)

2

!

+9

If we change the numerator we can use the first shift theorem (Proposition 2.2-1): !

Y (s) =

( s + 1) + 1

( s + 1)

2

+9

=

s +1

( s + 1)

2

+9

+

1

3

3 ( s + 1) + 9

( ) ( )

Applying the initial conditions: !

⎡⎣ s 2 + 4 s + 2 ⎤⎦ Y ( s ) = s + 5

and so the Laplace transform of y ( t ) is:

2

!

Y (s) =

From the transform table in Appendix E we obtain the

s+5 s + 4s+2 2

The denominator is an irreducible quadratic, and so the next

inverse transform: !

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + 4 ⎡ s Y ( s ) − y 0 − ⎤ + 2Y ( s ) = 0 ⎣ ⎦ ⎣ ⎦

1 y ( t ) = e− t cos ( 3t ) + e− t sin ( 3t ) 3

step is to complete the square: !

Y (s) =

s+5

( s + 2 )2 − 2

Example 9.2-12

If we change the numerator we can use the first shift theorem

Determine the solution of the differential equation:

(Proposition 2.2-1):

!

y′′ ( t ) + 4 y′ ( t ) + 2y ( t ) = 0

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 1

We first transform the entire ODE: !

{ y′′ (t )} + 4 L

Y (s) =

( s + 2) + 3

( s + 2)

2

−2

=

s+2

( s + 2)

2

−2

+

3

2

2 ( s + 2) − 2 2

From the transform table in Appendix E we obtain the

Solution:

L

!

{ y′ (t )} + 2 L

{ y (t )} = 0

inverse transform: !

y ( t ) = e− 2 t cosh

(

)

2t +

3 2

e− 2t sinh

(

2t

)

From Propositions 3.1-1 and 3.1-2 we have:

135

From the transform table in Appendix E we obtain the Example 9.2-13

inverse transform:

Determine the solution of the differential equation:

!

!

y′′ ( t ) − 2 y′ ( t ) + y ( t ) = e t

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 1

y (t ) = t et +

t 2 et 2

Example 9.2-14

Solution:

Determine the solution of the differential equation:

We first transform the entire ODE:

!

!

L

{ y′′ (t )} − 2 L { y′ (t )} + L { y (t )} =

{ }

L et

From Propositions 3.1-1 and 3.1-2 we have: !

( ) ( )

Applying the initial conditions: !

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ − 2 ⎡ s Y ( s ) − y 0 − ⎤ +Y ( s ) = 1 ⎣ ⎦ ⎣ ⎦ s −1

1 ⎡⎣ s 2 Y ( s ) − 1⎤⎦ − 2 s Y ( s ) +Y ( s ) = s −1

with initial conditions: y ( 0 ) = 0 , y′ ( 0 ) = 0 Solution: We first transform the entire ODE: !

1 ⎡⎣ s 2 − 2 s + 1⎤⎦ Y ( s ) = 1+ s −1

and so the Laplace transform of y ( t ) is: !

Y (s) =

1

( s − 1)

2

+

1

( s − 1)

L

{ y′′ (t )} + L { y′ (t )} − 2 L { y (t )} = 4 L {e2 t }

From Propositions 3.1-1 and 3.1-2 we have: !

or !

y′′ ( t ) + y′ ( t ) − 2y ( t ) = 4 e2 t

( ) ( )

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + ⎡ s Y ( s ) − y 0 − ⎤ − 2Y ( s ) = 4 ⎣ ⎦ ⎣ ⎦ s−2

Applying the initial conditions: !

4 ⎡⎣ s 2 + s − 2 ⎤⎦ Y ( s ) = s−2

and so the Laplace transform of y ( t ) is:

3

136

!

Y (s) =

4 ( s − 1)( s + 2 )( s − 2 )

Expanding as partial fractions: !

4 A B C = + + ( s − 1)( s + 2 )( s − 2 ) s − 1 s + 2 s − 2

! !

Y (s) = −

4 1 1 1 1 + + 3 s −1 3 s + 2 s − 2

or

inverse transform:

!

y (t ) = −

4 t 1 − 2t 2t e + e +e 3 3

( ) ( )

y′′′ ( t ) − y′′ ( t ) + y′ ( t ) − y ( t ) = 0

with initial conditions: y ( 0 ) = 2 , y′ ( 0 ) = 2 , y′′ ( 0 ) = 4

⎡⎣ s 3 Y ( s ) − 2 s 2 − 2 s − 4 ⎤⎦ − ⎡⎣ s 2 Y ( s ) − 2 s − 2 ⎤⎦ + ⎡⎣ s Y ( s ) − 2 ⎤⎦

−Y ( s ) = 0

⎡⎣ s 3 − s 2 + s − 1⎤⎦ Y ( s ) = 2 s 2 + 4

Y (s) = 2

s2 + 2

( s − 1)( s 2 + 1)

Dividing by s 2 + 1 : !

⎡ 1 1 Y (s) = 2 ⎢ + ⎢⎣ s − 1 ( s − 1) s 2 + 1

(

Solution:

Expanding as partial fractions:

We first transform the entire ODE:

!

!

L

{ y′′′ (t )} − L { y′′ (t )} + L { y′ (t )} − L { y (t )} = 0

( )

− ⎡⎣ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤⎦ + ⎡⎣ s Y ( s ) − y 0 − ⎤⎦ −Y ( s ) = 0

Example 9.2-15.

!

( )

and so the Laplace transform of y ( t ) is: !

Determine the solution of the differential equation:

( )

!

From the transform table in Appendix E we obtain the

!

( )

⎡ s 3 Y ( s ) − s 2 y 0 − − s y′ 0 − − y′′ 0 − ⎤ ⎣ ⎦

Applying the initial conditions: !

we find: !

From Propositions 3.1-1, 3.1-2, and 3.1-3 we have:

1

( s − 1)( s 2 + 1)

=

)

⎤ ⎥ ⎥⎦

A Bs + C + 2 s −1 s +1

137

we find: !

Y (s) =

⎡⎣ s 2 + 1⎤⎦ Y ( s ) = 2 + F ( s )

! 2 s −1

+

1 s +1 3 s +1 − 2 = − 2 s −1 s +1 s −1 s +1

and so the Laplace transform of y ( t ) is:

From the transform table in Appendix E we obtain the

Y (s) =

!

inverse transform: !

y ( t ) = 3 et − cos ( t ) − sin ( t )

2 s2 + 1

+

F (s)

s2 + 1

The second term on the right is the product of two transforms. Therefore using the transform tables in Appendices E and F we obtain the inverse transform:

Example 9.2-16 Determine the solution of the differential equation: !

or

y′′ ( t ) + y ( t ) = f ( t )

Solution: We first transform the entire ODE: L

{ y′′ (t )} + L { y (t )} =

L

!

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ +Y ( s ) = F ( s ) ⎣ ⎦

Applying the initial conditions: ! or

⎡⎣ s Y ( s ) − 2 ⎤⎦ +Y ( s ) = F ( s ) 2

9.3!

{ f (t )}

From Proposition 3.1-2 we have:

y ( t ) = sin ( 2t ) +

!

with initial conditions: y ( 0 ) = 2 , y′ ( 0 ) = 0

!

y ( t ) = sint + f ( t ) ∗sint

!

!

∫

t

f ( τ ) sin ( t − τ ) dt

0

SOLVING LINEAR ODES HAVING DISCONTINUOUS FORCING FUNCTIONS Laplace transforms provide an easy way to evaluate linear

nonhomogeneous ODEs that have a discontinuous forcing function. This is one of the great advantages of using the Laplace transform to solve ODEs. The unit step function and the unit impulse function can both be used to represent some discontinuous forcing functions. 138

y′ ( t ) − 3y ( t ) = U ( t )

Example 9.3-1

!

Determine the solution of the differential equation:

with initial condition: y ( 0 ) = 1 .

!

a f ′′ ( t ) + b f ′ ( t ) + c f ( t ) = U ( t )

Solution:

with initial conditions: f ( 0 ) = f ′ ( 0 ) = 0

We first transform the entire ODE:

Solution:

!

We first transform the entire ODE: !

aL

{ f ′′ (t )} + b L { f ′ (t )} + c L { f (t )} =

L {U ( t )}

!

Y (s) =

a s2 + b s + c

+

L {U ( t )} a s2 + b s + c

=

L {U ( t )} a s2 + b s + c

The Laplace transform of f ( t ) is then: !

L

{ f (t )} =

(

1

s a s2 + b s + c

)

!

!

f (t ) = L

⎧ 1 ⎨ 2 ⎪⎩ s a s + b s + c

−1⎪

(

)

⎫⎪ ⎬ ⎪⎭

L {U ( t )}

( )

⎡ s Y ( s ) − y 0 − ⎤ − 3Y ( s ) = 1 ⎣ ⎦ s

Applying the initial condition: !

1

[ s − 3 ] Y ( s ) = 1+ s

and so the Laplace transform of y ( t ) is: !

and so the inverse transform is:

{ y′ (t )} − 3 L { y (t )} =

From Proposition 3.1-1 we have:

From Proposition 9.2-1 we have:

( a s + b ) y ( 0 ) + a y′ ( 0 )

L

Y (s) =

1 1 + s − 3 s ( s − 3)

Expanding as partial fractions: !

1 A B = + s ( s − 3) s − 3 s

we find: Example 9.3-2 Determine the solution of the differential equation:

!

Y (s) =

1 1 3 1 3 + − s−3 s−3 s 139

From the transform table in Appendix E, we obtain the inverse transform: !

y (t ) =

1 3t 3t 1 4 1 e + e − U ( t ) = e3t − 3 3 3 3

!

Y (s) =

e− π s 2 1 2 e− π s 2 = s2 + 4 2 s2 + 4

From Proposition 5.2-1 and the transform table in Appendix E, we obtain the inverse transform: !

Example 9.3-3

y (t ) =

1 1 ⎡ ⎛ π⎞ ⎤ ⎛ π⎞ ⎛ π⎞ sin ⎢ 2 ⎜ t − ⎟ ⎥ U ⎜ t − ⎟ = − sin ( 2t ) U ⎜ t − ⎟ ⎝ 2⎠ 2 2 ⎣ ⎝ 2⎠⎦ ⎝ 2⎠

Determine the solution of the differential equation: !

⎛ π⎞ y′′ ( t ) + 4 y ( t ) = δ ⎜ t − ⎟ ⎝ 2⎠

Example 9.3-4 Determine the solution of the differential equation:

with initial conditions: y ( 0 ) = y′ ( 0 ) = 0 .

!

Solution:

with initial conditions: y ( 0 ) = y′ ( 0 ) = 0 .

We first transform the entire ODE: !

Solution:

⎧ ⎛ π⎞ ⎫ L { y′′ ( t )} + 4 L { y ( t )} = L ⎨δ ⎜ t − ⎟ ⎬ ⎩ ⎝ 2⎠⎭

We first transform the entire ODE: !

From Proposition 3.1-2 we have: !

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + 4Y ( s ) = e− π s ⎣ ⎦

Applying the initial conditions: !

⎡⎣ s + 4 ⎤⎦ Y ( s ) = e 2

−πs 2

and so the Laplace transform of y ( t ) is:

y′′ ( t ) + 3 y′ ( t ) + 2 y ( t ) = δ ( t − 4 )

2

L

{ y′′ (t )} + 3 L { y′ (t )} + 2 L { y (t )} =

L {δ ( t − 4 )}

From Propositions 3.1-1 and 3.1-2 we have:

( ) ( )

( )

! ⎡⎣ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤⎦ + 3 ⎡⎣ s Y ( s ) − y 0 − ⎤⎦ + 2Y ( s ) = e− 4 s Applying the initial conditions: !

⎡⎣ s 2 + 3s + 2 ⎤⎦ Y ( s ) = e− 4 s

and so the Laplace transform of y ( t ) is: 140

!

Y ( s ) = e− 4 s

⎡ 1 1 ⎤ = e− 4 s ⎢ − ⎥ ( s + 2 )( s + 1) ⎣ ( s + 1) ( s + 2 ) ⎦ 1

Y ( s ) = e− s

!

1 s + 2s + 5 2

From Proposition 5.2-1 and the transform table in Appendix

The denominator is an irreducible quadratic, and so the next

E we obtain the inverse transform:

step is to complete the square:

!

y ( t ) = ⎡⎣ e− (t−4 ) − e− 2 (t−4 ) ⎤⎦ U ( t − 4 )

with initial conditions: y ( 0 ) = y′ ( 0 ) = 0 .

9.4!

Solution: !

We first transform the entire ODE:

{ y′′ (t )} + 2 L { y′ (t )} + 5 L { y (t )} =

L {δ ( t − 1)}

From Propositions 3.1-1 and 3.1-2 we have:

( ) ( )

( )

! ⎡⎣ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤⎦ + 2 ⎡⎣ s Y ( s ) − y 0 − ⎤⎦ + 5Y ( s ) = e− s Applying the initial conditions: ! or

⎡⎣ s 2 + 2 s + 5 ⎤⎦ Y ( s ) = e− s

y (t ) =

!

y′′ ( t ) + 2 y′ ( t ) + 5 y ( t ) = δ ( t − 1)

L

( s + 1)2 + 4

E we obtain the inverse transform:

Determine the solution of the differential equation:

!

1

From Proposition 5.2-1 and the transform table in Appendix

Example 9.3-5

!

Y ( s ) = e− s

!

1 U ( t − 1) e t sin ⎡⎣ 2 ( t − 1) ⎤⎦ 2

SOLVING SIMULTANEOUS ODES USING LAPLACE TRANSFORMS The Laplace transform is also very useful for obtaining the

solution of a system of simultaneous linear nonhomogeneous ODEs having constant coefficients and closed-form solutions. The general procedure for obtaining such solutions consists of the following: 1.! The entire system of ODEs including forcing functions are first transformed using the Laplace transform. 2.! The initial conditions are applied to the transformed equations. 141

From Proposition 3.1-1 we have:

3.! The system of algebraic equations is solved for the

4.! The

transformed

dependent

( )

⎡ s X ( s ) − x 0− ⎤ + X ( s ) − Y ( s ) = 0 ⎣ ⎦

!

transformed dependent variables. variable

solutions

are

!

rewritten using partial fractions if necessary. 5.! The inverse Laplace transform of each of the terms in the partial fraction expansions of the transformed solutions is determined. 6.! By this process the solutions to the original ODEs are

( )

⎡ s Y ( s ) − y 0− ⎤ − 2 X ( s ) = 0 ⎣ ⎦

Applying the initial conditions: !

s X (s) + X (s) − Y (s) = 0

!

⎡⎣ s Y ( s ) − 2 ⎤⎦ − 2 X ( s ) = 0

or

obtained.

!

( s + 1) X ( s ) − Y ( s ) = 0

Example 9.4-1

!

− 2 X (s) + s Y (s) = 2

Solve the following system of differential equations:

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

!

x′ (t ) + x (t ) − y (t ) = 0

!

y′ ( t ) − 2 x ( t ) = 0

with initial conditions: x ( 0 ) = 0 , y ( 0 ) = 2

!

X (s) =

0 −1 2 s s + 1 −1 −2 s

=

2

s ( s + 1) − 2

=

2 2 = s + s − 2 ( s − 1) ( s + 2 ) 2

Solution: We first transform the entire ODEs: !

L

{ x′ (t )} + L { x (t )} − L { y (t )} = 0

!

L

{ y′ (t )} − 2 L { x (t )} = 0

!

Y (s) =

s +1 0 −2 2 s + 1 −1 −2 s

=

2 ( s + 1) s2 + s − 2

=

2 ( s + 1)

( s − 1)( s + 2 )

142

with initial conditions: x ( 0 ) = −1 , y ( 0 ) = 1

Expanding as partial fractions: !

!

2

( s − 1)( s + 2 )

=

A B + s −1 s + 2

Solution: We first transform the entire ODEs:

2 ( s + 1)

A B = + ( s − 1)( s + 2 ) s − 1 s + 2

we find: ! !

X (s) =

2 1 2 1 − 3 s −1 3 s + 2

4 1 2 1 Y (s) = + 3 s −1 3 s + 2

!

L

{ x′ (t )} + L { y (t )} =

!

L

{ y′ (t )} − L { x (t )} = 0

From Proposition 3.1-1 we have:

( )

⎡ s X ( s ) − x 0− ⎤ + Y ( s ) = 2 ⎣ ⎦ s

! !

From the transform table in Appendix E we obtain the inverse transforms:

( )

⎡ s Y ( s ) − y 0− ⎤ − X ( s ) = 0 ⎣ ⎦

Applying the initial conditions:

2 2−s −1 = s s

!

x (t ) =

2 t 2 − 2t e − e 3 3

!

s X (s) + Y (s) =

!

y (t ) =

4 t 2 − 2t e + e 3 3

!

s Y (s) − X (s) = 1

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

Example 9.4-2 Solve the following system of differential equations: !

x′ (t ) + y (t ) = 2

!

y′ ( t ) − x ( t ) = 0

L {2}

!

X (s) =

2−s s 1 s 1 −1 s

1 s

=

2 − s −1 1 s = 2 − 2 2 s +1 s +1 s +1

143

!

Y (s) =

s

2−s s

−1

1

s 1 −1 s

with initial conditions: x ( 0 ) = 0 , y ( 0 ) = 0 2−s s+ s

Solution:

s 2 1 = 2 = 2 + − 2 s +1 s + 1 s s2 + 1 s + 1

(

)

We first transform the entire ODEs:

Expanding as partial fractions: !

(

2

)

s s +1 2

=

A s

+

Bs + C

L

!

2L

Y (s) =

s 2 2s 1 2 s 1 + − 2 − 2 = − 2 − 2 s +1 s s +1 s +1 s s +1 s +1 2

{ y′ (t )} − L { x (t )} =

L {1}

From Proposition 3.1-1 we have:

s2 + 1

( )

⎡ s X ( s ) − x 0 − ⎤ + 2Y ( s ) = 0 ⎣ ⎦

!

we find: !

{ x′ (t )} + 2 L { y (t )} = 0

!

!

( )

1 2 ⎡⎣ s Y ( s ) − y 0 − ⎤⎦ − X ( s ) = s

From the transform table in Appendix E we obtain the

Applying the initial conditions:

inverse transforms:

!

s X ( s ) + 2Y ( s ) = 0

!

− X (s) + 2 s Y (s) =

! !

x ( t ) = sint − cost y ( t ) = 2 − cost − sint

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

Example 9.4-3 Solve the following system of differential equations: ! !

x′ (t ) + 2 y (t ) = 0 2 y′ ( t ) − x ( t ) = 1

1 s

!

X (s) =

0 2 1 2s s s 2 −1 2 s

=

−

(

2 s

=

−1

) s (s

2 s2 + 1

2

)

+1

144

!

0 1 −1 s

!

s

Y (s) =

s 2 −1 2 s

with initial conditions: x ( 0 ) = 1 , y ( 0 ) = 0

=

(

1

)

Solution:

2 s2 + 1

We first transform the entire ODEs: !

L

{ x′ (t )} − 2 L { x (t )} + L { y (t )} = 0

!

L

{ y′ (t )} − 2 L { y (t )} − L { x (t )} = 0

Expanding as partial fractions: !

(

−1

)

s s2 + 1

=

A Bs + C + 2 s s +1

From Proposition 3.1-1 we have:

( )

⎡ s X ( s ) − x 0 − ⎤ − 2 X ( s ) +Y ( s ) = 0 ⎣ ⎦

!

we find: !

y′ ( t ) − 2y ( t ) − x ( t ) = 0

1 s X (s) = − + 2 s s +1

!

( )

⎡ s Y ( s ) − y 0 − ⎤ − 2Y ( s ) − X ( s ) = 0 ⎣ ⎦

From the transform table in Appendix E we obtain the

Applying the initial conditions:

inverse transforms:

!

( s − 2) X ( s) + Y ( s) = 1

!

− X ( s) + ( s − 2)Y ( s) = 0

! !

x ( t ) = − 1+ cost y (t ) =

1 sint 2

Example 9.4-4 Solve the following system of differential equations: !

x′ (t ) − 2 x (t ) + y (t ) = 0

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

!

X (s) =

1 1 0 s−2 s−2 1 −1 s − 2

=

s−2

( s − 2 )2 + 1

145

!

Y (s) =

s−2 1 −1 0 s−2 1 −1 s − 2

From Proposition 3.1-1 we have:

=

1

( s − 2)

( )

⎡ s X ( s ) − x 0 − ⎤ − 2 X ( s ) +Y ( s ) = 0 ⎣ ⎦

! 2

+1 !

( )

⎡ s Y ( s ) − y 0 − ⎤ − 2Y ( s ) − X ( s ) = 0 ⎣ ⎦

From the transform table in Appendix E we obtain the

Applying the initial conditions:

inverse transforms:

!

( s − 2) X ( s) + Y ( s) = 1

!

x ( t ) = e2t cos ( t )

!

− X ( s) + ( s − 2)Y ( s) = 0

!

y ( t ) = e2t sin ( t )

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

Example 9.4-5

!

X (s) =

Solve the following system of differential equations: !

x′ (t ) − 2 x (t ) + y (t ) = 0

!

y′ ( t ) − 2y ( t ) − x ( t ) = 0

with initial conditions: x ( 0 ) = 1 , y ( 0 ) = 0

!

Solution: We first transform the entire ODEs: !

L

{ x′ (t )} − 2 L { x (t )} + L { y (t )} = 0

!

L

{ y′ (t )} − 2 L { y (t )} − L { x (t )} = 0

Y (s) =

1 1 0 s−2 s−2 1 −1 s − 2 s−2 1 −1 0 s−2 1 −1 s − 2

=

=

s−2

( s − 2 )2 + 1

1

( s − 2 )2 + 1

From the transform table in Appendix E we obtain the inverse transforms: !

x ( t ) = e2t cos ( t )

!

y ( t ) = e2t sin ( t ) 146

or Example 9.4-6

!

⎡⎣ s 2 − 1⎤⎦ X ( s ) −Y ( s ) = 4 s X ( s ) + ⎡⎣ s 2 + 1⎤⎦ Y ( s ) = − 2 s

Solve the following system of differential equations: !

x ′′ ( t ) − x ( t ) − y ( t ) = 0

!

!

y′′ ( t ) + y ( t ) + x ( t ) = 0

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

with initial conditions: x ( 0 ) = 4 , y ( 0 ) = − 2 , x ′ ( 0 ) = 0 , and

y′ ( 0 ) = 0 .

4s !

X (s) =

Solution: We first transform the entire ODEs: !

L

{ x′′ (t )} − L { x (t )} − L { y (t )} = 0

!

L

{ y′′ (t )} + L { y (t )} + L { x (t )} = 0

!

From Proposition 3.1-2 we have:

( ) ( )

⎡ s 2 X ( s ) − s x 0 − − x ′ 0 − ⎤ − X ( s ) −Y ( s ) = 0 ⎣ ⎦

! !

( ) − y′ ( 0 )⎤⎦ + Y ( s ) + X ( s ) = 0

⎡s Y (s) − s y 0 ⎣ 2

−

−

Applying the initial conditions: !

⎡⎣ s 2 X ( s ) − 4 s ⎤⎦ − X ( s ) −Y ( s ) = 0

!

⎡⎣ s 2 Y ( s ) + 2 s ⎤⎦ + Y ( s ) + X ( s ) = 0

Y (s) =

−1

− 2 s s2 + 1 s2 − 1

−1

1

s2 + 1

s2 − 1

4s

1

−2s

s2 − 1

−1

1

s2 + 1

=

=

(

)

4 s s2 + 1 − 2 s

(s

2

)(

)

−1 s +1 +1

(

2

)

2 − 2 s s − 1 − 4s

(s

2

)(

)

−1 s +1 +1 2

=

4 2 + s s3

=−

2 2 − s s3

From the transform table in Appendix E we obtain the inverse transforms: !

x (t ) = 4 + t 2

!

y (t ) = − 2 − t 2

147

!

2 ⎡⎣ s 2 X ( s ) − 2 s ⎤⎦ + s 2 Y ( s ) = 3 s

!

4 ⎡⎣ s 2 X ( s ) − 2 s ⎤⎦ − s 2 Y ( s ) = 2 s

Example 9.4-7 Solve the following system of differential equations: !

x ′′ ( t ) + y′′ ( t ) = t 2

!

x ′′ ( t ) − y′′ ( t ) = 4 t

Subtracting the second equation from the first:

with initial conditions: x ( 0 ) = 2 , y ( 0 ) = 0 , x ′ ( 0 ) = 0 , and

!

y′ ( 0 ) = 0 .

or

Solution:

!

We first transform the entire ODEs: !

2 L { x ′′ ( t )} + L { y′′ ( t )} = 3 s

!

L

!

{ x′′ (t )} − L { y′′ (t )} = s42

1 2 − s5 s4

Y (s) =

1 2 2 ⎡⎣ s 2 X ( s ) − 2 s ⎤⎦ + s 2 ⎛⎜ 5 − 4 ⎞⎟ = 3 ⎝s s ⎠ s

or !

( ) ( )

( ) ( )

( ) ( )

( ) ( )

⎡ s 2 X ( s ) − s x 0 − − x ′ 0 − ⎤ + ⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ = 2 ⎣ ⎦ ⎣ ⎦ s3 !

2 4 − s3 s2

We then also have:

From Proposition 3.1-2 we have: !

2 s2 Y ( s) =

⎡ s 2 X ( s ) − s x 0 − − x′ 0 − ⎤ − ⎡ s 2 Y ( s ) − s y 0 − ⎣ ⎦ ⎣

4 − y′ 0 − ⎤⎦ = 2 s

X (s) =

2 1 2 2 1 2 2 − + + = + + s5 s5 s4 s s5 s4 s

From the transform table in Appendix E we obtain the inverse transforms: !

t4 2 3 x (t ) = + t + 2 4! 3!

!

y (t ) =

Applying the initial conditions:

t4 2 3 − t 4! 3! 148

s 2 X ( s ) − X ( s ) + 5 sY ( s ) =

!

s 2 Y ( s ) − 4Y ( s ) − 2 s X ( s ) = −

Example 9.4-8 Solve the following system of differential equations: ! !

x ′′ ( t ) − x ( t ) + 5 y′ ( t ) = 2t

2 s2

!

4 s

or

y′′ ( t ) − 4y ( t ) − 2 x ′ ( t ) = − 4

!

2 ⎡⎣ s 2 − 1⎤⎦ X ( s ) + 5 sY ( s ) = 2 s

y′ ( 0 ) = 0 .

!

− 2 s X ( s ) + ⎡⎣ s 2 − 4 ⎤⎦ Y ( s ) = −

Solution:

Solving for X ( s ) and Y ( s ) using Cramer’s rule:

with initial conditions: x ( 0 ) = 0 , y ( 0 ) = 0 , x ′ ( 0 ) = 0 , and

We first transform the entire ODEs: ! !

2 L { x ′′ ( t )} − L { x ( t )} + 5 L { y′ ( t )} = 2 s 4 L { y′′ ( t )} − 4 L { y ( t )} − 2 L { x ′ ( t )} = − s

!

X (s) =

From Propositions 3.1-1 and 3.1-2 we have:

( ) ( )

( )

⎡ s 2 X ( s ) − s x 0 − − x ′ 0 − ⎤ − X ( s ) + 5 ⎡ sY ( s ) − y 0 − ⎤ = 2 ⎣ ⎦ ⎣ ⎦ s2

! !

( ) ( )

( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ − 4Y ( s ) − 2 ⎡ s X ( s ) − x 0 − ⎤ = − 4 ⎣ ⎦ ⎣ ⎦ s

Applying the initial conditions:

2 s2 4 − s

!

Y (s) =

5s s2 − 4

s2 − 1

5s

−2s

s2 − 4

s2 − 1

2 s2 4 − s

− 2s

4 s

s2 − 1

5s

− 2s

s2 − 4

( )(

)

2 2 2 s − 4 + 20 s = 2 2 s 2 − 1 s − 4 + 10 s

(

=

( ( s − 1)( s

)

) − 4 ) + 10 s

4 2 4 − s s −1 + s

2

2

2

149

or !

!

X (s) =

Y (s)

(

22 s 2 − 8

)(

s2 s2 + 1 s2 + 4 − 4s 2 + 8

)(

(

s s2 + 1 s2 + 4

)

)

Therefore: !

X (s) = −

!

Y (s) =

2 s

2 s

2

−

+

10 8 − 2 s +1 s + 4 2

4s 2s + s2 + 1 s2 + 4

From the transform table in Appendix E we obtain the inverse transforms: !

x ( t ) = − 2 t + 10sin ( t ) − 4 sin ( 2t )

!

y ( t ) = 2 − 4 cos ( t ) + 2 cos ( 2t )

150

Chapter 10 Solution of Partial Differential Equations

F ( x, s ) =

∫

∞

0−

f ( x, t ) e− s t dt

151

!

In this chapter we will consider the application of Laplace

higher than two, the Laplace transform can be repetitively

transform techniques to the solution of constant coefficient

applied to obtain solutions.

linear partial differential equations.

!

10.1! !

change in some variable known as a dependent variable. If the dependent variable is a function of more than one independent variable, the differential equation is called a partial differential equation (PDE). A solution of a partial differential equation is any function of the independent variables that satisfies the differential equation. !

a function of two independent variables v ( x, t ) is:

PARTIAL DIFFERENTIAL EQUATIONS

A differential equation describes a process involving the

The Laplace transform of a linear PDE is accomplished by

The most general form of a second order linear PDE that is

!

∂v ∂v ∂2 v ∂2 v ∂2 v c1 2 + 2 c2 + c3 2 + c4 + c5 + c6 v = f ! ∂x ∂t ∂x ∂t ∂x ∂t

where f ( x, t ) is the forcing function or driving function. The coefficients ci are functions of the two independent variables x and t . Second order linear PDE are classified according as follows: !

Elliptic!

c22 − 4 c1 c3 < 0 !

(10.1-2)

!

Hyperbolic!

c22 − 4 c1 c3 > 0 !

(10.1-3)

!

Parabolic!

c22 − 4 c1 c3 = 0 !

(10.1-4)

considering the PDE to be a function of only one independent variable so that the Laplace transform operates only upon that

(10.1-1)

one variable. The other independent variables are then treated as parameters, and so are considered constants for the given

!

transformation. The Laplace transform reduces the number of

PDE, it is necessary that:

independent variables in the PDE. !

For a second order PDE only one independent variable

will remain after solution of the Laplace transformed PDE, and so the PDE has become an ODE. For some linear PDEs of order

For the Laplace transform to be useful in the solution of a

1.! The PDE must be linear. 2.! The coefficients of the PDE must be constants. 3.! One or more of the independent variables must have a range from 0 to ∞ . 152

Proposition 10.1-1: If f ( x, t ) is a continuous function of exponential order over

!

[ 0, ∞ ) and if the first partial derivatives of f ( x, t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) ,

and so:

then:

!

⎧ ∂ f ( x, t ) ⎫ − L ⎨ ⎬ = s F ( x, s ) − f x, 0 ! ⎩ ∂t ⎭

(

!

where L

)

(10.1-5)

!

0−

∂t

∫

e

−st

dt !

(10.1-6)

)

parts considering x as a parameter. Letting:

! we have:

∂ f ( x, t )

du = − s e

∂t

dt !

)

∫

0−

f ( x, t ) e− s t dt (10.1-10)

⎧ ∂ f ( x, t ) ⎫ − L ⎨ ⎬ = s F ( x, s ) − f x, 0 ! ⎩ ∂t ⎭

(

)

(10.1-11)

)

functions of exponential order over ⎡⎣ 0 − , ∞ , and if the second partial derivatives of f ( x, t ) are piecewise-smooth functions of

)

exponential order over ⎡⎣ 0 − , ∞ , then:

)

−st

(

∞

(10.1-9)

If f ( x, t ) and its first partial derivatives are continuous

⎡⎣ 0 − , ∞ , and its first partial derivatives are piecewise-smooth functions of exponential order over ⎡⎣ 0 − , ∞ , we can integrate by

u = e− s t !

0−

Proposition 10.1-2:

Since f ( x, t ) is a continuous function of exponential order over

!

∫

f ( x, t ) e− s t dt !

■

From the definition of the Laplace transform we can write:

∂ f ( x, t )

+s

⎧ ∂ f ( x, t ) ⎫ L ⎨ lim f ( x, t ) e− s t − f x, 0 − + s ⎬ = t→ ∞ ⎩ ∂t ⎭

!

∞

t = 0−

∞

Therefore:

{ f ( x, t )} = F ( x, s ) .

⎧ ∂ f ( x, t ) ⎫ L ⎨ ⎬= ⎩ ∂t ⎭

∞

! !

Proof: !

⎧ ∂ f ( x, t ) ⎫ −st L ⎨ ⎬ = f ( x, t ) e ⎩ ∂t ⎭

= dv !

v = f ( x, t ) !

! (10.1-7)

(

where L (10.1-8)

(

)

∂ f x, 0 − ⎧⎪ ∂2 f ( x, t ) ⎫⎪ 2 − ! (10.1-12) L ⎨ ⎬ = s F ( x, s ) − s f x, 0 − 2 ∂t ⎩⎪ ∂t ⎭⎪

)

{ f ( x, t )} = F ( x, s ) .

Proof: !

From the definition of the Laplace transform we can write: 153

⎧⎪ ∂2 f ( x, t ) ⎫⎪ L ⎨ ⎬= 2 ⎩⎪ ∂t ⎭⎪

!

∞

∂2 f ( x, t )

0−

∂t 2

∫

e

−st

or dt !

(10.1-13)

Since f ( x, t ) and its first partial derivatives are continuous

)

functions of exponential order over ⎡⎣ 0 − , ∞ , and the second partial derivatives of f ( x, t ) are piecewise-smooth functions of

)

−

exponential order over ⎡⎣ 0 , ∞ , we can integrate by parts. Letting: −st

!

u=e

!

dv =

!

du = − s e− s t dt !

v=

∂2 f ( x, t ) ∂t 2

∂f ( x, t ) ∂t

!

(10.1-14)

!

(10.1-15)

(

)

(10.1-18)

We can use Proposition 10.1-1 to write:

∂ f x, 0 − ⎧⎪ ∂2 f ( x, t ) ⎫⎪ − ⎤ ⎡ ! L ⎨ ! ⎬ = s ⎣ s F ( s ) − f x, 0 ⎦ − 2 ∂t ⎩⎪ ∂t ⎭⎪

(

)

(10.1-19)

(

)

2 ∂ f x, 0 − ⎪⎧ ∂ f ( x, t ) ⎪⎫ 2 − ! L ⎨ ! (10.1-20) ⎬ = s F ( x, s ) − s f x, 0 − 2 ∂t ⎪⎩ ∂t ⎪⎭

(

)

■

Proposition 10.1-3:

⎧⎪ ∂2 f ( x, t ) ⎫⎪ ∂ f ( x, t ) − s t L ⎨ e ⎬= 2 ∂t ⎩⎪ ∂t ⎭⎪

∞

+s t = 0−

∞

∂ f ( x, t )

0−

∂t

∫

! !

e

−st

If f ( x, t ) is a continuous function of exponential order over

[ 0, ∞ ) and if the first partial derivatives of f ( x, t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) ,

dt

(10.1-16)

then:

and so:

(

− ∂f ( x, t ) − s t ∂f x, 0 ⎧⎪ ∂2 f ( x, t ) ⎫⎪ lim e − ! L ⎨ ⎬ = t→ 2 ∞ ∂t ∂t ⎩⎪ ∂t ⎭⎪

!

)

and so:

dt !

we have: !

(

− ⎧⎪ ∂2 f ( x, t ) ⎫⎪ ⎧ ∂f ( x, t ) ⎫ ∂ f x, 0 ! L ⎨ ⎬= s L ⎨ ⎬− 2 ∂t ⎩ ∂t ⎭ ⎩⎪ ∂t ⎭⎪

+s

∫

∞

∂f ( x, t )

−

∂t

0

) e

−st

dt !

(10.1-17)

!

⎧ ∂ f ( x, t ) ⎫ dF ( x, s ) L ⎨ ! ⎬= ∂x dx ⎩ ⎭

(10.1-21)

!

2 2 ⎪⎧ ∂ f ( x, t ) ⎪⎫ dF ( x, s ) L ⎨ ! ⎬= 2 dx 2 ⎩⎪ ∂x ⎭⎪

(10.1-22) 154

Therefore:

Proof: !

From the definition of the Laplace transform we can write:

⎧ ∂ f ( x, t ) ⎫ L ⎨ ⎬= ⎩ ∂x ⎭

!

∞

∂ f ( x, t )

0−

∂x

∫

e− s t dt !

(10.1-23)

10.2!

or since x is independent of t :

⎧ ∂ f ( x, t ) ⎫ d L ⎨ ⎬= ⎩ ∂x ⎭ dx

!

∫

∞

0−

f ( x, t ) e− s t dt =

dF ( x, s ) dx

!

(10.1-24)

!

!

SOLVING PDES USING LAPLACE TRANSFORMS

The Laplace transform is very useful for obtaining

solutions to linear PDEs having constant coefficients and closed-form solutions. The general procedure for using the

Similarly we have: 2 ⎪⎧ ∂ f ( x, t ) ⎪⎫ d 2 L ⎨ ⎬= 2 2 ⎩⎪ ∂x ⎭⎪ dx

!

⎧ ∂cos ( x t ) ⎫ s x2 L ⎨ −1 = − 2 ⎬= s 2 s + x2 s + x2 ⎩ ∂t ⎭

∫

∞

0−

f ( x, t ) e

−st

dt =

■

dF 2 ( x, s ) dx 2

Laplace transform to obtain solutions of linear PDEs consists of ! (10.1-25)

the following: 1.! The entire PDE is first transformed using the Laplace transform for one of the independent variables that has a

Example 10.1-1

range from 0 to ∞ .

Determine: !

2.! The initial and boundary conditions are applied to the

⎧ ∂ f ( x, t ) ⎫ ⎧ ∂cos ( x t ) ⎫ L ⎨ ⎬= L ⎨ ⎬ ⎩ ∂t ⎭ ⎩ ∂t ⎭

transformed equation. 3.! The new PDE or ODE is solved using classical methods for

Solution:

the transformed dependent variable.

From Proposition 10.1-1 we have: !

⎧ ∂ f ( x, t ) ⎫ − L ⎨ ⎬ = s F ( x, s ) − f x, 0 ⎩ ∂t ⎭

(

4.! The transformed dependent variable solution is rewritten

)

using partial fractions if necessary.

155

5.! The inverse Laplace transform of each of the terms in the partial fraction expansion of the transformed solution is determined.

!

We also have from Proposition 10.1-3:

6.! The initial and boundary conditions are applied to the !

solution. 7.! For PDE of order higher than two, the above procedure can be repeated. 8.! In this way the solution to the original PDE is obtained.

Determine the solution of the partial differential equation: !

∂x

+

∂v ( x, t ) ∂t

= 2x

v ( x, 0 ) = 0 !

We then have the transformed PDE: !

dV ( x, s ) dx

+ s V ( x, s ) =

2x s

!

dV ( x, s ) dx

es x + s V ( x, s ) es x = 2 es x

s dx

= es x :

x s

or

where the initial and boundary conditions are: !

⎧ ∂v ( x, t ) ⎫ ∂V ( x, s ) dV ( x, s ) L ⎨ = ⎬= ∂x dx ⎩ ∂x ⎭

Multiplying by the integrating factor µ = e ∫

Example 10.2-1 ∂v ( x, t )

⎧ ∂v ( x, t ) ⎫ L ⎨ ⎬ = sV ( x, s ) ⎩ ∂t ⎭

v ( 0, t ) = 0

!

(

)

d x V ( x, s ) es x = 2 es x dx s

Integrating, we obtain:

V ( x, s ) es x =

2 s

∫ xe

sx

dx + c ( s )

Solution:

!

From Proposition 10.1-1 we have:

where c ( s ) is an integration constant. Integrating by parts,

!

⎧ ∂v ( x, t ) ⎫ − L ⎨ ⎬ = sV ( x, s ) − v x, 0 ⎩ ∂t ⎭

(

Applying the initial condition:

)

we have: !

2 e− s x ⎡ x e s x e s x ⎤ V ( x, s ) = − 2 ⎥ + c ( s ) e− s x ⎢ s ⎣ s s ⎦ 156

!

or !

V ( x, s ) =

2x 2 − 3 + c ( s ) e− s x 2 s s

Since a boundary condition is v ( 0, t ) = 0 , we must have 2 V ( 0, s ) = 0 , and so c ( s ) = 3 . Therefore: s −s x 2 x 2 2e ! V ( x, s ) = 2 − 3 + 3 s s s

v ( x, t ) = 2 x t − t 2 + ( t − x ) U ( t − x )

!

!

Determine the solution of the partial differential equation: =

∂v ( x, t ) ∂t

− 2 v ( x, t )

where lim v ( x, t ) = 0 and where the initial condition is: x→ ∞

!

⎧ ∂v ( x, t ) ⎫ ∂V ( x, s ) dV ( x, s ) L ⎨ = ⎬= ∂x dx ⎩ ∂x ⎭

!

dV ( x, s ) dx

= s V ( x, s ) − e− x − 2V ( x, s ) = ( s − 2 ) V ( x, s ) − e− x

or

Example 10.2-2

∂x

⎧ ∂v ( x, t ) ⎫ −x L ⎨ ⎬ = sV ( x, s ) − e ∂t ⎩ ⎭

We also have from Proposition 10.1-3:

2

∂v ( x, t )

)

We then have the transformed PDE:

solution:

!

(

Applying the initial condition:

From the transform tables in Appendix E and F we obtain the !

⎧ ∂v ( x, t ) ⎫ − L ⎨ ⎬ = sV ( x, s ) − v x, 0 ⎩ ∂t ⎭

v ( x, 0 ) = e− x

Solution: From Proposition 10.1-1 we have:

!

dV ( x, s ) dx

+ ( 2 − s ) V ( x, s ) = − e− x

Multiplying by the integrating factor µ = e ∫ ( !

2−s ) dx

= e( 2−s ) x :

dV ( x, s ) ( 2−s ) x e + ( 2 − s ) V ( x, s ) e( 2−s ) x = − e− x e( 2−s ) x dx

or !

(

)

d V ( x, s ) e( 2−s ) x = − e(1−s ) x dx

Integrating this equation, we obtain: 157

V ( x, s ) e

( 2−s ) x

!

e(1−s ) x = + c(s) s −1

!

where c ( s ) is an integration constant. We then have: V ( x, s ) =

!

e (1−s ) x e− ( 2−s ) x + c ( s ) e ( s−2 ) x s −1

Since the boundary condition is lim v ( x, t ) = 0 , we must have

lim V ( x, s ) = 0 , and so c ( s ) = 0 :

x→ ∞

!

!

v ( x, t ) = e− x e t

Example 10.2-3 Determine the solution of the one-dimensional wave equation: !

∂x 2

(

∂v x, 0 − ⎧⎪ ∂2 v ( x, t ) ⎫⎪ 2 − L ⎨ ⎬ = s V ( x, s ) − s v x, 0 − 2 ∂t ⎩⎪ ∂t ⎭⎪

(

)

)

Applying the initial condition: !

⎧⎪ ∂v 2 ( x, t ) ⎫⎪ 2 L ⎨ ⎬ = s V ( x, s ) 2 ⎩⎪ ∂t ⎭⎪

We also have from Proposition 10.1-3:

solution:

∂v 2 ( x, t )

v ( 0, t ) = f ( t )

= 0!

From Proposition 10.1-2 we have:

From the transform table in Appendix E we obtain the !

∂t

Solution:

x→ ∞

e− x V ( x, s ) = s −1

∂v ( x, 0 )

v ( x, 0 ) = 0 !

2 1 ∂v ( x, t ) = 2 k ∂t 2

where lim v ( x, t ) = 0 and where the initial and boundary x→ ∞

conditions are:

!

∂v 2 ( x, t ) ∂x 2

=

∂V 2 ( x, s ) ∂x 2

We then have the transformed PDE: !

∂V 2 ( x, s ) ∂x 2

=

s2 V ( x, s ) k2

or !

V ( x, s ) = c1 ( s ) e ( x k ) s + c2 ( s ) e− ( x k ) s

We also have: !

lim V ( x, s ) = lim

x→ ∞

x→ ∞

∫

∞

0

−

v ( x, t ) e

−st

dt =

∫

∞

lim v ( x, t ) e− s t dt = 0

0 x→ ∞ −

158

Therefore we must have c1 ( s ) = 0 so that: !

V ( x, s ) = c2 ( s ) e− ( x k ) s

!

Since the boundary condition is v ( 0, t ) = f ( t ) we have

V ( 0, s ) = F ( s ) , and so c2 ( s ) = F ( s ) : !

V ( x, s ) = F ( s ) e− ( x k ) s

have:

Example 10.2-4 Determine the solution of the equation: !

∂t

=k

∂x 2

boundary conditions are for x > 0 :

Solution:

v ( x, 0 ) = 0 !

⎧ ∂v ( x, t ) ⎫ L ⎨ ⎬ = sV ( x, s ) ⎩ ∂t ⎭ ∂v 2 ( x, t ) ∂x 2

=

∂V 2 ( x, s ) ∂x 2

v ( 0, t ) = v0

s V ( x, s ) = k

∂V 2 ( x, s ) ∂x 2

or !

∂v 2 ( x, t )

where v ( x, t ) is bounded as x → ∞ , and where the initial and !

)

We then have the transformed PDE: !

∂v ( x, t )

(

We also have from Proposition 10.1-3: !

x⎞ ⎛ x⎞ ⎛ v ( x, t ) = f ⎜ t − ⎟ U ⎜ t − ⎟ ⎝ k⎠ ⎝ k⎠

⎧ ∂v ( x, t ) ⎫ − L ⎨ ⎬ = sV ( x, s ) − v x, 0 ⎩ ∂t ⎭

Applying the initial condition: !

From the second shift theorem (Proposition 5.2-1) we then

!

From Proposition 10.1-1 we have:

V ( x, s ) = c1( s ) e x

s k

+ c2 ( s ) e− x

s k

Since v ( x, t ) is bounded as x → ∞ , we must have c1( s ) = 0 so that: !

V ( x, s ) = c2 ( s ) e −x

s k

Since the boundary condition is v ( 0, t ) = v0 we have:

159

!

V ( 0, s ) =

∫

∞

0−

v ( 0, t ) e− s t dt = v0

∫

∞

0−

e− s t dt =

v0 = c2 ( s ) s

Determine the Laplace transform of t cos at using the Laplace transform F ( a, s ) of f ( a, t ) = sin at .

Therefore: !

Example 10.3-1

e −x s k V ( x, s ) = v0 s

Solution: We have:

From the transform table in Appendix E we obtain the !

solution: !

⎛ x ⎞ v ( x, t ) = v0 erfc ⎜ ⎟ ⎝ 2 kt ⎠

OBTAINING LAPLACE TRANSFORMS USING PARTIAL DIFFERENTIATION

!

From equation (10.1-21) we have:

!

⎧ ∂ f ( x, t ) ⎫ dF ( x, s ) ! L ⎨ ⎬= dx ⎩ ∂x ⎭ ⎧ ∂ f ( a, t ) ⎫ ∂F ( a, s ) ! L ⎨ ⎬= ∂a ⎩ ∂a ⎭

∂ ( sin ( at )) ∂a

= t cos ( at )

(10.3-1)

F ( a, s ) =

a s + a2 2

From equation (10.3-2) we have: !

⎧⎪ ∂ ( sin ( at )) ⎫⎪ ∂ ⎛ a ⎞ L ⎨ ⎜⎝ 2 ⎬= 2⎟ ⎠ ∂a ⎪⎩ ⎪⎭ ∂a s + a

or !

or letting x = a : !

∂a

=

and !

10.3!

∂ f ( a, t )

L {t cos ( at )} =

1 s2 + a2

−

(

2 a2 s2 + a2

=

s2 − a2

) (s 2

2

+ a2

)

2

(10.3-2)

We can use this equation to obtain the Laplace transform of new functions. 160

Chapter 11 Complex Variables

!∫

C

()

f z dz = 2 π i

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ n

n=1

161

()

()

In the next chapter we will consider the inversion of the

where x t and y t are real-valued continuous functions. The

Laplace transform using residue integration. Since residue

parameter t has definite limits a ≤ t ≤ b such that the curve

integration uses complex analysis methods, in this chapter we

proceeds from a point z1 = z a

will present a brief review of some complex variable definitions

parameterization of the curve provides a direction to the curve.

and propositions. We will not prove most of the propositions.

!

Proofs of these propositions can be found in any good book on

continuous derivative z ′ t ≠ 0 at all points along the curve

complex variables. We will assume a knowledge of complex

between the end-points. A piecewise-smooth integration path

numbers.

C in the complex plain is known as a contour, and consists of a connected set of points z = x + i y . The integral of a complex

11.1! CONTOURS IN THE COMPLEX PLAIN

function along such a path is referred to as a contour integral.

!

!

The integral of a complex function is not restricted to an

interval of the x-axis, but can be evaluated over some path in the complex plane. The integral of a complex function is similar, therefore, to a line integral of a real function in the Cartesian plane. !

We will consider the integration path in a complex plane

to be a piecewise continuous curve. Such a curve can be defined as a set of points that can be represented by a piecewise

()

()

to a point zn = z b . The

A smooth curve is a continuous curve that has a

()

!

Open sets of points do not include any points that form a

boundary to the set. We will assume that contours are all within regions of the complex plane consisting of open sets of points that are simply connected (any two points in the region can be connected by a polygonal path). Such regions are called domains. A simply connected domain is a domain without holes. !

A δ neighborhood of a point z0 is taken to be an open set

continuous complex function (see Section 1.4.2). This function

of points in the form of a disk centered at the point z0 such that:

can be given in the form of a parameterization (such as z t ,

!

()

where t is a real parameter). We then have: !

() ()

()

z t = x t +iy t !

z − z0 < δ !

(11.1-2)

( )

The notation for the open δ -disk centered at z0 is Dδ z0 . An (11.1-1)

open disk is defined to be a disk that does not include any of its boundary points (see Figure 11.1-1). 162

!

Curve C4 is smooth, simple, and closed.

Figure 11.1-1! Graphical depiction of the set of points forming the open δ -disk Dδ z0 about the point z0 . The disk’s boundary circle is dashed to indicate that points on this circle are not in the disk.

( )

!

Since contours are piecewise-smooth curves, each segment

of a contour will be continuous and will have continuous derivatives at all points within the segment. Curves in the complex plain can be categorized as follows: !

Curve C1 is smooth, simple, and not closed.

! !

Curve C2 is piecewise-smooth, simple, and not closed.

!

Curve C3 is closed but not simple.

Figure 11.1-2! Curves in the complex plane: ! !

Contours can then be defined as follows: 1.! A simple contour is a piecewise-smooth curve in the complex plane that does not intersect itself except possibly at its end points (see curves C1 and C4 in Figure 11.1-2). 163

!

2.! A closed contour is a contour in the complex plane whose initial and terminal points are the same point (see curves C3 and C4 in Figure 11.1-2).

!

that is not closed (see curves C1 and C2 in Figure 11.1-2).

11.2! DERIVATIVES OF COMPLEX FUNCTIONS

where we will define:

( )

f ′ z0 = lim

(11.2-1)

) (

)

()

(11.2-2)

If f z is a complex function defined in the neighborhood of

( ) = lim f ( z ) − f ( z0 ) = lim f ( z0 + Δz ) − f ( z0 )

Δf z Δz

Δz →0

z → z0

z − z0

Δ z→ 0

Δz

(11.2-4)

()

() said to be differentiable at the point z0 . The derivative of f ( z )

be bounded in a neighborhood of z0 . If the limit exists, f z is

() () ( ) ( ) ( ) the limit f ′ ( z0 ) to exist at the point z0 . Therefore f ( z ) must be continuous at point z0 . !

()

For a function f z to be complex differentiable at a point

z0 , there must then exist for any real number ε > 0 a real number δ > 0 such that: !

z0 , then the change in the function f ( z ) due to the change from

or

point z0 to the point z is Δ f z , where:

!

()

( )

provided that the limit exists. For the limit to exist f z must

so that z = z0 + Δz . From equation (11.2-1) we have:

(

(11.2-3)

have Δ f z = f z − f z0 = f z0 + Δz − f z0 → 0 as Δz → 0 for

consider another point z located within a neighborhood of z0

Δz = x − x0 + i y − y0 = Δx + i Δ y !

( )

is then specified with reference to the given point z0 . We must

Given some point z0 in the complex z-plane, we will now

!

)

! !

4.! An open contour is any contour in the complex plane

Δz = z − z0 !

(

()

3.! A simple closed contour is a simple contour in the

!

( )

a complex function f z is the limit:

same point (see curve C4 in Figure 11.1-2).

!

()

By definition, the derivative f ′ z0 at a given point z0 of

!

complex plane whose initial and terminal points are the

!

()

Δ f z = f z − f z0 = f z0 + Δz − f z0 !

!

()

( )− f′

f z − f z0 z − z0

( )− f′

Δf z Δz

( z0 ) < ε !

( z0 ) < ε !

when z − z0 < δ ! (11.2-5)

when Δ z < δ !

(11.2-6) 164

!

The number δ will generally depend on ε , but the

( )

derivative f ′ z0

is independent of both ε and Δz since the

derivative represents the actual spatial rate of change in the

()

function f z at the given point z0 . From equation (11.2-6) we have for some ε > 0 :

()

( )

Δ f z = f ′ z0 Δ z + ε Δ z !

!

(11.2-7)

()

!

( ) (

) ( ) (

)

(11.2-8)

From equations (11.2-4), (11.2-1), and (11.2-3) with Δz → 0 ,

( )

at a given point z0 in the

form: !

( )

f ′ z0 =

()

df z dz

= lim z = z0

Δz → 0

( ) = lim f ( z0 + Δ z ) − f ( z0 ) !

Δf z Δz

Δz

Δz → 0

(11.2-9)

We will now omit the subscript 0 from z0 , and we will use simply z to represent the point at which we wish to obtain the derivative:

! !

we must have both Δx → 0 and Δ y → 0 . The limiting process for complex functions of a complex variable then involves both

Δx and Δ y . How Δz → 0 is dependent upon how both Δx → 0 and Δ y → 0 . The limiting process for complex functions of a

point z0 as Δz → 0 . !

( )

For the limit f ′ z0

to exist at a point z0 , it is necessary

that the same limit be determined as z0 is approached from any direction in the z-plane. That is, the same limit must be obtained independent of the argument of Δz .

! !

!

Since Δz = Δx + i Δ y is a complex number, to have Δz → 0 ,

infinity of directions of approach in the complex plane to a

f z = f z0 + z − z0 f ′ z0 + ε z − z0 !

we can write the derivative f ′ z0

!

complex variable is therefore two-dimensional. There are an

which gives us the mean value theorem: !

11.2.1! EXISTENCE OF THE DERIVATIVE

()

f′ z =

( ) = lim

df z dz

Δz → 0

( ) = lim f ( z + Δ z ) − f ( z )

Δf z Δz

Δz → 0

Δz

(11.2-10)

!

()

Since the derivative of a complex function f z at a point

z0 is independent of the direction to z0 along which the

()

derivative is taken, f z will be differentiable at z0 only if the

()

spatial rate of change in f z is the same no matter from what direction z0 is approached. !

The two-dimensional existence requirement of direction

independence for the derivative of a complex function is a much more stringent condition that must be satisfied than is the analogous one-dimensional existence requirement for the derivative of a real-valued function. As a result, many relatively 165

simple complex functions are not differentiable, and those that

If this derivative exists, then the limits of the real and

are differentiable have properties very different from their real

imaginary parts of w = f z must exist:

()

counterparts. !

In particular, complex functions have one property not

()

()

! f ′ z = lim

commonly found in real functions. If a complex function f z

Δz → 0

(

)

(

)

()

()

⎡⎣u z + Δz + i v z + Δz ⎤⎦ − ⎡⎣ u z + i v z ⎤⎦ ! (11.2-13) Δz

has one derivative at a point, it will have an infinity of higher

or

order derivatives at the same point. By comparison, if a real

⎡⎣u ( x + Δx, y + Δy ) + i v ( x + Δx, y + Δy ) ⎤⎦ ! f ′ ( z ) = lim Δx→ 0 Δx + iΔy Δy → 0

()

function f x has one derivative at a point, it may have no, some, or many higher order derivatives at the same point.

11.2.2! CAUCHY-RIEMANN EQUATIONS !

( )

( )

⎡⎣u x, y + i v x, y ⎤⎦ ! − Δlim x→ 0 Δ x + iΔ y Δy→0

!

(11.2-14)

We will now develop two partial differential equations

known as the Cauchy-Riemann equations that together provide

11.2.2.1!

the necessary and sufficient conditions for the existence of a

NECESSARY CONDITIONS FOR THE EXISTENCE OF A DERIVATIVE

()

If the derivative f ′ z given in equation (11.2-14) exists at

derivative of a complex function at a point. We will consider

!

the derivative f ′ z of a complex function w = f z at a given

a point z x, y in a domain D , the limits in this equation cannot

point z = x + i y in a domain D . Writing the function w = f z in

depend upon the path of approach to z x, y as Δz → 0 since

the form:

each limit must be unique. To examine the necessary conditions

()

()

() ()

() ( )

( )

w = f z = u z + i v z = u x, y + i v x, y !

!

()

(11.2-11)

()

( ) = lim

df z dz

Δz →0

( ) = lim f ( z + Δz ) − f ( z ) !

Δf z Δz

Δz → 0

Δz

( )

()

for the existence of the derivative f ′ z , we will choose two different paths of approach within the domain D to the point

( )

the derivative is given by equation (11.2-4): !f′ z ≡

( )

z x, y as shown in Figure 11.2-1: a path parallel to the real axis (11.2-12)

and a second path parallel to the imaginary axis.

166

!

Now lettings Δz → 0 parallel to the y-axis so that Δx = 0

and Δz = Δ y → 0 (see Figure 11.2-1), equation (11.2-14) then becomes:

(

) ( )

(

) ( )

⎡⎣u x, y + Δ y − u x, y ⎤⎦ + i ⎡⎣v x, y + Δ y − v x, y ⎤⎦ Δy →0 i Δy ! ! (11.2-17) or, from the definition of real partial derivatives:

()

f ′ z = lim

! !

( )

Figure 11.2-1! Two paths of approach to a point z x, y . !

Letting Δz → 0 parallel to the x-axis so that Δ y = 0 and

Δz = Δx → 0 (see Figure 11.2-1), equation (11.2-14) becomes:

()

f ′ z = lim

!

(

Δx → 0

) ( )

(

) ( )

⎡⎣u x + Δ x, y − u x, y ⎤⎦ + i ⎡⎣v x + Δ x, y − v x, y ⎤⎦ Δx ! (11.2-15)

()

Since the derivative f ′ z

()

f′ z =

∂u 1 ∂u ∂v ∂v ! + = −i i ∂y ∂y ∂y ∂y

(11.2-18)

()

( )

If the function w = f z is differentiable at a point z x, y ,

the derivatives given in equations (11.2-16) and (11.2-18), which were calculated for two different directions of approach, must exist and be equal. We then can write: !

()

f′ z =

( ) = ∂u + i ∂v = ∂v − i ∂u !

df z dz

∂x

∂x

∂y

∂y

(11.2-19)

Equating the real and imaginary parts, we obtain: !

exists, the limits of the real and

∂u ∂v ! = ∂x ∂ y

∂v ∂u =− ! ∂x ∂y

(11.2-20)

imaginary parts of equation (11.2-15) exist. From the definition

These two linear partial differential equations are known as the

of real partial derivatives, we then have:

Cauchy-Riemann equations.

!

()

f′ z =

∂v ∂u ! +i ∂x ∂x

(11.2-16)

!

()

() ( )

( )

If the derivative f ′ z of a function f z = u x, y + i v x, y

( )

exists at a point z x, y , then the first order partial derivatives 167

( )

( )

of u x, y and v x, y with respect to x and y must exist and

!

must satisfy the Cauchy-Riemann equations at this point. The

and are continuous in a neighborhood of the point z x, y , we

Cauchy-Riemann equations therefore constitute necessary

can write for Δw = Δu + i Δv :

conditions for a complex function to be differentiable at a point. 11.2.2.2! !

SUFFICIENT CONDITIONS FOR THE EXISTENCE OF A DERIVATIVE

Given that we considered only two paths of approach to

( )

the point z x, y , it is not obvious that the Cauchy-Riemann equations constitute sufficient conditions for any complex

!

() ()

() ( )

( )

w = f z = u z + i v z = u x, y + i v x, y !

(11.2-21)

()

to have a derivative f ′ z at a given point so that we have:

()

f′ z =

! !

∂u ∂v ∂v ∂u ! +i = −i ∂x ∂x ∂ y ∂y

(11.2-22)

We will now show that if the partial derivatives ∂u ∂x ,

∂u ∂ y , ∂v ∂x , and ∂v ∂ y exist and are continuous at a point

( )

z x, y

within a domain D , and if the Cauchy-Riemann

()

equations hold at this point, then the derivative f ′ z given in equation (11.1-22) exists.

( )

⎛ ∂u ⎞ ⎛ ∂u ⎞ Δu = ⎜ + ε1 ⎟ Δx + ⎜ + η1 ⎟ Δ y ! ⎝ ∂x ⎠ ⎝ ∂y ⎠

(11.2-23)

⎛ ∂v ⎞ ⎛ ∂v ⎞ Δv = ⎜ + ε 2 ⎟ Δx+ ⎜ + η2 ⎟ Δ y ! ⎝ ∂x ⎠ ⎝ ∂y ⎠

(11.2-24)

and !

where ε1 , ε 2 → 0 and η1 , η2 → 0 as Δx → 0 and Δ y → 0 . We then

( )

have at the point z x, y :

function: !

Since we are assuming that the partial derivatives exist

!

⎛ ∂u ⎛ ∂u ∂v ⎞ ∂v ⎞ Δw = Δu + i Δv = ⎜ + i ⎟ Δx+ ⎜ + i ⎟ Δy ∂x ⎠ ∂y⎠ ⎝ ∂x ⎝ ∂y

(

)

(

)

+ ε1 + i ε 2 Δx+ η1 + i η2 Δ y !

!

(11.2-25)

or ⎛ ∂u ⎛ ∂u ∂v ⎞ ∂v ⎞ ! Δw = ⎜ + i ⎟ Δx + ⎜ + i ⎟ Δ y+ ε Δx+ η Δ y ! (11.2-26) ∂x ⎠ ∂y⎠ ⎝ ∂x ⎝ ∂y

where ε = ε1 + i ε 2 and η = η1 + i η 2 . !

Using the Cauchy-Riemann equations, we can rewrite

equation (11.1-26) as:

168

!

⎛ ∂u ⎛ ∂v ∂v ⎞ ∂u ⎞ Δw = ⎜ + i ⎟ Δx + ⎜ − + i ⎟ Δ y + ε Δx + η Δ y ∂x ⎠ ∂x ⎠ ⎝ ∂x ⎝ ∂x

!!

(11.2-27) ⎛ ∂u ∂v ⎞ Δw = ⎜ + i ⎟ Δx+ i Δ y + ε Δx+ η Δ y ! ∂x ⎠ ⎝ ∂x

(

)

(11.2-28)

Δw ⎛ ∂u ∂v ⎞ Δx Δy =⎜ +i ⎟ +ε +η ! Δz ⎝ ∂x ∂x ⎠ Δz Δz

lim

Δz →0

(11.2-29)

() z ( x, y ) . The Cauchy-Riemann

with respect to x and y , and if f z satisfies the CauchyRiemann equations at the point

Δw ⎛ ∂u ∂v ⎞ Δx Δy ! − ⎜ + i ⎟ = lim ε +η Δz →0 Δz ⎝ ∂x ∂x ⎠ Δz Δz

()

()

( )

point z x, y . 11.2.2.3!

(11.2-30)

!

DERIVATIVE OF A COMPLEX FUNCTION IN RECTANGULAR FORM

From equations (11.2-33) and (11.2-20), we then have the

following equivalent expressions for the derivative of a

( )

complex function at a point z x, y :

or

Δw ⎛ ∂u ∂v ⎞ Δx Δy lim − ⎜ + i ⎟ ≤ lim ε +η ! Δ z → 0 Δz ∂x ⎠ Δ z → 0 Δz Δz ⎝ ∂x

We have Δx

Δz ≤ 1 and Δy

(11.2-31)

Δw ∂u ∂v ! lim = +i Δ z → 0 Δz ∂x ∂x

!

dw ∂u ∂v ∂v ∂u ∂u ∂u ∂v ∂v = f′ z = +i = −i = −i = +i ! dz ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x

()

! !

Δz ≤ 1 . Since ε → 0 and η → 0

as Δz → 0 , we have:

or

has continuous first partial derivatives

complex function f z to have a derivative f ′ z at a given

and so:

!

()

equations therefore constitute sufficient conditions for a

Using Δz = Δ x + i Δ y , we have: !

(11.2-33)

independent of how Δz → 0 . The derivative f ′ z then exists

()

!

!

dw ∂u ∂v = +i ! dz ∂x ∂x

and is unique if f z

or

!

()

f′ z =

!

(11.2-34)

We also have from equation (11.2-34): !

(11.2-32)

dw dz

2

()

= f′ z

2

2

2

!

2

2

⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ =⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ ∂x ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂y ⎠ 2

2

2

2

⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ =⎜ ⎟ +⎜ ⎟ =⎜ ⎟ +⎜ ⎟ ! ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂ y ⎠

(11.2-35) 169

!

We can summarize the results of the last two sections in

11.2.4! HOLOMORPHIC FUNCTIONS !

Proposition 11.2-1.

We will now consider a more restrictive set of functions

that are differentiable not only at some single point z0 , but also

11.2.3! CAUCHY-RIEMANN EQUATIONS IN RECTANGULAR FORM

at every point z of a δ neighborhood of z0 . A complex function

()

w= f z

()

is said to be holomorphic at a point z0 if f z

is

defined and differentiable in an open disk z − z0 < δ centered

Proposition 11.2-1, Cauchy-Riemann Equations:

()

at z0 , where δ > 0 . For f z to be holomorphic at a point z0 in a

()

The necessary and sufficient conditions for a complex function

domain D requires then that the function f z be differentiable

w = f z = u x, y + i v x, y

in a whole neighborhood Dδ z0 that exists within D . Therefore

( ) ( ) ( ) to be differentiable z ( x, y ) are that the Cauchy-Riemann equations: ∂u ∂v = ! ∂x ∂ y

!

∂u ∂v =− ! ∂y ∂x

at a point

( )

z0 must be an interior point of D . (11.2-36)

( )

hold at the point z x, y , and that the partial derivatives ∂u ∂x , ∂u ∂ y , ∂v ∂x , and ∂v ∂ y all exist and are continuous in a

( )

neighborhood of the point z x, y .

!

() f ( z ) to be holomorphic

While it is possible for a derivative f ′ z to exist at just a

single point, it is then not possible for

just at a single point. For this reason holomorphicity is considered to be a global property of a function, while differentiability is considered to be a local property of a function. We can conclude then that, for a function to be

Proof:

holomorphic along a curve, the function must be differentiable

!

Given in Sections 11.2.2.1 and 11.2.2.2.

!

The Cauchy-Riemann equations are a consequence of the

not only at points on the curve, but at points in some region

■

()

requirement that the derivative of a complex function f z at a point z0 must be independent of the path in the complex plane taken as Δz → 0 during the calculation of the derivative.

surrounding the curve. !

()

A complex function w = f z is said to be holomorphic on

()

a domain if f z is defined and differentiable at every point in

()

the domain. If f z

is holomorphic over the entire finite

complex plane, the function is referred to as an entire function. 170

Functions such as e z , sin z , and cos z are differentiable

Proposition 11.3-2, Independent of Path:

everywhere in the finite complex plane, and so are entire

()

If a complex function f z is holomorphic in a simply connected

functions. Polynomials are also entire functions. !

domain D , then the integral:

Since we will consider the inversion of the Laplace

transform using residue integration in the next chapter, we will

!

now define residues and discuss theorems related to finding

Proposition 11.3-3, Contour Integral Inequality:

11.3! IMPORTANT THEOREMS FOR DETERMINING RESIDUES

()

If C is a simple contour and f z is a complex function continuous over C , then:

Proposition 11.3-1, Fundamental Theorem of Calculus for Complex Functions:

()

connected domain D , then !

∫

z1

()

f z dz =

()

∫

z1

( ) dz = F ( z ) − F ( z ) !

dF z dz

∫

!

2

1

()

f z dz ≤

C

If a complex function f z is holomorphic in a simply

z2

(11.3-2)

C

is independent of path C within D .

residues.

z2

∫ f ( z ) dz !

∫

()

f z

C

dz !

(11.3-3)

Proposition 11.3-4, ML-Inequality:

()

Let C be a simple contour of length L , and let f z be a

()

continuous complex function on C . If f z is bounded by a real

(11.3-1)

number M along C :

()

where F z is the antiderivative of f z .

()

!

f z ≤ M!

(11.3-4)

∫

(11.3-5)

then: !

C

()

f z dz ≤ M L !

171

The arrow on the integral sign indicates that the direction of

Proof: ! !

integration along the contour is counterclockwise.

From Proposition 11.3-3 we can write:

∫

C

()

f z dz ≤

∫

()

f z

C

dz ≤ M

∫

Proposition 11.3-6, Deformation Theorem:

dz !

(11.3-6)

C

Let C1 and C2 be two simple closed contours within a domain

D , and let C2 be interior to C1 (see Figure 11.3-1). If f ( z ) is a

In parametric form we have: !

∫

C

()

f z dz ≤ M

∫

b

a

()

dz t dt

dt = M

complex function holomorphic at all points within or on C1 , and

b

∫ ds ! a

outside or on C2 , then:

(11.3-7)

where t = a and t = b are the end points of the contour C , and

!

!∫

C1

()

f z dz =

!∫

C2

()

f z dz !

(11.3-10)

ds is differential arc length. We then have:

!

∫

C

()

f z dz ≤ M L !

(11.3-8)

which is the upper bound for moduli of contour integrals. This equation is known as the ML-inequality.

■

Proposition 11.3-5, Cauchy-Goursat Theorem:

()

If f z is a holomorphic complex function in a simply connected domain D , then along every simple closed contour C within D we must have: !

!∫

C

()

f z dz = 0 !

(11.3-9)

Figure 11.3-1! Arrows show direction of integration. 172

Proposition 11.3-8, Cauchy’s Integral Formula for Derivatives: Proposition 11.3-7, Cauchy’s Integral Formula:

()

Let f z be a holomorphic complex function in a simply

()

Let f z be a holomorphic complex function in a simply

connected domain D . If C is a simple closed contour within the

connected domain D . If C is a simple closed contour that lies

domain D , and if z0 is any given point inside the contour C ,

within D , and if z0 is a fixed point inside the contour C , then: !

!

( )

f z0 =

1 2π i

!∫

C

( ) dz !

f z

(11.3-11)

z − z0

Cauchy’s integral formula states that if a complex

()

()

then f z has derivatives of all orders at the point z0 given by: ! !

n! n f ( ) z0 = 2π i

( )

( ) dz ! !∫C ( z − z0 )n+1 f z

(11.3-12)

where n is a nonnegative integer.

function f z is holomorphic at all points inside and on a

()

simple closed contour C , then the value of f z at any point

Proposition 11.3-9, Higher Order Derivatives of a

z = z0 inside the contour C is completely determined by just

Holomorphic Function Exist and are

the values of f z on this contour. This is a special property of

Holomorphic:

()

holomorphic functions that provides a method of computing

()

equivalent property for real-valued functions of a real variable.

If a complex function f z is holomorphic at a point z = z0 , then its derivatives f ( n) ( z ) of all orders n exist and are

!

holomorphic at the point z0 .

their value at any point within the contour C . There is no Holomorphic functions possess derivatives of all orders.

They differ in this regard from real-valued functions that often have derivatives only up to a certain order. If a complex

()

function f z

()

has a derivative at a point z0 , then f z has

derivatives of all orders at z0 , and these derivatives can be

Proposition 11.3-10: If C is a simple closed contour within a domain D , and if z = z0 is a point inside the contour C , then:

expressed in integral form using an integral formula of Cauchy. 173

!

!∫

C

1

( z − z0 )

n

⎧⎪ 2 π i n = 1 dz = ⎨ ! 0 n ≠ 1 ⎪⎩

(11.3-13)

{ }

terms of an infinite complex sequence zn is known as the k th

A sequence of complex numbers

{ zn }

is obtained if a

infinite set of complex numbers zn can be put in one-to-one correspondence with the set of positive integers, and if the complex numbers zn have a definite order of arrangement:

{ }

zn = z1 , z2 , z3 , ! !

(11.4-1)

A complex sequence then forms a countably infinite set of complex numbers. !

partial sum: !

{ }

zn . Complex series therefore require the

existence of a sequence and the algebraic operation of addition. An infinite complex series is defined as the sum of the terms zn

{ }

of an infinite complex sequence zn :

∑ z = z + z + z + !! n

1

2

3

n

1

2

3

k

(11.4-3)

n =1

As additional terms of the sequence are summed, new partial sums are formed: !

()

S1 z =

1

∑z = z n

1

n =1

! ! !

()

2

S2 z =

∑z = z + z ! n

1

2

n =1

!

(11.4-4)

()

S3 z =

3

∑z = z + z + z n

1

2

3

n =1

(11.4-2)

n =1

!

∑z = z + z + z + ! + z !

()

∞

!

k

Sk z =

A complex series is defined as the sum of the terms zn of a

complex sequence

()

sequence of partial sums is used. The sum Sk z of the first k

11.4! SERIES

!

must be determined in the form of a limit. To obtain a limit that can represent the sum of an infinite complex sequence, a

where n is an integer.

!

sequence. Instead the sum of an infinite complex sequence

Since it is not possible to add an infinity of terms, it is not

possible to directly obtain the sum of an infinite complex

!

! !

()

Sk z =

k

∑z = z + z + z + ! + z n

1

2

3

k

n =1

174

()

and so Sk z

represents the summation of the terms of the

{ }

infinite complex sequence zn truncated at the k th term.

{ zn } , the set of partial sums forms a complex sequence of partial sums { S ( z ) } : !

For any given complex sequence

k

{S ( z ) } = S ( z ) , S ( z ) , S ( z ) ,!, S ( z ) !

!

k

1

2

3

k

(11.4-5)

(11.4-6) {S ( z ) } = z , ( z + z ) , ( z + z + z ) , !! If the sequence of partial sums { S ( z ) } has a finite limit, then k

1

1

2

1

2

3

k

the infinite series given in equation (11.4-3) is said to converge to a sum. For the purpose of determining a limit, therefore, an infinite series is treated as a sequence of partial sums. If the series terms zn are not simply constant complex

!

()

numbers, but are functions f n z of complex numbers, then the series is a complex function series. Such a function series is

()

defined as the sum of terms f n z

{ f ( z )} of complex functions:

of an infinite sequence

n

n=0

0

1

2

n

becomes a sequence of complex numbers. If the sequence of partial sums

{S ( z ) } k

()

of the function f n z converges for all

points in a region R , then R is called a region of convergence If for the same value of N ∈! , a sequence of partial sums

!

{S ( z ) } k

()

converges to f z

for all points z in its region of

convergence R when n ≥ N , then the sequence of partial sums

()

is said to converge uniformly to f z , and we will have for any real number ε > 0 : !

()

()

Sk z − f z < ε !

for all k > N !

(11.4-8)

For uniform convergence the number N can depend on ε , but

()

( )

not z . We therefore have N = N ε , but not N = N ε , z . Proposition 11.4-1, Term-by-Term Integration:

()

()

converges

uniformly to the sum f z in a domain D , then for any simple

∞

∑ f ( z ) = f ( z ) + f ( z ) + f ( z ) + !!

{ f ( z )}

the region R , the sequence of complex functions

If a series of continuous complex functions f k z

n

!

a region R of the complex plane, then for any given point z in

for the function series.

or !

()

If a complex function f n z is defined and single-valued in

!

(11.4-7)

contour C that lies within D the integral of the sum along C can be determined using term-by-term integration of the series:

175

n

lim

!

n→ ∞

n

∫ ∑ f ( z ) dz = lim ∑ ∫ C

k

n→ ∞

k=0

k=0

C

∞

()

fk z dz !

(11.4-9)

n

n

0

2

0

1

0

2

0

where an , z, z0 ∈! . This series is also known as a power series

}

If fk ( z ) is a sequence of holomorphic functions in a simply

in z − z0 . The point z0 is a fixed point in the complex plane, and is called the center of the series or the point of expansion of the series. The coefficients an are complex constants that are

connected domain D , and if the series:

generally dependent on the particular center of the series z0 .

∞

!

∑ a ( z − z ) = a + a ( z − z ) + a ( z − z ) !! (11.5-1) n=0

Proposition 11.4-2:

{

!

∑ f (z) !

(11.4-10)

k

k= 0

()

The coefficients of a power series for any given point of expansion z0 determine the series.

()

converges uniformly to f z in D , then f z is holomorphic

!

on D .

circle centered at z0 and having a radius z − z0 = r ≥ 0 such that

For every complex power series there is always a unique

the series converges absolutely at every point inside the circle, Proposition 11.4-3, Term-by-Term Differentiation:

()

and diverges at all points outside of the circle. This circle is converges

known as the circle of convergence or disk of convergence,

uniformly to the sum f z in a domain D , then the derivative

and its radius r is known as the radius of convergence. A

of f z at any point z0 in D can be determined by term-by-

power series may converge on some or none of the points that

term differentiation of the series.

are on its circle of convergence. Each power series can have

If a series of continuous complex functions fn z

()

()

only one radius of convergence.

11.5! POWER SERIES !

A complex power series is an infinite series of complex

functions having the form:

Proposition 11.5-1, Circle of Convergence of a Power Series: Every power series:

176

∞

!

∑a (z− z ) ! n

n

(11.5-2)

0

Proposition 11.5-4, Power Series is Holomorphic:

n=0

()

If a function f z can be represented by a power series:

has a circle of convergence z − z0 = r with radius given by: !

an ! r = lim n→ ∞ an+1

(11.5-3)

!

()

f z =

∞

∑

(

an z − z0

)n !

(11.5-5)

n=0

()

if this limit exists. Within the circle of convergence the power

then f z is holomorphic at every point inside its circle of

series will converge absolutely, and outside this circle it will

convergence z − z0 = r

diverge. Proposition 11.5-5, Derivatives of All Orders: Proposition 11.5-2, Uniform Convergence of a Power Series:

Derivatives of all orders exist for a power series:

()

Every function f z that can be represented by a power series: !

()

f z =

∞

∑ a (z − z ) ! n

n

0

!

()

f z =

∞

∑

(11.5-4)

!

k f ( )( z) =

∞

∑ ( n n!− k )! a ( z − z ) n

0

n−k

!

(11.5-7)

n= k

Proposition 11.5-3, Uniqueness of Power Series Representation:

()

(11.5-6)

at every point inside its circle of convergence so that:

converges uniformly inside its circle of convergence z − z0 = r .

f z within its circle of convergence z − z0 = r .

)n

n=0

n=0

A power series provides a unique representation of a function

(

an z − z0 !

11.6! TAYLOR SERIES !

()

If a function f z can be represented by a power series

having the form: 177

()

f z =

!

n f ( ) z0

( )

∞

∑

n!

n=0

!

( z − z0 )n !

(11.6-1)

coefficients of the series exist for all points within D . From Proposition 11.6-1 we know that the coefficients of a Taylor

Proposition 11.6-1, Taylor Series:

()

()

series are the derivatives of f z , and so derivatives of all can be represented by the power

series:

()

orders of f z

must exist at all points within D . From the

definition of a holomorphic function, we then see that a Taylor

()

f z =

!

()

complex function f z , we see that a necessary and sufficient condition for the Taylor series to exist on a domain D is that the

then the power series is known as a Taylor series.

If a complex function f z

From the definition of a Taylor series representation of a

∞

∑

(

)

an z − z0

n

()

series representation of a complex function f z will exist on a !

(11.6-2)

n=0

()

domain D if and only if f z is holomorphic on D .

■

inside its circle of convergence z − z0 = r , then the coefficients

!

of the series are given by:

series is known as an analytic function. A Taylor series is a

an =

!

n f ( ) z0

( )=

n!

1 2π i

( ) ds ! !∫C ( s − z0 )n+1 f s

A function that can be represented locally by a power

power series. If and only if a function is holomorphic on a (11.6-3)

domain D can the function be represented on D by a Taylor

()

series. Any function f z that is holomorphic on a domain D will be analytic on D . From Propositions 11.5-1 and 11.5-2 we

where n = 0, 1, 2, ! , and the series is a Taylor series.

know that the Taylor series will converge absolutely and Proposition 11.6-2, Existence of Taylor Series:

uniformly inside its circle of convergence. Within its circle of

()

A Taylor series representation of a complex function f z will

convergence, therefore, a complex function will be both

exist on a domain D if and only if f z is holomorphic on D .

holomorphic and analytic, and so these two designations can

()

be taken to be equivalent for our purposes.

Proof: 178

( )

the notation Dδ* z0 for an open punctured disk centered at a

11.7! LAURENT SERIES A point z = z0 in the complex plane is called a singular

!

( )

point z0 and having a radius of δ . The disk Dδ* z0 is defined

()

()

point or a singularity of a function f z if the function f z is

()

not holomorphic at z0 . A complex function

f z

by:

0 < z − z0 < δ !

!

is not

(11.7-1)

holomorphic at any of its singularities. If a point z0 is a

() () f ′ ( z0 ) will not exist.

singularity of a function f z , then f z discontinuous at z0 since

will always be

A point z0 is called an isolated singular point or an

!

isolated singularity of a function

()

()

f z

if

()

f z

is not

holomorphic at z0 , but f z is holomorphic throughout some neighborhood of z0 . We see then that, for an isolated singular

()

point z0 of a function f z , a δ neighborhood of z0 will always

()

exist in which there are no other singular points of f z .

()

If a complex function f z is holomorphic at a point z0 ,

!

()

then a Taylor series can be used to expand f z about the point

z0 . If the point z0 is an isolated singularity of a complex

()

function f z , then a Taylor series cannot be used to expand

()

f z about the point z0 . Instead we must use a different series known as a Laurent series. To describe the Laurent series we need to first define the concept of a punctured disk. !

An open δ -disk centered at a point z0 that does not

include the point z0 is called a punctured δ -disk. We will use

Figure 11.7-1! Graphical representation of the set of points in the open punctured disk centered at the point z0 and having a radius of δ . !

()

f z

A point z0 that is an isolated singular point of a function will have an open punctured δ -disk centered at z0 in

()

which f z is holomorphic. The inner radius of the punctured 179

()

disk in which f z

is holomorphic can be made arbitrarily

!

A Laurent series can include terms having negative

small (see Figure 11.7-1).

powers. That part of a Laurent series having only negative

!

powers of z − z0 is referred to as the principal part; the rest of

(

A Laurent series has the form:

()

∞

f z =

!

∞

∑ (z − z ) +∑ c (z − z ) ! bn

n

n=1

0

n

n

0

the series is referred to as the analytic part. The principal part of a Laurent series is given by:

(11.7-2)

n=0

!

bn =

cn =

1 2π i 1 2π i

( ) ds ! !∫C (s − z0 )− n +1 f s

( ) ds ! !∫ C (s − z0 )n +1 f s

()

∑ (z − z ) ! bn

n =1, 2, 3, ! !

( )

0

and the analytic part of a Laurent series is given by:

(11.7-3)

∞

()

f2 z =

!

n = 0, 1, 2, ! !

(11.7-5)

n

n =1

∑ c (z − z ) ! n

n

(11.7-6)

0

n=0

(11.7-4)

!

and where C is any simple closed contour lying within some

()

punctured disk Dδ* z0 which surrounds z0 , and in which f z

The Laurent series about z0 given in equation (11.7-2) can

be written in the form:

()

∞

f z =

!

is analytic. !

∞

f1 z =

!

where !

)

∑ a (z − z ) ! n

n

(11.7-7)

0

n=−∞

The Laurent series can be considered a generalization of

the Taylor series to expansions of functions about points at which the function is not analytic. The Laurent series can

()

therefore represent a function f z

that is analytic in an

annular domain D centered at an isolated singularity z0 , and defined by r < z − z0 < R , where 0 ≤ r < R ≤ ∞ .

where the coefficients an are: !

an =

1 2π i

( ) dz ! !∫C ( z − z0 )n+1 f z

n = 0, ± 1, + 2, + 3, ! ! (11.7-8)

and where C is any simple closed contour centered at z0 , and

( )

lying within some punctured disk Dδ* z0 which surrounds z0 ,

()

and in which f z is analytic. 180

()

∞

f z =

∑a (z− z ) ! n

11.8! SINGULARITIES

!

11.8.1! ISOLATED SINGULARITIES

which defines f z on the punctured disk Dδ* z0 . This series is

()

isolated singularity z0 of f z

( )

(11.8-1)

0

n=0

( )

()

()

The Laurent series expansion for a function f z about an

!

n

will converge in an open

()

punctured disk Dδ* z0 . Any isolated singularity of f z can be classified using the Laurent series expansion about the

()

singularity. An isolated singularity of f z at a point z0 can be

also a Taylor series that converges on the punctured disk

( ) f ( z ) at only the one point z0 so that: ! lim f ( z ) = f ( z ) = a !

Dδ* z0 . The singularity at z0 can be removed by redefining

0

z → z0

()

(11.8-2)

0

()

classified as either a removable singularity, a pole, or an

making f z then continuous at z0 . The function f z then

essential singularity according to the number of terms in the

becomes holomorphic on the disk Dδ z0 .

( )

principal part of its Laurent series expansion about z0 . 11.8.1.2! 11.8.1.1!

REMOVABLE SINGULARITY

()

!

!

If a complex function f z has an isolated singularity at a

()

point z0 , and if the Laurent series for f z about z0 has no principal part, then

()

f z

()

is said to have a removable

singularity at z0 . If f z has a removable singularity at z0 , then

( )

a finite lim f ( z ) exists, but this limit is not equal to f z0 . We z → z0

()

see, therefore, that f z is not continuous at z0 . !

()

()

The Laurent series expansion centered at z0 for f z when

f z has a removable singularity at z0 is:

POLE

()

If a function f z has an isolated singularity at a point z0

()

and, if the principal part of the Laurent series for f z about

z0 has a finite number of terms, then f ( z ) is said to have a

()

pole at the point z0 . If a pole exists at z0 for f z , we will have: !

()

lim f z = ∞ !

z → z0

(11.8-3)

()

and so the magnitude of the function f z increases without bound as z → z0 . Equation (11.8-3) holds independently of the

()

manner in which z → z0 . The Laurent series for f z about z0 is

181

( )

convergent in the punctured disk Dδ* z0 . Such a punctured disk always exists for a pole. !

()

If the principal part of the Laurent series for f z about

z = z0 can be written with k > 0 : a− k ≠ 0 !

!

a− ( k+1) = a− ( k+2) = ! = 0 !

(11.8-4)

()

()

f z =

∑

(

an z − z0

z − z0

(

)

(

)2

+ a0 + a1 z − z0 + a2 z − z0 +! !

(11.8-7)

11.8.1.3!

ESSENTIAL SINGULARITY

()

If a function f z has an isolated singularity at a point z0

()

and if the principal part of the Laurent series for f z about

)n !

(11.8-5)

n ≥ −k

z0 has an infinity of terms, then f ( z ) is said to have an

isolated essential singularity at the point z0 . For a function

()

f z with an essential singularity at z0 , the order of the pole

or !

a−1

and the pole is said to be a pole of order one or a simple pole.

!

so that the Laurent series for f z has the form: !

()

f z =

!

goes to infinity, and its Laurent series has the form:

()

f z =

a− k

( z − z0 )

k

+

a− ( k −1)

( z − z0 )

k−1

+!+

a−1 z − z0

(

)

+ a0 + a1 z − z0 + !

! ! (11.8-6) then z0 is said to be a pole of order k . Therefore the principal

()

part of the Laurent series expansion of f z is a polynomial of

(

degree k in z − z0

)

−1

. A pole can be seen to have an analytic

reciprocal that is zero. If z0 is a pole, then z = z0 makes the

()

! f z = !+ ! !

a−2

( z − z0 )2

+

a−1 z − z0

(

)

(

)2

+ a0 + a1 z − z0 + a2 z − z0 +! (11.8-8)

where an infinity of the coefficients an with n < 0 are nonzero, and where the series is convergent in the punctured disk

( )

Dδ* z0 .

denominators of the principal part of the Laurent series equal

11.8.1.4!

to zero. A pole can be real or complex.

!

!

singularity at a point z0 , a Laurent series expansion of f z

If the principal part of a Laurent series has exactly one

term, then the series can be written as:

SINGULARITY CLASSIFICATION

As noted above, if a function

()

f z

has an isolated

()

about z0 can be used to classify this isolated singularity: 182

!

Removable singularity – Laurent series has no

purpose of a branch cut is to demarcate branches, with each

!

!

edge of the curve belonging to different branches. The branch

!

Pole – principle part of Laurent series has a finite

principle part.

!

number of terms.

!

cut acts to transform a complex function that is multivalued into one that is single-valued and continuous within each of its branches.

Essential singularity – principle part of Laurent series

!

has an infinity of terms.

11.8.2! NONISOLATED SINGULARITIES

()

11.8.2.2! !

BRANCH POINTS

A point z0 is designated a branch point for a complex

()

()

multivalued function f z if f z does not return to its initial

If w = f z is a multivalued function of a complex variable

value after once traversing a closed circle of arbitrarily small

z , then more than one value of w will correspond to some

radius centered at z0 . Therefore a multivalued function will

values of z in the domain of f z . A number of points in the w-

never be single-valued in any δ neighborhood Dδ z0

plane can then be the images of a single point z0 in the z-plane.

branch point z0 , where z − z0 < δ . If a closed contour does not

!

encircle a branch point of a multivalued function, that function

!

()

If we consider a restricted part of the range of a

( )

of a

multivalued function in which the function assigns just one

will be single-valued when traversing such a contour.

value w for each z , and in which the function is continuous,

!

we refer to this part of the range as a branch of the multivalued

(and only two) branch points of the function. The location of

function. Within any given branch of a multivalued function,

any branch point depends on the function, and is independent

the function will be single-valued and continuous.

of the chosen branch cut. Therefore the location of branch

Every branch cut of a multivalued function will join two

points is unique for a given multivalued function. Branch 11.8.2.1! !

BRANCH CUTS

A branch cut of a function is a curve in the complex plane

across which the multivalued function is discontinuous. The

points of a function can be located at infinity. !

Branch points and all points on a branch cut are examples

of nonisolated singular points. Even the smallest circuit around 183

a branch point of a function will cross a branch cut, and so every neighborhood of a branch point will contain at least one singularity other than the singularity at the branch point. Since any δ neighborhood of z0 will contain other singularities of

!

C

)

(11.9-2)

We therefore have:

() () * Dδ ( z0 ) by a Laurent series centered at z0 . Some functions that

!

have nonisolated singularities have an infinity of poles that

or

converge to a limit point z0 .

!

f z , it is not possible to represent f z in a punctured disk

!∫ (

⎧⎪ 2 π i for n = −1 n z − z0 dz = ⎨ ! 0 for n ≠ −1 ⎩⎪

!∫

C

()

f z dz = 2 π i a−1 !

a−1 =

1 2π i

!∫

(11.9-3)

()

f z dz !

C

(11.9-4)

()

11.9! RESIDUES

Obviously if a−1 is known for the function f z , then the

!

integral of the function f z over the closed contour C can be

One of the coefficients in a Laurent series expansion of an

()

analytic function plays a key role both in evaluating integrals of

obtained immediately from equation (11.9-3).

the function, and in determining the location of isolated poles

!

of the function. This coefficient is known as the residue of the

expansion of f z about the singularity z0 is the only nonzero

function.

term left following term-by-term integration of the Laurent

!

()

If an analytic function f z has an isolated singularity at a

The term with coefficient a−1 in the Laurent series

()

series over the contour C . For this reason a−1 is called the

()

point z0 , then integrating its Laurent expansion term-by-term

residue of the function f z for the singularity z0 .

over a simple closed contour C encircling z0 , we can write:

!

!

!∫

C

()

f z dz =

∞

∑ a !∫ ( z − z ) dz n

n

n=−∞

C

0

Using Proposition 11.3-10 we obtain:

()

We will designate the residue of a function f z

at a

singularity z0 by: !

(11.9-1)

!

()

a−1 = Res ⎡⎣ f z , z0 ⎤⎦ !

(11.9-5)

Equation (11.9-3) can then be written:

184

!∫

!

C

!

()

()

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

(11.9-6)

If the Laurent series expansion for an analytic function

(

) + ( z − z ) ⎡⎣ a

Multiplying by z − z0 we have:

( z − z0 ) f ( z ) = a−1

!

0

0

(

)

+ a1 z − z0 +! ⎤⎦ !

f z about a singularity z0 can be found, then the coefficient

Taking the limit as z → z0 , we obtain a−1 :

a−1 of the Laurent series can be determined by inspection, and

!

()

the residue of the function f z at the singularity z0 obtained.

!

) ()

()

If an analytic function f z has the form:

()

function f z can be quite difficult. Fortunately there are some !

first having to find other terms of the Laurent series expansion.

()

f z =

()

P z

=

A

+

()

()

(

) ()

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ ! z → z0

Proof: !

()

(11.9-7)

()

Since z0 is a simple pole of f z , the Laurent series for

()

c are the poles of f z , then:

()

( ) ! ( a − b) ( a − c )

(11.9-12)

()

() ! ( b − a )( b − c )

(11.9-13)

()

() ! ( c − a ) ( c − b)

(11.9-14)

!

A = Res ⎡⎣ f z , a ⎤⎦ =

!

B = Res ⎡⎣ f z , b ⎤⎦ =

!

C = Res ⎡⎣ f z , c ⎤⎦ =

f z is: !

()

f z =

a−1 z − z0

(

C

where P z is a polynomial of degree ≤ 2 , and where a , b , and

If an analytic function f z has a simple pole at a point z = z0 , !

+

(11.9-11)

()

Proposition 11.9-1:

B

( z − a ) ( z − b) ( z − c ) ( z − a ) ( z − b) ( z − c )

!!

then:

(11.9-10)

Proposition 11.9-2:

Nevertheless, finding the Laurent series expansion of a

other ways of determining the single coefficient a−1 without

(

■

It is often necessary to calculate only a few terms of a Laurent series expansion to determine the term with coefficient a−1 .

()

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ = a−1 ! z → z0

(11.9-9)

)

(

+ a0 + a1 z − z0 + a2 z − z0

)

2

+!!

(11.9-8)

P a

P b

P c

185

Proof: !

Representing the terms in equation (11.9-11) as a Laurent

series about the point z = a , we have:

A

!

()

(

!

(11.9-19)

() ! ( b − a )( b − c ) P(c) ! C = Res ⎡⎣ f ( z ) , c ⎤⎦ = ( c − a ) ( c − b)

(11.9-15)

()

B = Res ⎡⎣ f z , b ⎤⎦ =

!

−B

∞

∑(

1 B = =− b − a 1− z − a z− b n=0 b − a b− a

) (

)

)

n+1

( z − a )n !

(11.9-16)

C

=

( z − c) (

−C

∞

∑(

1 C =− c − a 1− z − a n=0 c − a c−a

)

)

!

n ! z − a ( ) n+1

()

() ()

()

!f z =

( z − a)

−

∑ n=0

⎡ B ⎢ ⎢ b− a ⎣

(

⎤ ⎥ z − a n! + n+1 n+1 ⎥ c−a ⎦ C

)

and so using Proposition 11.9-1:

(

)

(

)

()

( )

both analytic at a point z = z0 . If p z0 ≠ 0 and

() q ( z ) has a

()

simple zero at z0 , then z0 is a simple pole of f z , and we have: !

()

Res ⎡⎣ f z , z0 ⎤⎦ =

(11.3-11) becomes: ∞

(11.9-21)

Let f z = p z q z where the functions p z and q z are

(11.9-17)

Using equations (11.9-15), (11.9-16), and (11.9-17), equation

A

(11.9-20)

Proposition 11.9-3:

which is a Taylor series analytic at the point z = a . !

P b

■

which is a Taylor series analytic at the point z = a , and: !

(

!!

which is a single term Laurent series. We also have: !

() )( )

Using a similar derivation for the other singularities, we obtain:

(z − a) B

() )( ) (

P z P a A = Res ⎡⎣ f z , a ⎤⎦ = lim = z→a z − b z − c a−b a−c

!

(11.9-18)

( )! q′ ( z0 ) p z0

(11.9-22)

Proposition 11.9-4:

()

()

()

Let f z = 1 q z where the function q z is analytic at a point

z = z0 . If q ( z ) has a simple zero at z0 , then z0 is a simple pole of

()

f z and we have: 186

!

()

Res ⎡⎣ f z , z0 ⎤⎦ =

1

( )

q′ z0

!

(11.9-23)

C

()

(11.10-1)

()

within D and enclosing z0 .

If an analytic function f z has a pole of order k ≥ 2 at a point

z = z0 , then:

Proof:

d n−1 ⎡ ! Res ⎡⎣ f z , z0 ⎤⎦ = lim z − z0 ⎢ z → z0 n − 1 ! dz n−1 ⎣

(

1

)

(

) () n

f z ⎤ ! (11.9-24) ⎦⎥

11.10! RESIDUE INTEGRATION !

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

where C is any positively oriented simple closed contour lying

Proposition 11.9-5:

()

!∫

!

Rather than the singularities of an analytic function

increasing the difficulty in evaluating integrals of the function, these singularities can actually aid in this evaluation. Residues provide a method for evaluating integrals of functions that are analytic inside a closed contour except for some isolated singularities. It is the residues of the integrand at singular

!

()

Since the function f z

can enclose z0 with a simple circular contour C . From equation (11.7-8) we can then write: !

1 2π i

a−1 =

( ) dz = 1 f z dz ! !∫C ( z − z0 )−1+1 2π i !∫C ( ) f z

!

!∫

C

()

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

Proposition 11.10-1, Cauchy’s Residue Theorem for an

!

an isolated singularity at a point z = z0 , then:

(11.10-2)

or

as given in equation (11.9-6).

()

at

point z0 using the Laurent series given in equation (11.7-7). We

contour integral.

If a complex function f z is analytic in a domain D except at

()

isolated singularity at a point z0 , we can represent f z

points within the contour that determine the value of the

Isolated Singularity:

is analytic in D except at an

(11.10-3)

■

The implications of this equation are rather profound and

far-reaching. If we are able by any method to determine the

()

coefficient a−1 of the Laurent series of a complex function f z

having an isolated singularity z0 , then we can evaluate the 187

()

integral of f z using a closed contour around z0 . This type of integration is known as residue integration, and it can be considered an extension of the method of integration using Cauchy’s integral formula. Residue integration can often make possible the evaluation of integrals even when an explicit antiderivative does not exist. Proposition 11.10-2, Cauchy’s Residue Theorem: Let C be a simple closed contour that lies within a simply

()

connected domain D . If a complex function f z is analytic in

D except at a finite number of isolated singular points z1 , z2 , ! , zk that lie inside C , then: !

!∫

C

()

f z dz = 2 π i

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ ! n

(11.10-4)

n =1

188

Chapter 12 Inverse Laplace Transforms from Residue Integration

1 f (t ) = 2πi

∫

σ0 + i ∞

F ( s ) e s t ds

σ0 − i ∞

189

In this chapter we will consider the inversion of the

F ( s ) . Once having obtained F ( s ) , however, there is then no

Laplace transform using residue integration. This is a general

longer a requirement that Re s > σ 0 , and so we can use the

!

inversion technique for the Laplace transform, and it is the

analytic properties of the function F ( s ) for residue integration

inversion method often used when a Laplace transform cannot

wherever F ( s ) may be analytic in the s-plane (essentially by

be found in a table.

means of analytic continuation).

12.1!

12.2!

!

INVERSE LAPLACE TRANSFORM

As given in equations (6.2-3) and (6.2-4), the direct Laplace Proposition 12.2-1:

transform is: !

F (s) = L

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt !

f (t ) = L

{F ( s )} = 21π i

−1

∫

The inverse Laplace transform of the unit step function is:

(12.1-1)

σ0 + i ∞

F ( s ) es t ds !

(12.1-2)

σ0 − i ∞

form a direct and an inverse pair of transforms. We will also present examples of the use of the inverse Laplace transform for certain elementary functions. We will begin by considering the inverse Laplace transforms of the unit step function and of the unit impulse function. The requirement that Re s > σ 0 is necessary to ensure

convergence of the Laplace integral from which we obtain

−1⎧

1⎫ 1 ⎨ ⎬= ⎩ s ⎭ 2πi

∫

σ 0 +i ∞

σ 0 −i ∞

es t ds ! s

(12.2-1)

Proof: !

We will now show that these two integral transforms do indeed

!

U (t ) = L

!

and the inverse Laplace transform is: !

INVERSE LAPLACE TRANSFORM OF A UNIT STEP FUNCTION

!

The Laplace transform of the unit step function is:

L {U ( t )} =

1 = F ( s)! s

(12.2-2)

as given in equation (5.1-10). We will now show that the inverse Laplace transform of the unit step function F ( s ) = 1 s is: !

1 U (t ) = 2πi

∫

σ0 + i ∞

1 F ( s ) e dss = 2πi σ0 − i ∞ st

∫

σ 0 +i ∞

σ 0 −i ∞

es t ds ! (12.2-3) s

The integration path for this integral is shown in Figure 12.2-1. 190

is comprised of two curves: C L and CR. which together enclose the pole at s = 0 (see Figure 12.2-2).

Figure 12.2-1! Integration path for the integral in equation (12.2-1). !

We can use residue integration to evaluate the integral in

equation (12.2-3). The only singularity of es t s is a simple pole located at the origin. Residue integration can be performed by constructing a closed contour C that includes part of the

Figure 12.2-2! Contour C = C L ∪ C R for the inverse Laplace transform of the unit step function.

integration path for equation (12.2-3) together with a semicircle of radius R centered at ( σ 0 , 0 ) and enclosing the pole at the

origin. The contour C is known as the Bromwich contour and

!

We can now write the integral in equation (12.2-3) as a

contour integral over C , where C consists of C = C L ∪ C R : 191

!∫

!

C

es t ds = s

∫

es t ds + CL s

∫

es t ds ! CR s

(12.2-4)

or using Proposition 11.3-3: iφ

Such a contour integral is known as a Bromwich contour

∫

e ds ≤ CR s

!

integral. We will evaluate each of the integrals on the right side of equation (12.2-4) as the semicircle CR radius R → ∞ . !

∫

e ds = CL s

!

∫

σ 0 +i R

σ 0 −i R

∫

es t ds ≤ eσ 0 t CR s

!

st

e ds ! s

(12.2-5)

Path C L is the line extending from σ 0 − i R to σ 0 + i R where the constant σ 0 is the abscissa of absolute convergence. As R → ∞

∫

es t ds = CL s

! !

σ0 − i ∞

es t ds ! s

(12.2-6)

To evaluate the integral along the path CR , we will let

s = σ 0 + R ei φ so that ds = i R ei φ dφ . We then have: !

∫

es t ds = CR s

iφ

∫

CR

eσ 0 t e Rt e i R ei φ dφ ! σ 0 + i R ei φ

(12.2-7)

∫

es t ds ≤ eσ 0 t CR s

!

∫

es t ds ≤ eσ 0 t CR s

!

∫

iφ

e Rt e dφ !

(12.2-10)

CR

∫

3π 2

e Rt cosφ ei Rt sin φ dφ !

π 2

(12.2-11)

∫

3π 2

e Rt cosφ dφ !

(12.2-12)

π 2

Letting φ = θ + π 2 :

∫

es t ds ≤ eσ 0 t CR s

! !

and so: !

(12.2-9)

and so:

σ0 + i ∞

∫

i R ei φ dφ !

We can integrate over the range π 2 ≤ φ ≤ 3π 2 :

the path C L becomes the integration path of the integral as given in equation 12.2-3:

σ0 + i R e

CR

iφ

For R >> σ 0 we have:

For the integral along path C L we have: st

eσ 0 t e Rt e

∫

st

!

∫

π

0

e

Rt cos( θ + π 2 )

dθ = 2 e

σ0 t

∫

π 2

e− Rt sin θ dθ !

0

(12.2-13)

where we have used the fact that e− Rt sinθ is symmetric about

∫

es t ds = CR s

∫

CR

iφ

eσ 0 t e Rt e iφ ! i φ i R e dφ σ0 + i R e

(12.2-8)

θ = π 2 . Over 0 ≤ θ ≤ π 2 we have sin θ greater than the straight

line connecting the end-points of this range (see Figure 12.2-3): 192

sin θ ≥

!

θ 2θ = ≥ 0! π π 2

(12.2-14)

This is known as Jordan’s inequality. Therefore we can write: !

∫

es t ds ≤ 2 eσ 0 t CR s

∫

π 2

0

e− Rt 2θ π dθ =

π eσ0 t 1− e− Rt ! (12.2-15) Rt

(

)

As R → ∞ we have: !

∫

es t ds → 0 ! CR s

!

From equations (12.2-6), (12.2-16), and (12.2-4) we then

have:

∫

!

σ0 + i ∞

σ0 − i ∞

es t ds = s

!∫

C

es t ds ! s

(12.2-17)

Since the integrand es t s has a simple pole at s = 0 , from Proposition 11.9-1 the residue of the integrand is: ⎡ est ⎤ ⎡ est ⎤ Res ⎢ , 0 ⎥ = lim ⎢ s ⎥ = lim est = 1! ⎣ s ⎦ s→ 0 ⎣ s ⎦ s→ 0

! (12.2-16) !

(12.2-18)

Using Cauchy’s residue theorem given in Proposition

11.10-1, we can now evaluate the contour integral in equation (12.2-17): ⎡ est ⎤ est ds = 2 π i Res ⎢ , 0 ⎥ = 2 π i ! C s ⎣ s ⎦

!∫

!

(12.2-19)

Therefore we have:

∫

!

σ0 + i ∞

σ0 − i ∞

es t ds = 2 π i ! s

(12.2-20)

and so the inverse Laplace transform of the unit step function is: Figure 12.2-3!

Straight line connecting the end-points of sin θ in the range 0 ≤ θ ≤ π 2 .

1 2πi

!

∫

σ0 + i ∞

σ0 − i ∞

es t 1 ds = ( 2 π i ) = 1 = U (t ) ! s 2πi

t > 0!

(12.2-21)

■

193

12.3! Proposition 12.2-2: The inverse Laplace transform of the shifted unit step function is: !

U (t − t0 ) = L

−1⎧ e

⎫ 1 ⎨ ⎬= ⎩ s ⎭ 2πi − s t0

∫

σ0 + i ∞

σ0 − i ∞

! !

Proposition 12.3-1: The inverse Laplace transform for the shifted unit impulse

e− s t0 s t e ds ! (12.2-22) s

function is: δ (t − t0 ) = L

!

Proof:

1 U ( τ − τ0 ) = 2πi

∫

σ0 + i ∞

e

s ( τ−τ 0 )

− s t0

1 = 2πi

∫

σ0 + i ∞

e− s t0 es t ds !

σ0 − i ∞

(12.3-1)

σ0 − i ∞

Proof:

1 ds ! s

(12.2-23)

!

Since the shifted unit impulse function δ ( t − t 0 ) is equal to

the derivative of the shifted unit step function: !

shifted unit step function is:

1 U (t − t0 ) = 2πi

∫

σ0 + i ∞

σ0 − i ∞

e− s t0 s t e ds = L s

−1⎧ e

⎫ ⎨ ⎬! ⎩ s ⎭ − s t0

(12.2-24)

where the Laplace transform of the shifted unit step function is: !

{e }

−1

Letting t = τ − τ 0 we can write equation (12.2-1) as:

or changing notation, the inverse Laplace transform of the

!

INVERSE LAPLACE TRANSFORM OF THE UNIT IMPULSE FUNCTION

L {U ( t − t 0 )} =

e− s t0 = F (s) ! s

as given in Proposition 5.1-1.

■

(12.2-25)

δ (t − t0 ) =

d U (t − t0 ) ! dt

(12.3-2)

we have from Proposition 12.2-2: !

δ (t − t0 ) =

d ⎡ 1 ⎢ dt ⎣ 2 π i

∫

σ0 + i ∞

σ0 − i ∞

⎤ e− s t0 s t e ds ⎥ ! s ⎦

(12.3-3)

Since the exponential function is continuous, and since s and t are independent, we can move the derivative inside the integral: !

1 δ (t − t0 ) = 2πi

∫

σ0 + i ∞

e− s t0 es t ds !

σ0 − i ∞

(12.3-4)

194

If f ( t ) is a piecewise continuous function of exponential order

and so the inverse Laplace transform of the shifted unit

over [ 0, ∞ ) with L

impulse function is:

{ e } = 21π i ∫

δ (t − t0 ) = L

!

−1

− s t0

σ0 + i ∞

e− s t0 es t ds !

σ0 − i ∞

transform exists where Re s > σ 0 , and is given by:

(12.3-5)

where the Laplace transform of the shifted unit impulse

{ f (t )} = F ( s ) , then the inverse Laplace

{F ( s )} =

f (t ) = L

!

−1

1 2πi

∫

σ0 + i ∞

F ( s ) es t ds !

σ0 − i ∞

(12.4-1)

function is:

L {δ ( t − t 0 )} = e− s t0 = F ( s ) !

!

as given in Proposition 5.4-1. !

(12.3-6)

■

the inverse Laplace transforms of unit step functions, shifted unit step functions, and unit impulse functions all have the form known as the Bromwich integral: !

∫

σ0 + i ∞

F ( s ) es t dss !

σ0 − i ∞

(12.3-7)

We will now show that the inverse Laplace transforms of functions of exponential order also have the same form.

12.4!

!

Since the function f ( t ) is taken to be of exponential order,

we have:

We see from equations (12.2-21), (12.2-24), and (12.3-5) that

1 f (t ) = 2πi

Proof:

INVERSE LAPLACE TRANSFORM OF A FUNCTION OF EXPONENTIAL ORDER

Proposition 12.4-1, Existence of the Inverse Laplace Transform:

f ( t ) ≤ M eσ0 t !

!

t ≥ T0 !

(12.4-2)

where M , σ 0 , and T0 are all non-negative real constants. The Laplace transform:

F (s) =

!

∫

∞

0

−

f ( t ) e− s t dt = L

{ f (t )}!

(12.4-3)

will then exist (see Section 1.4). All the singularities will fall to the left of the line segment specified by s = σ 0 . !

Multiplying equation (12.4-3) by es t 2 π i and integrating,

we obtain: ! ! !

1 2πi

∫

σ0 + i ∞

1 F ( s ) e ds = 2πi σ0 − i ∞ st

∫

σ0 + i ∞

e

σ0 − i ∞

st

∫

∞

0−

f ( τ ) e− sτ dτ ds

(12.4-4) 195

!

Since the integrals are uniformly convergent (as shown in

Proposition 1.4-7), we can change the order of integration: !

1 2πi

σ0 + i ∞

∫

F ( s ) es t ds =

σ0 − i ∞

∞

∫

0

−

1 2πi

f (τ)

∫

σ0 + i ∞

es t e− s τ ds dτ

σ0 − i ∞

! !

(12.4-5)

Using Proposition 12.3-1 we know that the inner integral exists and is given by: δ (t − τ ) =

!

1 2πi

∫

σ0 + i ∞

e− s t0 es t ds !

(12.4-6)

σ0 − i ∞

1 2πi

∫

σ0 + i ∞

F ( s ) es t ds =

σ0 − i ∞

∫

∞

0−

f (t ) = L

{F ( s )} =

−1

if

The Laplace transform will be finite, by definition, in its

region of convergence. Any singularities in the Laplace transform must therefore occur in the s-plain to the left of a vertical line through the abscissa of simple convergence.

12.5.1! FINITE NUMBER OF POLES

f ( τ ) δ ( t − τ ) dτ = f ( t ) !

If all singularities of the Laplace transform are in the form

with residue integration to determine the inverse Laplace (12.4-7)

transform. The contour will include a segment of the vertical line Re s = σ 0 and a semicircle that encloses all the poles.

1 2πi

∫

σ0 + i ∞

F ( s ) es t ds !

(12.4-8)

Proposition 12.5-1: !If f ( t ) is a piecewise continuous function of exponential order,

σ0 − i ∞

over [ 0, ∞ ) with L

and so the inverse Laplace transform of f ( t ) exists where Re s > σ 0

RESIDUE INTEGRATION OF THE INVERSE LAPLACE TRANSFORM

of a finite number of poles, a Bromwich contour can be used

Therefore: !

!

!

and so equation (12.4-5) becomes: !

12.5!

transform:

f ( t ) is a piecewise continuous function of

exponential order over [ 0, ∞ ) .

■

{ f (t )} = F ( s ) , then the inverse Laplace

!

f (t ) = L

{F ( s )} =

−1

1 2πi

∫

σ0 + i ∞

F ( s ) es t ds !

σ0 − i ∞

(12.5-1)

is given by:

196

f (t ) =

!

N

∑ n =1

Res ⎡⎣F ( s ) es t , sn ⎤⎦ !

(12.5-2)

where sn are the poles of F ( s ) e , and where Re s > σ 0 . st

constant Re s > σ 0 is the abscissa of absolute convergence. As

R → ∞ we have the desired integral given in equation (12.5-1). !

To evaluate the integral along the path C R , we will let

s = σ 0 + R ei φ so that ds = i R ei φ dφ . We then have:

Proof: !

Path C L is the line extending from σ 0 − i R to σ 0 + i R where the

We will use the semicircular contour C shown in Figure

12.2-2 to evaluate the integral in equation (12.5-1). The contour

∫

!

CR

C is composed of two curves: a segment C L of the vertical line

and so:

the Laplace transform F ( s ) has poles sn where n = 1, 2, !, N ,

!

Re s = σ 0 and a semicircle C R of radius R centered at ( σ0 , 0 ) . If

then R is taken to be large enough so that the contour C encloses all these poles. !

We can now write the integral in equation (12.5-1) as a

contour integral over C , where C consists of C = C L ∪ C R : !

!∫

F ( s ) es t ds =

C

∫

F ( s ) es tds +

CL

∫

F ( s ) es tds !

(12.5-3)

CR

We will evaluate each of the integrals on the right side of

F ( s ) es tds =

∫

∫

iφ

0

CR

F ( s ) es t ds ≤ e σ 0 t

CR

!

( σ +R e ) t F

e

∫

e Rt e

(σ iφ

0+

)

R ei φ i R ei φ dφ ! (12.5-5)

(

F σ 0 + R ei φ

CR

)

i R ei φ dφ

!

(12.5-6)

We have:

∫

!

F ( s ) es t ds ≤ R e σ 0 t

CR

!

∫

CR

!

e Rt e

iφ

(

F σ 0 + R ei φ

)

i ei φ dφ

(12.5-7)

For a function f ( t ) of exponential order, the Laplace transform

equation (12.5-3) as the radius R → ∞ .

F ( s ) is bounded such that F ( s ) ≤ MR where MR → 0 as R → ∞ .

!

We then have:

!

For the integral along path C L we have:

∫

F ( s ) e ds =

CL

st

∫

σ 0 +i R

σ 0 −i R

F ( s ) es t ds !

(12.5-4)

!

(

F σ 0 + R ei φ

)

≤ MR !

(12.5-8)

and equation (12.5-7) becomes: 197

∫

!

F ( s ) es t ds ≤ MR R e σ 0 t

CR

∫

e Rt cosφ ei Rt sin φ dφ !

(12.5-9)

CR

∫

F ( s ) e ds ≤ MR R e st

∫

σ0 t

3π 2

e

Rt cosφ

e

i Rt sin φ

! !

dφ

(12.5-10)

or

∫

F ( s ) es t ds ≤ MR R eσ 0 t

CR

e Rt cosφ dφ !

π 2

∫

F ( s ) es t ds ≤ MR R eσ 0 t

CR

∫

e Rt cos (θ + π 2 ) dθ !

(12.5-11)

(12.5-12)

or

∫

∫

CR

∫

e− Rt sin θ dθ !

(12.5-13)

0

where we have used the fact that e

)

(12.5-15)

es t F ( s ) ds → 0 !

(12.5-16)

∫

σ 0 +i R

F ( s ) es t ds =

!∫ F ( s ) e ds ! st

(12.5-17)

C

and from equations (12.5-17) and (12.4-8) we have: f (t ) =

∫

1 2πi

σ 0 +i R

σ 0 −i R

F ( s ) es t ds =

1 2πi

!∫ F ( s ) e ds ! st

(12.5-18)

C

Finally from Cauchy’s residue theorem (Proposition 11.10-2) we can write:

− Rt sinθ

is symmetric about

θ = π 2 . From Jordan’s inequality given in equation (12.2-14)

we have:

(

From equations (12.5-3), (12.5-4), and (12.5-16) we then have:

! π 2

π MR eσ 0 t 1− e− Rt ! t

Since MR → 0 as R → ∞ , we have:

!

σ 0 −i R

0

F ( s ) es t ds ≤ 2 MR R eσ 0 t

es t F ( s ) ds ≤

CR

! π

(12.5-14)

CR

Letting φ = θ + π 2 :

!

∫

!

3π 2

∫

e− Rt 2θ π dθ !

0

!

!

∫

π 2

or

π 2

CR

!

F ( s ) es t ds ≤ 2 MR R eσ 0 t

CR

We can integrate over the range π 2 ≤ φ ≤ 3π 2 : !

∫

!

1 ! f (t ) = 2πi

∫

σ0 + i ∞

F ( s ) e ds =

σ0 − i ∞

st

N

∑ Res ⎡⎣ F ( s ) e , s ⎤⎦ ! st

n

(12.5-19)

n =1

■

198

and so:

Proposition 12.5-2: !If f ( t ) is a piecewise continuous function of exponential order

!

f (t ) = L

with Laplace transform F ( s ) , and if F ( s ) es t has only N simple poles sn , then the inverse Laplace transform L f (t ) =

!

{F ( s )} is:

n =1

s → sn

n

st

⎤⎦ !

(12.5-20)

Proof: !

Follows from Propositions 12.5-1 and 12.4-1.

The Laplace transform has simple poles at s = 0 and s = − a .

1 . s+a

Solution:

!

⎡ ⎤ ⎡ ⎤ 1 1 f ( t ) = Res ⎢ es t , 0 ⎥ + Res ⎢ es t , − a ⎥ ⎣ s ( s + a) ⎦ ⎣ s ( s + a) ⎦

Using Proposition 12.3-1:

The Laplace transform has a simple pole at s = − a . From Propositions 12.5-1 and 12.5-2 we then have: ⎡ 1 st ⎤ f ( t ) = Res ⎢ e , − a⎥ ⎣ s+a ⎦

!

⎡ 1 f t = lim ⎢ s est s→ 0 ⎢⎣ s s + a

()

(

)

⎤ ⎥ + lim ⎥⎦ s→ − a

⎡ 1 est ⎢ s+a s s+a ⎢⎣

(

)

(

)

⎤ ⎥ ⎥⎦

and so:

Using Proposition 11.9-1:

!

⎡ 1 st ⎤ ⎡ 1 st ⎤ f t = Res ⎢ e , − a ⎥ = lim ⎢ s + a e ⎥ = e− a t s→ − a + a + a s s ⎣ ⎦ ⎣ ⎦

()

Determine the inverse Laplace transform f ( t ) of 1 . F (s) = s ( s + a)

From Propositions 12.5-1 and 12.5-2 we then have:

Determine the inverse Laplace transform f ( t ) of F ( s ) =

!

Example 12.5-2

Solution: ■

Example 12.5-1

!

1 ⎫ − at ⎨ ⎬=e ⎩s + a⎭

−1

N

∑ lim ⎡⎣ ( s − s ) F ( s ) e

−1⎧

(

)

f (t ) = L

−1⎧

1 ⎫ 1 − at ⎨ ⎬ = 1− e ⎩ s ( s + a) ⎭ a

(

)

199

!

Example 12.5-3 Determine the inverse Laplace transform f ( t ) of F ( s ) =

1 . s2

⎧⎪ ⎡ ⎤ 1 f ( t ) = Res ⎨ lim ⎢ ( s − i a ) es t ⎥ ( s − i a )( s + i a ) ⎦ ⎪⎩ s → i a ⎣ ⎡ ⎤ ⎫⎪ 1 lim ⎢ ( s + i a ) es t ⎥ ⎬ s→ −ia ( s − i a )( s + i a ) ⎦ ⎪⎭ ⎣

!

Solution: The Laplace transform has a pole of order 2 at s = 0 . From

or

Proposition 12.5-1 we then have: !

!

⎡ 1 ⎤ f ( t ) = Res ⎢ 2 es t, 0 ⎥ ⎣s ⎦

!

()

( s − 0)2 12 est s

⎤ ⎥ ⎦

1

( )

!

If the residue theorem is used to integrate a multivalued

function, it is important that the contour for the integral not enclose any branch point nor cross any branch cut of the

or !

⎫ e isa e− isa 1 e isa − e− isa 1 = − = = sin at ⎨ 2 2⎬ 2i a ⎩ s + a ⎭ 2i a 2i a a

−1⎧

12.5.2! BRANCH CUTS

Using Proposition 11.9-5: ⎡ 1 st ⎤ d 2−1 ⎡ f t = Res ⎢ 2 e , 0 ⎥ = lim 2−1 ⎢ ⎣s ⎦ s→ 0 ds ⎣

()

f t = L

()

f t = L

−1⎧

1⎫ d st lim e = lim t es t = t ⎨ 2 ⎬ = s→ 0 ds s→ 0 ⎩s ⎭

Example 12.5-4

1 Determine the inverse Laplace transform of F ( s ) = 2 . s + a2 Solution: The Laplace transform has simple poles at s ± i a . From Propositions 12.5-1 and 12.5-2 we then have:

function. If a Laplace transform F ( s ) is multivalued, then the

branch point and the branch cut for F ( s ) must be taken into account when constructing the integration contour used to calculate the inverse Laplace transform of F ( s ) . Both the branch point and the branch cut must be outside the contour, while any poles of F ( s ) included in the residue summation will be inside the contour. !

A branch for F ( s ) must first be selected in order to

perform the integration. Generally the branch selected and the 200

contour chosen for evaluating the integral will be dependent

Example 12.5-5

upon the particular Laplace transform integral.

Determine the inverse Laplace transform of F ( s ) =

!

An example of the determination of an inverse Laplace

transform when F ( s ) is multivalued is given in Example 12.5-5 for F ( s ) = 1

1 . s

Solution: The Laplace transform F ( s ) is multivalued (has two values)

s.

with a branch point at s = 0 . We will use the contour C shown in Figure 12.5-1 to determine the inverse Laplace transform: !

f (t ) = L

{F ( s )} =

−1

1 2πi

∫

σ0 + i ∞

σ0 − i ∞

1 st e ds s

The contour C has a branch cut along the negative real axis and is composed of the contours: !

C = C L ∪ C1 ∪ C2 ∪ L1 ∪ Cr ∪ L2 ∪ C3 ∪ C4 where Cr is the arc of a circle of radius r centered at ( 0, 0 ) , and C1 , C2 , C3 and C4 are arcs of a circle of radius R centered at

( 0, 0 ) .

Cauchy’s residue theorem (Proposition

11.10-2) for a function F ( s ) on the contour C shown in Figure 12.5-1 can be symbolically expressed as: ! Figure 12.5-1! Contour C = C L ∪ C1 ∪ C2 ∪ L1 ∪ Cr ∪ L2 ∪ C3 ∪ C4 around a branch point and a branch cut in the splane for F ( s ) = 1 s .

!∫ = ∫ + ∫ + ∫ + ∫ + ∫ + ∫ + ∫ + ∫ C

CL

C1

C2

L1

Cr

L2

C3

C4

k

!

= 2π i

∑ Res ⎡⎣ F (s) e , s ⎤⎦ st

j

j =1

201

Since f ( t ) is a function of exponential order, the Laplace

where sj are any poles of F ( s ) within the contour C . Since F (s) = 1

transform F ( s ) is bounded as R → ∞ . If we also have as

s has no poles within C , we have from the

R→∞ :

Cauchy-Goursat theorem (Proposition 11.3-5):

!∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ =

!

C

+

CL

+

+

C1

+

C2

+

L1

+

Cr

L2

+

C3

=0

(

C4

where M > 0 and m > 1 , then we can write:

We will now evaluate each of these integrals separately. ! !

∫

!

CL

σ0 + i ∞

∫

σ0 − i ∞

∫

1 st e ds ≤ M R − m s

∫

1 st e ds ≤ M R − m+1 s

C1

Along the contour C L we have: 1 st e ds = s

!

Along the contour C1 we will let s = R ei φ so that

ds = i R e dφ : !

∫

C1

∫

π 2

(

iφ

)

e Rt e F R ei φ i R ei φ dφ

β

!

!

∫

C1

∫

π 2

e

Rt ei φ

β

(

F Re

∫

C1

1 st e ds ≤ s

∫

π 2

β

e Rt e

iφ

(

iφ

i R ei φ dφ

β

∫

π 2

∫

π 2

e Rt cosφ ei Rt sin φ dφ

β

1 st e ds ≤ M R − m+1 s

e Rt cosφ dφ

β

Since e Rt cosφ ≤ eσ 0 t we can write: iφ

) i Re

iφ

dφ

!

F R ei φ

∫

C1

1 st e ds ≤ M R − m+1 eσ 0 t s

∫

π 2

dφ

β

and so:

Using Proposition 11.3-3, we have: !

∫

C1

where 0 < β < π 2 . We then have: 1 st e ds = s

e Rt e

and so:

iφ

1 st e ds = s

∫

π 2

or

1 st e ds s

C1

!

)

F R ei φ < M R − m

!

)

i R ei φ dφ

!

∫

C1

1 st ⎛π ⎞ e ds ≤ M R − m+1 eσ 0 t ⎜ − β⎟ ⎝2 ⎠ s 202

As R → ∞ we have:

∫

!

!

we will let s = R e

Along the contour C2 ds = i R ei φ dφ :

!

∫

C2

∫

π

(

iφ

iφ

so that

C2

)

∫

!

We then have: !

∫

C2

∫

π

e

Rt ei φ

π 2

(

F Re

iφ

) i Re

dφ

!

Using Proposition 11.3-3, we have:

∫

C2

1 st e ds ≤ s

∫

π

π 2

e Rt e

iφ

(

F R ei φ

)

i R ei φ dφ

!

π 2

∫

π

∫

π

e Rt cosφ ei Rt sin φ dφ

π 2

1 st e ds ≤ M R − m+1 s

e Rt cosφ dφ

π 2

∫

1 st e ds ≤ M R − m+1 s

∫

1 st e ds ≤ 2 M R − m+1 s

C2

∫

π

e Rt cos(θ + π 2 ) dθ

0

)

where M > 0 and m > 0 , then we can write:

∫

e− Rt sin θ dθ

0

θ = π 2 . From Jordan’s inequality we then have:

R→∞: F R ei φ < M R − m

π 2

where we have used the fact that e− Rt sinθ is symmetric about

transform F ( s ) is bounded as R → ∞ . If we also have as

(

i R ei φ dφ

or

Since f ( t ) is a function of exponential order, the Laplace

!

iφ

Letting φ = θ + π 2 we have: iφ

C2

!

e Rt e

and so:

e Rt e F R ei φ i R ei φ dφ

π 2

!

C2

1 st e ds = s

∫

π

or

!

1 st e ds = s

∫

1 st e ds ≤ M R − m+1 s

C2

1 st e ds → 0 s

C1

∫

1 st e ds ≤ M R − m s

∫

!

C2

1 st e ds ≤ 2 M R − m+1 s

∫

π 2

e− Rt 2θ π dθ

0

or

203

1 st πM e ds ≤ m 1− e− Rt s R t

(

∫

!

C2

)

!

Cr

As R → ∞ we have:

∫

!

!

!

∫

L1

∫

= i r1 2

r

e

st

1 s

R

ds =

= = −

!

∫

r R

∫

e− x t dx x 1 2e i π 2

r R

e− x t 1 1 2 dx = i ix

∫

C3

ds = ei π dx = − dx . We then have: !

∫

L2

∫

−R

−r

es t

1 s

ds = −

=−

!

∫

R

r

∫

R

r

!

π

as r → 0

∫

Along the contour C4 as R → ∞ we have:

e− x t dx i x1 2

R

r

!

Along the contour L 2 we will let s = x ei θ = x e− i π = −x and 1 st e ds = s

∫

iθ

er e t e i θ 2 dθ = 0 !

using a similar derivation as for contour C2

∫

1 st e ds → 0 s

using a similar derivation as for contour C1

ds = e− i π dx = − dx . We then have: !

−π

1 st e ds → 0 s

C4

!

π

iθ

er e t iθ 1 2 i θ 2 i r e dθ r e

Along the contour C3 as R → ∞ we have:

Along the contour L1 we will let s = x ei θ = x ei π = −x and 1 st e ds = s

∫

−π

!

1 st e ds → 0 s

C2

∫

1 st e ds = s

e− x t − i x1

e− x t dx ! x 1 2 e− i π 2 2

1 dx = i iθ

∫

R

r

We then have for the contour C : e− x t dx ! x1 2 iθ

Along the contour Cr we will let s = r e and ds = i re dθ :

!

!∫

C

1 s

st

e ds =

∫

σ0 + i∞

1

σ0 − i∞

1 e ds + i s st

∫

r

R −xt

e x1

2

1 dx + i

∫

r

R −xt

e x1

2

dx = 0

Multiplying by 1 2 π i : !

1 2πi

∫

σ0 + i ∞ σ0 − i ∞

1 st 1 e ds − π s

∫

r

R

e− x t x1

2

dx = 0

204

Letting R → ∞ and r → 0 : !

()

f t =

1 2π i

Let u = ( x t ) !

()

f t =

∫

12

1 2π i

∫

σ0 + i∞

1 s

σ0 − i∞

est ds =

so that du = σ0 + i∞ σ0 − i∞

1 s

number of poles (see Figure 12.5-2). The contour is then 1 π

∫

∞ −xt

e x1

0

2

dx

expanded to include an infinite number of poles, and residue integration takes the form of an infinite sum of residues.

1 −1 2 x t ) t dx . We then have: ( 2

est ds =

2

π t

∫

∞

2

e− u du

0

Using the integral identity: !

∫

∞

2

e − y dy =

0

π 2

we obtain: !

f (t ) =

1 2π i

∫

σ0 + i∞ σ0 − i∞

1 s

est ds =

1

πt

12.5.3! INFINITE NUMBER OF POLES !

If the Laplace transform of a function f ( t ) has an infinite

number of poles to the left of the abscissa of convergence, then it is not possible to use partial fraction expansion since the denominator of the transform is not a finite polynomial. Therefore transform tables cannot be used for inverting the

Figure 12.5-2! Infinite number of poles.

Laplace transform. For an infinite number of poles, residue integration uses an initial Bromwich contour enclosing a finite 205

where sn are the poles of F ( s ) within the contour C :

Example 12.5-6 Determine the inverse Laplace transform of F ( s ) =

1 . s sinh ( s )

The residue at s = 0 is:

Solution: The Taylor series expansion of sinh ( s ) is: s 3 s5 sinh s = s + + +! = 3! 5!

!

∑( n=0

s 2 n+1 2n +1 !

)

or

( )

()

s sinh s ≈ s 2 + O s 4

Laplace transform F ( s ) has an infinite number of simple

!

poles along the imaginary axis as can be seen from:

( )

()

!

{F ( s )} =

−1

1 2πi

∫

σ0 + i ∞

es t

σ0 − i ∞

1 ds s sinh ( s )

!

!

∫

1 f (t ) = e ds = 2 π i s sinh ( s ) σ0 − i R st

()

( )

cosh s ≈ 1+ O s 2

⎡ est st est s est ⎤ Res ⎡⎣ F s est , s = 0 ⎤⎦ = lim ⎢ + − 2 ⎥ = lim ⎡⎣t est ⎤⎦ = t s→ 0 s s s ⎦ s→ 0 ⎣

()

The residue at sn = ± n π i is:

From Proposition 12.5-1 we have: σ0 + i R

( )

()

sinh s ≈ s + O s3 ! we have:

determine the inverse Laplace transform: f (t ) = L

s cosh ( s ) es t ⎤ ⎡ es t s t es t Res ⎡⎣ F ( s ) , s = 0 ⎤⎦ = lim ⎢ + − ⎥ 2 s→0 sinh ( s ) ⎦ ⎣ sinh ( s ) sinh ( s )

Approximating the hyperbolic function near s = 0 with:

sinh iθ = i sin θ

We will use the contour C shown in Figure 12.5-2 to

!

d ⎡ s es t ⎤ Res ⎡⎣ F ( s ) , s = 0 ⎤⎦ = lim ⎢ ⎥ s → 0 ds ⎣ sinh ( s ) ⎦ and so:

and so F ( s ) has a pole of order 2 at s = 0 . In addition, the

!

d ⎡ es t ⎤ 2 Res ⎡⎣ F ( s ) , s = 0 ⎤⎦ = lim ⎢( s − 0 ) ⎥ s → 0 ds s sinh ( s ) ⎦ ⎣

!

!

or !

∞

n = 1, 2, 3, !

sn = ± n π i !

!

k

∑ Res ⎡⎣ F ( s ) e , s ⎤⎦ st

n

n =1

!

⎡ es t ⎤ st ⎡ ⎤ Res ⎣ F ( s ) e , s = sn ⎦ = lim ⎢( s − sn ) ⎥ s →sn s sinh ( s ) ⎦ ⎣ 206

or using L’Hôpital’s rule:

Res ⎡⎣ F ( s ) es t , s = sn ⎤⎦ = lim

!

s →sn

es t sinh ( s ) + s cosh ( s )

Since:

!

!

cosh ( ± n π i ) = cos ( n π ) = ( −1)

!

sinh ( ± n π i ) = i sin ( ± n π ) = 0

e± n π i t

am s m + am−1 s m−1 +!+ a1 s + a0 ( ) F ( s) = = ! bn s n + bn−1 s n−1 +!+ b1 s + b0 Q ( s) P s

(12.6-1)

( −1)n ( ± n π i )

Assuming this fraction is a proper rational fraction where

and so summing all residues in a Bromwich contour

to factor the denominator into first-order and quadratic factors

extending to infinity:

having real coefficients so that we obtain:

f (t ) = t +

!

∞

∑

( −1)n e+n π i t

n =1

nπi

∞

−

∑ n =1

n > m , the first step in obtaining an inverse Laplace transform is

( −1)n e− n π i t nπi

!

F (s) =

P (s)

( s − s1 )( s − s2 )( s − s3 )!( s − sn )

!

(12.6-2)

where sn are the roots of Q ( s ) .

or !

As noted in Chapter 6, a Laplace transform F ( s ) generally

denominator Q ( s ) that are polynomials in the variable s : !

Res ⎡⎣ F ( s ) es t , s = 0 ⎤⎦ =

POLE POSITION DETERMINES SYSTEM STABILITY

takes the form of a fraction having a numerator P ( s ) and a

n

we have: !

12.6!

f (t ) = t + 2

∞

∑(−1) n =1

n

sin ( n π t ) nπ

!

A rational fraction transform that has been expanded into

partial fractions will have a term for each pole in the contour C of the transform. The inverse transform will then depend upon both the nature and position of the poles in the s-plane as follows (see Figure 12.6-1):

207

2.! Conjugate simple poles on the imaginary axis result in stationary harmonic oscillation: ! !

()

f t = Acos ω t + Bsin ω t !

(12.6-4)

A physical system whose output is described by this function is considered to be stable. 3.! Conjugate poles with positive real parts result in increasing harmonic oscillations which tend towards infinity:

! !

()

(

)

f t = e α t Acos ω t + Bsin ω t !

(12.6-5)

A physical system whose output is described by equation (12.6-5) is considered to be unstable.

Figure 12.6-1! Pole positions indicated by ∞ .

4.! Simple or higher order poles on the negative real axis result in decreasing exponential functions which tend

1.! Conjugate poles with negative real parts result in damped

towards zero:

harmonic oscillations which tend towards zero: ! !

()

(

)

f t = e− α t Acos ω t + Bsin ω t !

(12.6-3)

A physical system whose output is described by equation (12.6-3) is considered to be stable.

! !

()

f t = Atn e

− α1 t

+ B tn e

−α2 t

!

(12.6-6)

A physical system whose output is described by equation (12.6-6) is considered to be stable.

208

5.! Simple or higher order poles on the positive real axis result in increasing exponential functions such as e α t , t e α t , t 2 e α t , t 3 e α t , ! which tend towards infinity:

! !

()

f t = Atn e

α1 t

+ B tn e

α2 t

!

(12.6-7)

A physical system whose output is described by equation (12.6-7) is considered to be unstable.

!

We see that a function f ( t ) will have a unique final value

only if all the poles of its Laplace transform are located left of the imaginary axis.

209

Chapter 13 Solution of Linear Ordinary Differential Equations having Variable Coefficients

L {t y′′ ( t )} = − s 2Y ′ ( s ) − 2 sY ( s ) + y ( 0 )

210

!

Most differential equations with variable coefficients are

insolvable using Laplace transforms. There are exceptions,

Proof: !

From Proposition 4.1-1, we have:

however, as we will show in this chapter. In all cases the Laplace transform of a differential equation with variable coefficients will not result in an algebraic equation, but in another differential equation.

L { t f ( t )} = −

!

!

DIFFERENTIAL EQUATIONS OF SECOND DEGREE

(13.1-3)

L { t f ′ ( t )} = −

d L ds

{ f ′ (t )} = − dsd ⎡⎣ s F ( s ) − f ( 0 )⎤⎦ !

(13.1-4)

or

A second order linear differential equation of f ( t ) with

variable coefficients in the form of t

{ f (t )} = − F ′ ( s ) !

Therefore we can write: !

13.1!

d L ds

may be Laplace

transformed into a first order linear differential equation of

F ( s ) with variable coefficients in the form of polynomials of s . This transformed differential equation must then be solved for the transformed dependent variable.

L

!

!

L { t f ′′ ( t )} = −

d L ds

L { t f ′′ ( t )} = −

d 2 ⎡ s F ( s ) − s f ( 0 ) − f ′ ( 0 ) ⎤⎦ ! ds ⎣

Therefore:

if L

!

!

L

{ t f ′ (t )} = − s F ′ ( s ) − F ( s ) !

(13.1-1)

!

L

{ t f ′′ (t )} = − s 2 F ′( s ) − 2 s F ( s ) + f ( 0 ) !

(13.1-2)

{ f ′′ (t )} !

(13.1-6)

and so:

If f ( t ) is a piecewise-smooth function of exponential order, and

{ f (t )} = F ( s ) , then:

(13.1-5)

We then have:

!

Proposition 13.1-1:

{ t f ′ (t )} = −s F ′ ( s ) − F ( s ) !

L

{ t f ′′ (t )} = − s 2 F ′( s ) − 2 s F ( s ) + f ( 0 ) !

(13.1-7)

(13.1-8)

■

Example 13.1-1 Determine the solution of the differential equation: 211

!

t y′′ ( t ) + 2 y′ ( t ) + t y ( t ) = sin ( t )

!

with initial condition: y ( 0 ) = 0 .

⎡ sin ( t ) ⎤ y ( t ) = lim ⎢ − cos ( t ) ⎥ = lim ⎡⎣ cos ( t ) − cos ( t ) ⎤⎦ = 0 t→0 ⎣ t ⎦ t→0

Solution: From Propositions 13.1-1, 3.1-1, and 4.1-1 we have:

Example 13.1-2

1 ! ⎡⎣ − s 2 Y ′( s ) − 2 sY ( s ) + y ( 0 ) ⎤⎦ + 2 ⎡⎣ sY ( s ) − y ( 0 ) ⎤⎦ − Y ′ ( s ) = 2 s +1

Determine the solution of the differential equation:

Applying the initial condition, we obtain: !

1 − s Y ′( s ) − Y ′( s ) = 2 s +1

Y ′(s) = −

(s

1

−

Solution:

2

)

+1

4 ! ⎡⎣ − s 2 Y ′( s ) − 2 sY ( s ) + y ( 0 ) ⎤⎦ − 2 ⎡⎣ − s Y ′ ( s ) − Y ( s ) ⎤⎦ + 2Y ( s ) = s

2

4.1-1:

Applying the initial conditions, we obtain: !

d 1 Y (s) = ds s2 + 1

(

)

2

⎧1 ⎫ = L ⎨ ( sin ( t ) − t cos ( t )) ⎬ = L {t y ( t )} ⎩2 ⎭

y (t ) =

1 ⎡ sin ( t ) ⎤ − cos ( t ) ⎥ ⎢ 2⎣ t ⎦

This is consistent with the initial condition since:

4 ⎡⎣ − s 2 Y ′( s ) − 2 sY ( s ) + 2 ⎤⎦ − 2 ⎡⎣ − s Y ′ ( s ) − Y ( s ) ⎤⎦ + 2Y ( s ) = s

or !

Therefore: !

with initial conditions: y ( 0 ) = 2 , y′ ( 0 ) = 1 .

From Proposition 13.1-1 we have:

Using the transform table in Appendix E and Proposition

!

t y′′ ( t ) − 2 t y′ ( t ) +2 y ( t ) = 4

2

and so: !

!

( 2 s − s ) Y ′( s ) + ( 4 − 2 s ) Y ( s ) + 2 = 4s 2

and so: !

s ( 2 − s ) Y ′( s ) + 2 ( 2 − s ) Y ( s ) =

2 (2 − s) 4 −2= s s 212

We then have: !

Y ′( s ) +

2 2 Y (s) = 2 s s

This is an ODE that has the solution: !

Y (s) =

2 c + s s2

Using the transform table in Appendix E: !

!

⎡⎣ − s 2 Y ′( s ) − 2 sY ( s ) + 1⎤⎦ + ⎡⎣ sY ( s ) − 1⎤⎦ − Y ′( s ) = 0

We then have: !

(s

2

)

+ 1 Y ′( s ) + s Y ( s ) = 0

or !

y ( t ) = 2 + ct

dY ( s ) Y (s)

+

s ds

(s

2

)

+1

=0

From the initial condition y′ ( 0 ) = 1 we have c = 1, and so:

This is an ODE that has the solution:

!

!

y (t ) = 2 + t

(

)

1 lnY ( s ) + ln s 2 + 1 = ln c 2

or Example 13.1-3

!

Determine the solution of the differential equation: !

t y′′ ( t ) + y′ ( t ) + t y ( t ) = 0

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 0 . Solution: From Propositions 13.1-1 and 4.1-1 we have: !

Y (s) =

c s2 + 1

Using the transform table in Appendix E: !

y (t ) = c J 0 (t )

From the initial condition y ( 0 ) = 1 we have c = 1 , and so: !

y (t ) = J 0 (t )

⎡⎣ − s 2 Y ′( s ) − 2 sY ( s ) + y ( 0 ) ⎤⎦ + ⎡⎣ s Y ′ ( s ) − y ( 0 ) ⎤⎦ − Y ′( s ) = 0

Applying the initial conditions, we obtain: 213

13.2!

DIFFERENTIAL EQUATIONS OF HIGHER DEGREE

L

!

{

}

t m f ( n )( t ) = ( −1)

If f ( t ) and f

!

L

{t

f

(n)

(t )

(t ) } = ( −1)

( )

m

!

( )

( )

dm ⎡ n n−1 f 0 − − s n−2 f ′ 0 − m ⎣s F (s) − s ds

( )

( )

where n and m are positive integers, and L

(13.2-1)

{ f (t )} = F ( s ) .

From Proposition 3.1-3, we have:

L

{ f ( ) (t )} = s F ( s ) − s f ( 0 ) − s

!

n

n

!

L

{

L

{t

n−1

−

n−2

( )

( )

f ′ 0 − − !− f ( n−1) 0 −

! !

{

(t ) , from Proposition 4.1-2 we can write:

}

t m g ( t ) = ( −1) G ( m ) ( s ) ! m

!

L

where L

{ g (t )} = G ( s ) . Therefore we have:

(13.2-3)

2

( )

d2 ⎡ s F ( s ) − f 0 − ⎤⎦ ! ds 2 ⎣

(13.2-5)

2

}

f ′ ( t ) = s F ′′ ( s ) + 2 F ′ ( s ) !

(13.2-6)

We also have: L

{t

2

L

{t

2

f ′′ ( t )

( )

(13.2-7)

f ′′ ( t ) = s 2 F ′′ ( s ) + 4 s F ′ ( s ) + 2 F ( s ) !

(13.2-8)

}

d2 ⎡ 2 − ⎤ = ( −1) 2 ⎣s F (s) − s f 0 ⎦! ds 2

or

(13.2-2)

Letting g ( t ) = f

}

t 2 f ′ ( t ) = ( −1)

or

!

! (n)

(13.2-4)

Using equation (13.2-1) we can write:

Proof:

!

( )

are piecewise-smooth functions of

− s n−3 f ′′ 0 − − !− f ( n−1) 0 − ⎤⎦ !

!

!

( )

■

(n)

exponential order over [ 0, ∞ ) , then: m

( )

dm ⎡ n n−1 f 0 − − s n−2 f ′ 0 − m ⎣s F (s) − s ds

− s n−3 f ′′ 0 − − !− f ( n−1) 0 − ⎤⎦ !

! Proposition 13.2-1:

m

! !

}

Laplace transforms of derivatives are summarized in Table

13.2-1.

214

L {functions }

functions ! !

f ′ (t ) !

!

f ′′ ( t ) !

!

f (n) (t ) !

!

t f (t ) !

!

t n f (t ) !

!

s2

( )

equation with variable coefficients in the form of polynomials

( ) F ( s) − s f (0 ) − f ′(0 ) s F (s) − f 0

−

−

−

( )

From Proposition 13.2-1 we see that a linear differential

( )

s n F ( s ) − s n−1 f 0 − − s n−2 f ′ 0 − − !− f ( n−1) 0 −

will be Laplace transformed into a differential equation having the order of the maximum degree of the polynomial coefficients. In general, therefore, only if the maximum degree of the polynomial coefficients is less than the order of the original linear differential equation will the method of Laplace transformation aid in solving the original differential equation.

− F′ ( s)

( −1) F (n ) ( s ) n

!

L

{ t f ′ (t )} !

− s F′ ( s) − F ( s)

!

L

{ t f ′′ (t )} !

− s 2 F ′( s ) − 2 s F ( s ) + f ( 0 )

!

L

{t

2

!

L

{t

2

!

L

{

!

!

}

s F ′′ ( s ) + 2 F ′ ( s )

f ′ (t ) !

}

s 2 F ′′ ( s ) + 4 s F ′ ( s ) + 2 F ( s )

f ′′ ( t ) !

}

t m f ( n )( t ) !

( −1)m

( ) ( ) f ′′ ( 0 ) − !− f ( ) ( 0 ) ⎤⎦

dm ⎡ n s F ( s ) − s n−1 f 0 − − s n−2 f ′ 0 − ds m ⎣

− s n−3

−

n−1

−

Example 13.2-1 Determine the solution of the differential equation: !

y′′ ( t ) + t 2 y ( t ) = 0

with initial conditions: y ( 0 ) = 1 , y′ ( 0 ) = 0 . Solution: From Proposition 13.1-1 we have: !

( ) ( )

⎡ s 2 Y ( s ) − s y 0 − − y′ 0 − ⎤ + Y ′′( s ) = 0 ⎣ ⎦

Applying the initial conditions, we obtain: !

⎡⎣ s 2 Y ( s ) − s ⎤⎦ + Y ′′( s ) = 0

We then have: Table 13.2-1! Laplace transforms of derivatives.

!

Y ′′( s ) + s 2 Y ( s ) = s 215

This differential equation is not simpler than the original differential equation. Therefore the method of Laplace transformation does not aid in solving the original differential equation.

216

Chapter 14 Evaluating Integral Equations with the Laplace Transform

y (t ) = f (t ) +

∫

t

y ( τ ) K ( t − τ ) dτ

0

217

!

In this chapter we will consider the application of Laplace

transform techniques to the solution of certain integral equations.

14.1! !

!

The Volterra equation of the second kind:

y (t ) = f (t ) + λ

!

!

Integral equations such as:

y (t ) = f (t ) +

!

∫ y (τ) K (t − τ) dτ !

(14.1-5)

are called integral equations of convolution type since their

The Fredholm equation of the first kind:

integral is a convolution integral.

b

∫ ϕ (t ) K ( y, t ) dt !

t

0

following are some examples of linear integral equations.

f ( y) =

(14.1-4)

where λ is an unknown parameter.

INTEGRAL EQUATIONS

An integral equation by definition is an equation having

!

∫ y (τ) K (t, τ) dτ ! a

an unknown function that appears under an integral sign. The !

t

(14.1-1)

a

where ϕ ( t ) is an unknown function and f ( y ) is a known

14.2!

LAPLACE TRANSFORM OF INTEGRAL EQUATIONS

function.

!

!

transforms. Exceptions are certain integral equations of

The Fredholm equation of the second kind:

ϕ ( y) = f ( y) + λ

!

∫

b

a

ϕ ( t ) K ( y, t ) dt !

(14.1-2)

convolution type. These equations have a Laplace transform of the form:

where λ is an unknown parameter.

!

!

or

The Volterra equation of the first kind:

!

f (t ) =

∫

t

a

ϕ ( τ ) K ( t, τ ) dτ !

(14.1-3)

where ϕ ( t ) is an unknown function and f ( t ) is a known

Most integral equations cannot be solved using Laplace

!

Y (s) = F (s) + Y (s) K (s) !

(14.2-1)

F (s) ! 1− K ( s )

(14.2-2)

Y (s) =

The inverse Laplace transformation then gives y ( t ) .

function. 218

Example 14.2-1

Example 14.2-2

Determine the solution of the integral equation: !

y ( t ) = 2t −

t

∫ y (τ) (t − τ) dτ 0

Solution: Laplace transforming this equation gives us: !

L

{ y (t )} =

L {2t } − L

{ y (t ) ∗t }

or !

Y (s) =

2 s2

− L

{ y (t )} L {t }

and so: !

Y (s) =

2 s2

− Y (s)

Therefore: !

2 Y (s) = 2 s +1

We then have: !

y ( t ) = 2sin ( t )

2 s2

1 = s 2 1+ 1 s2

Determine the solution of the integral equation: !

y (t ) = 2 − 2

t

∫ y (τ) cos(t − τ) dτ 0

Solution: Laplace transforming this equation gives us: !

L

{ y (t )} =

L {2} − 2 L

{ y (t ) ∗ cost }

or !

Y (s) + Y (s)

2s 2 = s +1 s 2

and so: !

⎛ s 2 + 2 s + 1⎞ 2 ⎜⎝ s 2 + 1 ⎟⎠ Y ( s ) = s

Therefore: !

Y (s) =

(

)=2−

2 s2 + 1 s ( s + 1)

2

s

4

( s + 1)2

We then have: !

y ( t ) = 2 − 4 t e−t

219

Example 14.2-3

Example 14.2-4

Determine the solution of the integral equation:

Determine the solution of the integral equation:

!

y ( t ) = 2sint − 2

∫

t

y ( τ ) cos ( t − τ ) dτ

!

y (t ) = t 2 +

0

t

∫ y (τ) sin (t − τ) dτ 0

Solution:

Solution:

Laplace transforming this equation gives us:

Laplace transforming this equation gives us:

!

L

{ y (t )} = 2 L {sint } − 2 L { y (t ) ∗ cost }

or !

Y (s) =

2 2s + Y (s) 2 s +1 s +1 2

2s ⎞ 2 ⎛ ⎜⎝ 1− 2 ⎟⎠ Y ( s ) = 2 s +1 s +1

Therefore: !

2s ⎞ 2 s2 + 1 2 ⎛ 1− Y s = = ( ) ⎜⎝ ⎟⎠ 2 2 2 s +1 ( s − 1) s + 1 ( s − 1)2

We then have: !

L

{ y (t )} =

L

{ t } + L { y (t ) ∗sint } 2

or

and so: !

!

y ( t ) = 2t et

!

Y (s) =

2 1 + Y (s) 2 3 s s +1

and so: !

s2 2 Y (s) = 3 2 s +1 s

Therefore: !

Y (s) =

2 2 + s3 s5

We then have: !

y (t ) = t 2 +

t4 12 220

Example 14.2-6 Example 14.2-5

Determine the solution of the integral equation:

Determine the solution of the integral equation: !

y ( t ) = 2 et −

∫

t

y ( τ ) sinh ( t − τ ) dτ

t

!

∫ y (τ) y (t − τ) dτ = 8sin (2t ) 0

0

Solution: Solution:

Laplace transforming this equation gives us:

Laplace transforming this equation gives us: !

L

{ }

{ y (t )} = L 2 e − L t

or !

Y (s) =

2 1 − Y (s) 2 s −1 s −1

and so: !

s

2

s −1 2

Y (s) =

Therefore: !

2 2 Y (s) = + 2 s s

We then have: !

y ( t ) = 2 + 2t

2 s −1

{ y (t ) ∗sinh (t )}

!

L { y ( t ) ∗ y ( t )} = L { y ( t )} L { y ( t )} = L {8sin ( 2t )}

or !

Y (s) = 2

16 s +4 2

and so: !

Y (s) = ±

4 s2 + 4

Therefore: !

y ( t ) = ± 4J 0 ( 2t )

Example 14.2-7 Determine the solution of the integral equation: !

y (t ) +

t

∫ y (τ) (t − τ) dτ = sin ( 3t ) 0

221

Solution:

Solution:

Taking the Laplace transform:

Taking the Laplace transform:

!

L { y ( t )} + L { y ( t ) ∗t } =

3 2 s +9

3 L { y ( t )} + L { y ( t )} L {t } = 2 s +9

Y (s) +

1 3 2 Y (s) = 2 s s +9

Y (s) =

2

!

Y (s) =

y (t ) =

!

2 1 − Y s ( ) s3 s2

1⎞ 2 ⎛ ⎜⎝ 1+ 2 ⎟⎠ Y ( s ) = 3 s s

3 s 9 3 3 1 = − 2 2 s + 9 s + 1 8 s + 9 8 s2 + 1

!

Y (s) =

2

2 s2 2 2 2s = = − 2 3 2 2 s s +1 s s +1 s s +1

(

)

Therefore:

and so: !

{ y (t ) ∗t }

or

Therefore: !

{ }

L t2 − L

Therefore:

or !

{ y (t )} =

L

and so:

and so: !

!

9 3 sin ( 3t ) − sin ( t ) 8 8

!

y ( t ) = 2 − 2 cos ( t )

Example 14.2-9 Example 14.2-8

Determine the solution of the integral equation:

Determine the solution of the integral equation: !

y (t ) = t 2 −

∫

t

y ( τ ) ( t − τ ) dτ

!

∫

t

y ( τ ) y ( t − τ ) dτ = t e− 2t

0

0

222

Solution:

!

Taking the Laplace transform: !

L { y ( t ) ∗ y ( t )} = L { y ( t )} L { y ( t )} = ⎡⎣Y ( s ) ⎤⎦ =

( s + 2)

2 s2 + 22

or

1

2

Y (s) = ±

2

!

y ( t ) = ± 2 J 0 ( 2t )

and so: !

Y (s) = ±

1 s+2

or !

y ( t ) = ± e− 2t

Example 14.2-10 Determine the solution of the integral equation: t

!

∫ y (τ) y (t − τ) dτ = 2sin 2t 0

Solution: Taking the Laplace transform: !

L { y ( t ) ∗ y ( t )} = L { y ( t )} L { y ( t )} = ⎡⎣Y ( s ) ⎤⎦ = 2 2

2 s + 22 2

and so:

223

Appendix A

The Greek Alphabet

Kappa!

κ!

Κ

Lambda!

λ!

Λ

Mu!

µ!

Μ

Nu!

ν!

Ν

Xi!

ξ!

Ξ

Omicron!

ο!

Ο

Pi!

π!

Π

Rho!

ρ!

Ρ

Sigma!

σ!

Σ

Tau!

τ!

Τ

Upsilon!

υ!

ϒ

Phi!

φ, ϕ !

Φ

Alpha!

α!

Α

Beta!

β!

Β

Chi!

χ!

Χ

Gamma!

γ!

Γ

Psi!

ψ!

Ψ

Delta!

δ!

Δ

Omega!

ω!

Ω

Epsilon!

ε!

Ε

Zeta!

ζ!

Ζ

Eta!

η!

Η

Theta!

θ!

Θ

Iota!

ι!

Ι 224

∫ ( f g)′dt = ∫ ( f ′ g + f g′) dt !

!

Appendix B

(B-2)

or

fg=

!

∫ f ′ g dt + ∫ f g′ dt !

(B-3)

The constant resulting from integration of the left side of this equation is omitted since it will be combined with the constants from the remaining integrals. Rewriting equation (B-3), we have:

Integration by Parts

!

∫ f g′ dt = f g − ∫ f ′ g dt !

(B-4)

Now we can let: !

In this Appendix we describe the process of changing a

function that we wish to integrate into another function that

!

!

!

Let f ( x ) and g ( x ) be two functions of x , and let both

functions have first order derivatives. We can the write: !

( f g )′ = f ′ g + f g′ !

Integrating this equation, we have:

(B-1)

dv = g′ ( t ) dt !

(B-5)

We then have:

can be easier to integrate. This process is known as integration by parts.

u = f (t ) !

du = f ′ ( t ) dt !

v=

∫ g′ (t ) dt = g (t )!

(B-6)

Equation (B-4) becomes: ! !

∫ u dv = u v − ∫ v du !

(B-7)

This is the equation resulting from integration by parts.

For any particular equation it is necessary to select u and dv . 225

After dv is integrated to obtain v , equation (B-7) can be used to obtain the solution of the original integral. Note that the constant of integration in equation (B-6) is taken to be zero since, if it is nonzero, it will be incorporated in the constant of integration of the integral in equation (B-7). Example B-1 Evaluate the integral: !

L

{ f (t )} = lim

T →∞

∫

T

0

−

e− s t f ( t ) dt

Solution: Let: !

u = f (t ) !

dv = e− s t dt

!

du = f ′ ( t ) dt !

v=−

e− st s

We then have: !

lim

T →∞

⎡ 1 e− s t f ( t ) dt = lim ⎢ − f ( t ) e− s t T →∞ ⎢ s 0− ⎣

∫

T

∫

T

and so: !

lim

T →∞

0

−

e

−st

f ( t ) dt =

( )+1

f 0− s

s

∫

∞

0

−

∞ 0−

+

1 s

⎤ e− s t f ′ ( t ) dt ⎥ 0− ⎥⎦

∫

T

e− s t f ′ ( t ) dt 226

A proper rational fraction is a rational fraction F ( s ) having n > m . A proper rational fraction can be converted into partial

Appendix C

fractions by taking the following steps: ! 1! –! Factor the denominator Q ( s ) into its prime factors having the forms listed below in steps 2, 3, 4, and 5. ! 2! –! Each linear factor of Q ( s ) will have the form ( s + a ) , and can be written as the partial fraction:

Partial Fractions

A ! ( s + a)

! ! !

!

!

(C.1-2)

where A is a constant, and where − a can be real or complex, and is a simple pole of F ( s ) .

!

In this Appendix we describe the process of changing a

proper rational fraction into partial fractions.

! 3! –! Each repeated linear factor of Q ( s ) will have the form

C.1!

!

!

( s + a )n , and can be written as the partial fractions:

GENERAL PROCEDURE

denominator Q ( s ) is a polynomial of degree n :

am s m + am−1 s m−1 +!+ a1 s + a0 ( ) ! F ( s) = = bn s n + bn−1 s n−1 +!+ b1 s + b0 Q ( s)

! !

+

B2

( s + a ) ( s + a )2

A rational fraction is defined to be an algebraic fraction in

which the numerator is a polynomial P ( s ) of degree m , and the

!

B1

!

+!+

Bn

( s + a)

n

!

(C.1-3)

where B1, B2, ! Bn are constants, and where − a can be real or complex, and is a higher order pole of F ( s ) .

P s

(C.1-1)

! 4! –! Each irreducible quadratic factor of Q ( s ) will have the form

(s

2

)

+ a s + b , and can be written as the partial

fraction: 227

Cs+D ! 2 s + as +b

! ! ! !

!

!

(C.1-4)

where C and D are constants, and where each such

Solution:

quadratic factor can be expressed as linear factors that

We have:

are complex conjugates, and are poles of F ( s ) .

!

! 5! –! Each repeated irreducible quadratic factor of Q ( s ) will

(

The form of the partial fractions is:

partial fractions:

!

s2 + a s + b

! ! !

)

1 1 = s − 2 s − 3 ( s + 1) ( s − 3) 2

have the form s 2 + a s + b , and can be written as the

E1 s + F1

!

1 s − 2s − 3 2

+

(s

E 2 s + F2 2

+ as +b

)

2

n

+!+

(s

E n s + Fn 2

+ as +b

)

n

!

(C.1-5)

where E1, E 2, !, E n and F1, F2, !, Fn are constants, and where each quadratic factor can be expressed as linear factors that are complex conjugates, and are poles of

Example C.1-2 Determine the form of the partial fractions for: !

F ( s ) . Repeated pairs of complex conjugate poles will then exist. ! 6! –! The final step is to determine the constants of all the partial fractions. Example C.1-1

1 A1 A2 = + s − 2 s − 3 ( s + 1) ( s − 3) 2

s +1 s − 4s+ 4 2

Solution: We have: !

s +1 1 = s − 4 s + 4 ( s − 2 )2 2

The form of the partial fractions for repeated factors is:

Determine the form of the partial fractions for: 228

!

s +1 B B2 = 1 + s − 4 s + 4 s − 2 ( s − 2 )2

!

2

(

) (s

s 2 ( s + 1) s 2 + 1 3

+

!

Example C.1-3

1

B3

( s + 1)

3

+

2

2

C1 s + D1 s +1 2

)

+ s +1 +

2

=

A1 s

+

C2 s + D2

(s

2

+1

)

2

A2 s +

2

+

B1 B2 + s + 1 ( s + 1)2

C3 s + D3 s2 + s + 1

+

C4 s + D4

(s

2

)

+ s +1

2

Determine the form of the partial fractions for: !

(s

s2 + 1 2

C.2!

)

+ s + 1 ( s − 2)

!

The denominator is irreducible. The form of the partial fractions is:

s +1 2

(s

2

The constants associated with the partial fractions can be

determined by first multiplying the original rational fraction

Solution:

!

CALCULATING PARTIAL FRACTIONS

)

+ s + 1 ( s − 2)

=

C1 s + D1 s + s +1 2

and its partial fraction expansion by the denominator Q ( s ) of the original rational fraction. The resulting equation can be considered to be an identity. The constants of the partial

+

A s−2

fractions can be determined by selecting values for s or by comparing the coefficients of s on both sides of the equation.

Example C.1-4

Example C.2-1

Determine the form of the partial fractions for:

Determine the partial fractions for:

!

(

1

) (s

s ( s + 1) s 2 + 1 2

2

2

)

+ s +1

2

Solution: The form of the partial fractions is:

!

1 s2 − 2 s − 3

Solution: From Example C.1-1 we have:

229

A1 A2 1 = + s − 2 s − 3 ( s + 1) ( s − 3)

!

2

Clearing fractions by multiplying this equation by

(s

!

2

)

− 2 s − 3 = ( s + 1) ( s − 3) , we obtain:

1 ≡ A1 ( s − 3) + A2 ( s + 1)

!

1 = 4 A2 !

⇒!

A2 =

1 = − 4 A1 ! ⇒ !

1 4

1 A1 = − 4

Therefore: !

(s

2

)

− 4 s + 4 we obtain:

s + 1 ≡ B1 ( s − 2 ) + B2

Equating coefficients of equal powers of s we have:

Letting s = −1 we have: !

2

Clearing fractions by multiplying this equation by

!

Letting s = 3 we have:

s +1 B B2 = 1 + s − 4 s + 4 s − 2 ( s − 2 )2

!

!

B1 = 1

Letting s = 2 we have: !

B2 = 3

Therefore: !

1 −1 1 = + s − 2 s − 3 4 ( s + 1) 4 ( s − 3)

s +1 1 3 = + s − 4 s + 4 s − 2 ( s − 2 )2 2

2

Example C.2-3 Example C.2-2 Determine the partial fractions for: !

s +1 s2 − 4 s + 4

Determine the form of the partial fractions for: !

(s

s2 + 1 2

)

+ s + 1 ( s − 2)

Solution: Solution:

From Example C.1-3 we have:

From Example C.1-2 we have: 230

!

(s

s2 + 1 2

)

+ s + 1 ( s − 2)

=

Cs+D s2 + s + 1

+

A

!

s−2

Clearing fractions by multiplying this equation by

(s

!

2

)

+ s + 1 ( s − 2 ) we obtain:

(

)

s 2 + 1 ≡ ( Cs + D ) ( s − 2 ) + A s 2 + s + 1

5 = 7A !

5 ! 7

D=−

⇒!

Equating coefficients of s :

1= C + A = C +

5 ! 7

⇒!

C=

2 7

(s

s2 + 1 2

A1 A2 A3 + + 3 2 s − 2 ( s − 2) ( s − 2)

1 7

s 2 = A1 ( s − 2 ) + A2 ( s − 2 ) + A3 2

Letting s = 2 we have: !

A3 = 4

Differentiating s 2 = A1 ( s − 2 )2 + A2 ( s − 2 ) + A3 we have: !

2 s = 2 A1 ( s − 2 ) + A2

Equating coefficients of s :

Therefore: !

=

obtain:

2

!

( s − 2)

3

3

!

1= −2D + A = −2D +

s2

Clearing fractions by multiplying this equation by ( s − 2 ) we

Letting s = 0 we have: !

Solution:

!

5 A= 7

⇒!

( s − 2 )3

We have:

Letting s = 2 we have: !

s2

)

+ s + 1 ( s − 2)

=

(

2s −1

)

7 s2 + s + 1

+

5

7( s − 2)

!

Letting s = 2 we have: !

Example C.2-4

A1 = 1

A2 = 4

Therefore:

Determine the partial fractions for: 231

s2

!

( s − 2)

C.3! !

3

=

1 4 4 + + 3 s − 2 ( s − 2 )2 ( s − 2 )

HEAVISIDE COVER-UP METHOD

In the special case where the prime factors of the

denominator of a rational fraction are all linear and nonrepeating, a method known as the Heaviside cover-up can be used to easily determine the partial fraction constants. The Heaviside cover-up method consists of covering up each factor

!

1 1 B1 B2 = = + s − 2 s − 3 ( s + 1) ( s − 3) ( s + 1) ( s − 3) 2

Covering up s + 1 and letting s + 1→ 0 , the rational fraction becomes: !

1 1 =− s → −1 ( s − 3) 4 lim

From Example C.2-1 we see that we do have: !

B1 = −

1 4

in turn in the rational fraction, and in determining the limit of

Covering up s − 3 and letting s − 3 → 0 , the rational fraction

what remains of the rational function as the covered-up factor

becomes:

approaches zero. The limits obtained are the constants for the partial fractions corresponding to the covered up factors.

!

1 1 = s → 3 ( s + 1) 4 lim

Example C.3-1

From Example C.2-1 we see that we do have:

Using the Heaviside cover-up method determine the partial

!

B2 =

fractions for: !

1 2 s − 2s − 3

1 4

Therefore: !

1 −1 1 = + s − 2 s − 3 4 ( s + 1) 4 ( s − 3) 2

Solution: We have:

232

C.4! !

HEAVISIDE EXPANSION FORMULA

Heaviside developed a formula for evaluating inverse

Laplace transformations. We will consider the case where Laplace transform has the form: !

()

F s =

() ! Q ( s) P s

(C.4-1)

polynomials. If Q ( s ) has a finite number of roots, we then have:

F (s) =

is present in the numerator and

denominator making this equation indeterminate, we use L’Hôpital’s rule: !

Aj =

( ) ! Q′ ( s ) P sj

(C.4-5)

j

where F ( s ) is a rational fraction and where P ( s ) and Q ( s ) are

!

Since the factor s − sj

P (s)

( s − s1 )( s − s2 )( s − s3 )!( s − sn )

!

Equation (C.4-3) can now be written: !

F (s) =

P sj

j =1

j

(C.4-6)

j

The inverse Laplace transform of F ( s ) is then:

(C.4-2)

where sj are the roots of Q ( s ) and are called poles or

( ) 1 ! ∑ Q′ ( s ) ( s − s ) n

( ) {F ( s )} = ∑ Q′ ( s ) n

!

L

−1

P sj

j =1

j

L

−1

⎧⎪ 1 ⎨ ⎪⎩ s − sj

(

)

⎫⎪ ⎬! ⎪⎭

(C.4-7)

singularities of F ( s ) as discussed in Chapter 11.

or from the Laplace transform table in Appendix E and

!

Proposition 2.2-1:

If Q ( s ) does not have any repeating factors, we can write

equation (C.4-2) in the form of partial fractions as: !

A1 A2 An F (s) = + +!+ = ( s − s1 ) ( s − s2 ) ( s − sn )

n

∑ (s − s ) ! j =1

Aj

! !

The coefficients of the partial fractions are then given by: !

P (s) ⎤ ⎡ ⎡ s − sj ⎤ ! Aj = lim ⎢ s − sj = P (s) ⎢ ⎥ ⎥ s→sj Q s Q s ( ) ( ) ⎣ ⎦ ⎣ ⎦ s = sj

(

)

( )e ∑ Q′ ( s ) j =1

(C.4-3)

j

f (t ) =

n

P sj

sj t

!

(C.4-8)

j

This is Heaviside’s expansion formula when there are no

repeated factors. This is the same result obtained by calculating the residues for simple poles given by Proposition 11.9-3.

(C.4-4)

!

If Q ( s ) has repeating factors, formulas similar to equation

(C.4-7) can be developed. Using Proposition 11.9-5 the same 233

results can be obtained by evaluating higher order poles, and so we will not present additional Heaviside expansion formulas. Example C.4-1 Determine the inverse Laplace transform of F ( s ) =

s +1 . s +s−6 2

Solution: We have: !

F (s) =

s +1 s +1 = s 2 + s − 6 ( s + 3) ( s − 2 )

and so the roots are s1 = − 3 and s2 = 2 . We also have: !

P (s) = s + 1

!

Q ( s) = s2 + s − 6

!

Q′ ( s ) = 2 s + 1

We then have from equation (C.4-8): !

f (t ) =

( )e ∑ Q′ ( s ) 2

j =1

P sj

j

sj t

=

2 − 3t 3 2t e + e 5 5

(see Example 9.2-10).

234

Appendix D

Proposition 1.4-3, Existence of the Laplace Transform: If f ( t ) is a piecewise continuous function for t ≥ 0 , and if

f ( t ) ≤ M eσ 0 t where Re s > σ 0 with σ 0 ≥ 0 and t ≥ T0 , then the

Laplace transform:

L

!

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s )

exists.

Summary of Propositions

Proposition 1.4-4, Uniqueness of the Laplace Transform: If the Laplace transform L

{ f (t )} = F ( s ) exists, then f (t ) has

only the one Laplace transform F ( s ) . Proposition 1.4-1: If f ( t ) is a bounded function such that f ( t ) ≤ M for t ≥ T0 ,

Proposition 1.4-5, Riemann Lebesque Lemma: If L

then f ( t ) is of exponential order.

Proposition 1.4-2, Absolute Convergence of the Laplace Transform: If f ( t ) is a piecewise continuous function for t ≥ 0 , and if f ( t ) ≤ M eσ 0 t , then the Laplace transform:

!

L

{ f (t )} =

∫

∞

0

−

f ( t ) e− s t dt = F ( s )

converges absolutely for Re s > σ 0 where σ 0 ≥ 0 .

{ f (t )} = F ( s )

where f ( t ) is a piecewise continuous

)

function of exponential order over ⎡⎣ 0 − , ∞ , then: !

lim F ( s ) → 0

Re s→ ∞

uniformly. Proposition 1.4-6: If L

{ f (t )} = F ( s )

where f ( t ) is a piecewise continuous

)

function of exponential order over ⎡⎣ 0 − , ∞ , then s F ( s ) is bounded as Re s → ∞ . 235

Proposition 1.4-7, Uniform Convergence of the Laplace Transform:

Proposition 2.1-1, Linearity Property of Laplace Transforms: The Laplace transform is a linear operator. If f ( t ) and g ( t ) are

If f ( t ) is a piecewise continuous function having f (t ) ≤ M e

!

L

σ0 t

for t ≥ T0 , then the Laplace transform: ∞

real-valued functions having Laplace transforms that exist, then:

{ f (t )} = ∫ f (t ) e− s t dt = F ( s )

Re s > σ 0 .

Proposition 2.1-2, Convergence of a Sum of Laplace Transforms: If f ( t ) and g ( t ) are piecewise continuous functions of

Proposition 1.4-8: f (t ) ≤ M e

!

L

{ f (t )} converges for σ > σ f and {g (t )} converges for σ > σ g , then L {c1 f (t ) + c2 g (t )}

exponential order, and if L

If f ( t ) is a piecewise continuous function having σ0 t

{ f (t )} + c2 L {g (t )}

where c1 and c2 are arbitrary constants.

0−

converges uniformly with respect to s in the open half-plane

L {c1 f ( t ) + c2 g ( t )} = c1 L

!

L

for t ≥ T0 , then the Laplace transform:

(

∞

{ f (t )} = ∫ f (t ) e− s t dt = F ( s ) 0−

Proposition 2.2-1, First Shift Theorem (s-Shifting Property): If L

converges absolutely and uniformly with respect to s in the open half-plane Re s > σ 0 .

)

converges absolutely for σ > σ m where σ m = max σ f , σ g .

{ f (t )} = F ( s ) , then: L

!

{ f (t ) e } = F ( s − b ) bt

Proposition 1.4-9: If the Laplace transform of a function f ( t ) with f ( t ) ≤ M eσ 0 t converges absolutely with respect to s for Re s ≥ σ1 > σ 0 , then it converges uniformly and absolutely with respect to s for Re s = σ1 where σ1 ≥ σ 0 .

Proposition 2.2-2: If L L

{ f (t )} = F ( s ) is absolutely convergent for Re s > σ 0 , then

{ f (t ) e } is absolutely convergent for Re s > σ bt

0

+ Reb .

236

Proposition 2.3-1, Scaling Property (s Scaling): If L

Proposition 3.1-2, Laplace Transforms of Second Derivatives:

{ f (t )} = F ( s ) , then:

If f ( t ) and f ′ ( t ) are both continuous functions of exponential

!

)

have:

where b > 0 . Proposition 2.3-2, Scaling Property (t Scaling): If L

L

!

{ f (t )} = F ( s ) , then: F (b s ) =

!

1 ⎧ ⎛ t ⎞⎫ L ⎨f ⎜ ⎟⎬ b ⎩ ⎝ b⎠ ⎭

If f ( t ) , f ′ ( t ) , f ′′ ( t ) ,!, f ( n−1) ( t ) , are continuous functions of

)

{ f ′ (t )} = s F ( s ) − f ( 0 − )

where L

{

Re s > σ 0 , and we have:

and f ′ ( t ) is a piecewise-smooth function over [ 0, ∞ ) , then

L

)

exponential order over ⎡⎣ 0 − , ∞ , and if f ( n ) ( t ) is a piecewisesmooth function over ⎡⎣ 0 − , ∞ , then L f ( n ) ( t ) exists for all

If f ( t ) is a continuous function of exponential order over [ 0, ∞ )

!

{ f (t )} = F ( s ) .

Proposition 3.1-3, Laplace Transforms of Higher Order Derivatives:

Proposition 3.1-1, Laplace Transforms of Derivatives:

{ f ′ (t )} exists for all Re s > σ 0 , and we have:

{ f ′′ (t )} = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − )

where L

where b > 0 .

L

)

order over ⎡⎣ 0 − , ∞ , and if f ′′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then L { f ′′ ( t )} exists for all Re s > σ 0 , and we

1 ⎛ s⎞ L { f ( bt )} = F ⎜ ⎟ b ⎝ b⎠

!

L

{ f ( ) (t )} = s F ( s ) − s f ( 0 ) − s n

where L

n

n−1

−

n−2

( )

}

( )

f ′ 0 − − !− f ( n−1) 0 −

{ f (t )} = F ( s ) .

{ f (t )} = F ( s ) .

237

Proposition 3.2-1:

Proposition 3.3-2:

If f ( t ) is a piecewise continuous function of exponential order

)

over ⎡⎣ 0 − , ∞ with a single finite discontinuity at t = t 0 , and if f ′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then: !

L

)

{ f ′ (t )} = s F ( s ) − f ( 0− ) − ⎡⎣ f (t0+ ) − f (t0− )⎤⎦ e− s t

where L

If f ( t ) is a piecewise-smooth function of exponential order, then the Laplace tranform of the integral of f ( t ) exists. Proposition 3.3-3, Laplace Transform of Integrals: If f ( t ) is a piecewise-smooth function of exponential order, and

0

if L

{ f (t )} = F ( s ) and Re s > σ 0 .

⎧ L ⎨ ⎩

!

Proposition 3.2-2: If f ( t ) is a piecewise continuous function of exponential order

)

over ⎡⎣ 0 − , ∞ with finite discontinuities at t = t 0 , t1, t 2 , !, t n , and if f ′ ( t ) is a piecewise-smooth function over ⎡⎣ 0 − , ∞ , then: L

k

k=0

where L

{ f (t )} = F ( s ) and Re s > σ 0 .

0−

⎫ F (s) f ( τ ) dτ ⎬ = s ⎭

If f ( t ) is a piecewise-smooth function of exponential order, and

)

{ f ′ (t )} = s F ( s ) − f ( 0 − ) − ∑ ⎡⎣ f (t k+ ) − f (t k− )⎤⎦ e− s t

∫

t

Proposition 3.3-4, Laplace Transform of Double Integrals: if L

n

!

{ f (t )} = F ( s ) , then:

{ f (t )} = F ( s ) , then: ⎧ L ⎨ ⎩

!

t

τ

0−

0−

∫∫

⎫ F (s) f ( γ ) dγ dτ ⎬ = 2 s ⎭

Proposition 3.3-5, Laplace Transform of Multiple Integrals: If f ( t ) is a piecewise-smooth function of exponential order, and

Proposition 3.3-1: If f ( t ) is a piecewise-smooth function of exponential order, then

if L

the integral of f ( t ) is of exponential order.

!

{ f (t )} = F ( s ) , then:

⎧ L ⎨ ⎩

t

τ1

τ2

0−

0−

∫∫ ∫ ∫ 0−

!

τn

0−

⎫ F (s) f ( γ n ) dγ n dγ n −1 ! dγ 1 dτ ⎬ = n s ⎭ 238

Proposition 4.1-4:

Proposition 4.1-1: If f ( t ) is a piecewise-smooth function of exponential order, and if L

{ f (t )} = F ( s ) , then we have: L { t f ( t )} = −

!

If f ( t ) is a piecewise-smooth function of exponential order, and

d L ds

{ f (t )} = − F ′ ( s )

where F ′ ( s ) exists for Re s > σ 0 .

L

!

⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists, then: t →0 ⎝ t ⎠

{ f (t )} = F ( s ) , then F (n) ( s ) exists for Re s > σ 0 , and is

given by: L

{t

n

( )

If f ( t ) is a piecewise-smooth function of exponential order, and

If f ( t ) is a piecewise-smooth function of exponential order, and

!

dF ( s ) − 2 s F ( s ) + f 0− ds

L {t f ′′ ( t )} = −s 2

Proposition 4.2-1:

Proposition 4.1-2:

L

{ f (t )} = F ( s ) , then for all Re s > σ 0 :

}

!

f ( t ) = ( −1) F ( n ) ( s ) n

where L

where n is a positive integer. Proposition 4.1-3, Holomorphic Laplace Transform:

⎧ f (t ) ⎫ L ⎨ ⎬= ⎩ t ⎭

∫

∞

F ( u ) du

s

{ f (t )} = F ( s ) .

Proposition 4.2-2: If f ( t ) is a piecewise-smooth function of exponential order, and

If f ( t ) is a piecewise-smooth function of exponential order, and

⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists where a is a constant, then: t →0 ⎝ t + a ⎠

L { f ( t )} = F ( s ) , then F ( s ) is holomorphic in its region of

absolute convergence Re s > σ 0 . !

⎧ f (t ) ⎫ s a L ⎨ ⎬=e ⎩t + a⎭

where L

∫

∞

e−u a F ( u ) du

s

{ f (t )} = F ( s ) . 239

⎧ f (t ) ⎫ L ⎨ n ⎬= ⎩ t ⎭

!

Proposition 4.2-3: If f ( t ) is a piecewise-smooth function of exponential order, and

!

⎧ f (t ) ⎫ L ⎨ ⎬= ⎩ t ⎭

where L

∫

∞

If f ( t ) is a piecewise-smooth function of exponential order, and ⎛ f (t ) ⎞ if lim ⎜ 2 ⎟ exists, then: t → 0+ ⎝ t ⎠

where L

∫∫ s

s

u1

u2

un−1

F ( un ) dun dun−1 !du1

If L

∞

u1

Proposition 4.2-5:

s→∞

Proposition 4.3-2, Initial Value Theorem for Derivative:

{ f (t )} = F ( s ) where f (t ) and f ′ (t ) are continuous functions of exponential order over [ 0, ∞ ) , and f ′′ ( t ) is a piecewise-smooth function of exponential order on [ 0, ∞ ) , then: If L

F ( u 2 ) du 2 du1

{ f (t )} = F ( s ) .

( )

lim ( s F ( s )) = f 0 +

!

∞

∫∫∫ ∫

∞

!

{ f (t )} = F ( s ) where f (t ) is a continuous function of exponential order over [ 0, ∞ ) and f ′ ( t ) is a piecewise-smooth function of exponential order over [ 0, ∞ ) , then:

F ( u ) du

0

{ f (t )} = F ( s ) .

⎧ f (t ) ⎫ L ⎨ 2 ⎬= ⎩ t ⎭

∞

Proposition 4.3-1, Initial Value Theorem:

Proposition 4.2-4:

!

∞

{ f (t )} = F ( s ) .

where L

⎛ f (t ) ⎞ if lim+ ⎜ ⎟ exists, then: t →0 ⎝ t ⎠

∞

(

( )) = f ′ ( 0 )

lim s 2 F ( s ) − s f 0 −

!

s→∞

+

Proposition 4.3-3, Final Value Theorem:

{ f (t )} = F ( s ) where f (t ) is a continuous function of exponential order over [ 0, ∞ ) and f ′ ( t ) is a piecewise-smooth function of exponential order on [ 0, ∞ ) , then: lim ( s F ( s )) = lim f ( t ) s→0 t→∞ If L

If f ( t ) is a piecewise-smooth function of exponential order, and ⎛ f (t ) ⎞ if lim+ ⎜ n ⎟ exists, then: t →0 ⎝ t ⎠

!

240

Proposition 5.1-1, Laplace Transform of the Shifted Unit Step Function:

Proposition 5.3-1, Laplace Transform of a Rectangular Pulse: If f ( t ) is a rectangular pulse defined by:

The Laplace transform of the shifted unit step function is: e− s t0 ! L {U ( t − t 0 )} = s

!

t0 ≥ 0

then

Proposition 5.1-2, Laplace Transform of a Unit Step Function: The Laplace transform of a unit step function is:

L {U ( t )} =

!

1 s

!

{ f (t )} = F ( s ) where f (t ) is a piecewise continuous function of exponential order, and if U ( t − t 0 ) is the shifted unit If L

step function, then:

L

e − s t1 − e − s t 2 L { f ( t )} = L {U ( t − t1 )} − L {U ( t − t 2 )} = s

Proposition 5.4-1, Laplace Transform of the Unit Impulse Function:

Proposition 5.2-1, Second Shift Theorem (t-Shifting Property):

!

⎪⎧ 1 t1 ≤ t ≤ t 2 f (t ) = ⎨ ⎪⎩ 0 otherwise

!

{ f (t − t0 ) U (t − t0 )} = e

− s t0

L

{ f (t )}

Proposition 5.2-2, Third Shift Theorem:

{ f (t )} = F ( s ) where f (t ) is a piecewise continuous function of exponential order, and if U ( t − t 0 ) is the shifted unit If L

The Laplace transform of the unit impulse function δ ( t − t 0 ) is: !

L {δ ( t − t 0 )} = e− s t0

Proposition 5.4-2, Laplace Transform of the Unit Impulse Function: The Laplace transform of the unit impulse function δ ( t ) is: !

L {δ ( t )} = 1

step function, then: !

L

{ f (t ) U (t − t0 )} = e− s t

0

L

{ f (t + t0 )} 241

Proposition 5.4-3, Sifting Property: If f ( t ) is continuous at t 0 , then the unit impulse function has

Proposition 6.2-1, Uniqueness of the Inverse Laplace Transform:

)

If the functions f ( t ) and g ( t ) are continuous over ⎡⎣ 0 − , ∞ and of exponential order with Laplace transforms L { f ( t )} = F ( s )

the property: !

∫

∞

−∞

and L { g ( t )} = G ( s ) , and if F ( s ) = G ( s ) for some Re s > σ 0 ,

f ( t ) δ ( t − t 0 ) dt = f ( t 0 )

Proposition 5.4-4, Laplace Transform of the Derivative of a Unit Impulse Function:

then f ( t ) ≡ g ( t ) for all t > 0 .

Proposition 6.2-2, Linearity Property of Inverse Laplace Transforms: If f ( t ) and g ( t ) are continuous functions having Laplace

If δ ( t ) is a unit impulse function, then the Laplace transform of

transforms L

its derivative is given by: !

L

c1 F ( s ) + c2 G ( s ) is a linear operator:

If f ( t ) is a piecewise-smooth periodic function of exponential order with period T , then:

L

{ f (t )} =

1 1− e− sT

L { g ( t )} = G ( s ) ,

respectively, then the inverse Laplace transform of

{ δ′ (t )} = s

Proposition 5.5-1, Laplace Transform of a Periodic Function:

!

{ f (t )} = F ( s ) and

∫

T

0−

f ( t ) e− s t dt

! !

L

{c1 F ( s ) + c2 G ( s )} = c1 L −1{F ( s )} + c2 L −1{G ( s )}

−1

where c1 and c2 are arbitrary constants.

Proposition 7.1-1, Convergence of the Convolution: If f ( t ) and g ( t ) are piecewise continuous functions of exponential order, then their convolution: !

f (t ) ∗ g (t ) =

∫

t

f ( τ ) g ( t − τ ) dτ

0

is also a continuous function of exponential order, and so converges absolutely. 242

Proposition 7.1-2:

Proposition 7.2-3:

If f ( t ) and g ( t ) are piecewise continuous functions of

The convolution of any function f ( t ) with the unit impulse function δ ( t ) is f ( t ) :

exponential order, then their convolution: !

f (t ) ∗ g (t ) =

∫

t

f ( τ ) g ( t − τ ) dτ

!

f ( t ) ∗δ ( t ) =

0

Proposition 8.1-1, Laplace Transform of a Power Series:

)

If f ( t ) and g ( t ) are piecewise continuous functions of

If a function f ( t ) is continuous over ⎡⎣ 0 − , ∞ and is of exponential order, and if f ( t ) can be represented by a

exponential order with Laplace transforms of L

converging power series:

Proposition 7.2-1, Laplace Convolution Theorem:

and L L

{ g (t )} = G ( s ) , respectively, then:

{ f (t )} = F ( s )

{ f (t ) ∗ g (t )} = F ( s ) G ( s )

!

If f1 ( t ) , f2 ( t ) , !, fn ( t ) are piecewise continuous

!

n = 0, 1, 2, !

n

n

σ 0n an ≤ M ! n!

M >0

and Re s > σ 0 , then L

F1 ( s ) , F2 ( s ) , !, Fn ( s ) , respectively, then:

{ f1 (t ) ∗ f2 (t ) ∗!∗ fn (t )} = F1 ( s ) F2 ( s ) ! Fn ( s )

∑a t !

where

functions of exponential order with Laplace transforms of

L

f (t ) =

∞

n=0

Proposition 7.2-2:

!

f ( τ ) δ ( t − τ ) dτ = f ( t )

0

has a Laplace transform.

!

∫

t

{ f (t )} can be represented by the

Laurent series: ∞

!

L

{ f (t )} = ∑ an sn! n+1 n=0

243

!

Proposition 8.1-2, Watson’s Lemma:

)

If a function f ( t ) is continuous over ⎡⎣ 0 − , ∞ and is of

where M > 0 , then the series L

exponential order, and if f ( t ) can be represented by a

f (t ) =

Proposition 8.1-4:

∞

∑a t

n+ν

n

!

ν > −1

If f ( t ) , f ′ ( t ) , f ′′ ( t ) ,! are continuous functions of exponential

n=0

)

order over ⎡⎣ 0 − , ∞ , then:

where !

∞

σ 0n an ≤ M ! n!

M >0

{ f (t )} can be represented by the series:

∞

L

!

Γ n + ν +1 { f (t )} = ∑ an (s n + ν +1 )

Proposition 8.1-3, Convergence of the Laplace Transform of a Power Series:

represented by a series: ∞

!

∑a t ! n

L

1 { f (t )} = F ( s ) = ∑ an s n+1 n=0

that converges for Re s > σ0 with lim F ( s ) = 0 , then F ( s ) can

∞

n

s→∞

be inverted term-by-term to obtain:

n = 0, 1, 2, !

n=0

and if:

s n+1

If the Laplace transform F ( s ) of a function f ( t ) can be

If f ( t ) can be represented by a converging Taylor series: f (t ) =

{ f (t )} = ∑

Proposition 8.2-1, Inverse of a Laplace Transform Series:

n=0

!

L

f (n) ( 0 )

n=0

and Re s > σ 0 , then L !

{ f (t )} will be uniformly

convergent in Re s > σ 0 .

converging power series: !

M σ 0n an ≤ n!

!

f (t ) =

∞

∑ n! t an

n

n=0

244

where f ( t ) converges absolutely and uniformly in its circle of convergence.

!

{ f (t )} = F ( s ) = ∑ an s n +1ν +1 !

and y′ ( t ) are piecewise-smooth functions of exponential order

n=0

over [ 0, ∞ ) . Given the second-order ODE:

that converges for Re s > σ0 with lim F ( s ) = 0 , then F ( s ) can s→∞

!

f (t ) =

∞

∑a

n

n=0

!

convergence.

a y′′ ( t ) + b y′ ( t ) + c f ( t ) = f ( t ) where a , b , and c are constants, then if y ( 0 ) = 0 and y′ ( 0 ) = 0

1 t n + ν +1 Γ ( n + ν + 1)

where f ( t ) converges absolutely and uniformly in its circle of

a s2 + b s + c

{ f (t )} = F ( s ) where y (t ) is a continuous function of exponential order over [ 0, ∞ ) , and f ( t )

ν > −1

be inverted term-by-term to obtain:

F (s)

Let L { y ( t )} = Y ( s ) and L

∞

L

a s2 + b s + c

+

Proposition 9.2-2:

represented by a series: !

( a s + b ) y ( 0 ) + a y′ ( 0 )

where a , b , and c are constants.

Proposition 8.2-2, Inverse of a Laplace Transform Series: If the Laplace transform F ( s ) of a function f ( t ) can be

Y (s) =

we have: !

y (t ) =

t

∫ h (t − τ) f (τ) dτ 0

where: Proposition 9.2-1: If L { y ( t )} = Y ( s ) and L

{ f (t )} = F ( s ) , where y (t ) is a

continuous function of exponential order over [ 0, ∞ ) , and f ( t ) and y′ ( t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) , the Laplace transform of the second-order ODE: !

a y′′ ( t ) + b y′ ( t ) + c f ( t ) = f ( t ) is then:

!

H (s) =

Y (s) 1 = 2 F (s) a s + b s + c

Proposition 10.1-1: If f ( x, t ) is a continuous function of exponential order over

[ 0, ∞ ) and if the first partial derivatives of f ( x, t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) , then: 245

⎧ ∂ f ( x, t ) ⎫ − L ⎨ ⎬ = s F ( x, s ) − f x, 0 ⎩ ∂t ⎭

(

!

where L

!

)

{ f ( x, t )} = F ( x, s ) .

Proposition 11.2-1, Cauchy-Riemann Equations: The necessary and sufficient conditions for a complex function

Proposition 10.1-2:

( ) ( ) ( ) to be differentiable z ( x, y ) are that the Cauchy-Riemann equations: w = f z = u x, y + i v x, y

If f ( x, t ) and its first partial derivatives are continuous

)

functions of exponential order over ⎡⎣ 0 − , ∞ , and if the second partial derivatives of f ( x, t ) are piecewise-smooth functions of

)

exponential order over ⎡⎣ 0 − , ∞ , then: !

(

∂ f x, 0 − ⎧⎪ ∂2 f ( x, t ) ⎫⎪ 2 − L ⎨ ⎬ = s F ( x, s ) − s f x, 0 − 2 ∂t ⎩⎪ ∂t ⎭⎪

(

where L

2 2 ⎪⎧ ∂ f ( x, t ) ⎪⎫ dF ( x, s ) L ⎨ ⎬= 2 dx 2 ⎩⎪ ∂x ⎭⎪

)

!

hold at the point z x, y , and that the partial derivatives ∂u ∂x , ∂u ∂ y , ∂v ∂x , and ∂v ∂ y all exist and are continuous in a

( )

neighborhood of the point z x, y . Proposition 11.3-1, Fundamental Theorem of Calculus for

Proposition 10.1-3:

Complex Functions:

If f ( x, t ) is a continuous function of exponential order over

()

If a complex function f z is holomorphic in a simply

[ 0, ∞ ) and if the first partial derivatives of f ( x, t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) , then: !

connected domain D , then !

⎧ ∂ f ( x, t ) ⎫ dF ( x, s ) L ⎨ ⎬= ∂x dx ⎩ ⎭

∂u ∂v =− ∂y ∂x

( )

)

{ f ( x, t )} = F ( x, s ) .

∂u ∂v ! = ∂x ∂ y

at a point

∫

z2 z1

()

f z dz =

()

∫

z2 z1

( ) dz = F ( z ) − F ( z )

dF z dz

2

1

()

where F z is the antiderivative of f z .

246

Proposition 11.3-2, Independent of Path:

()

If a complex function f z is holomorphic in a simply connected

Proposition 11.3-5, Cauchy-Goursat Theorem:

()

If f z is a holomorphic complex function in a simply

domain D , then the integral: !

connected domain D , then along every simple closed contour C

∫ ()

f z dz

within D we must have:

C

is independent of path C within D .

!

!∫

()

f z dz = 0

C

Proposition 11.3-3, Contour Integral Inequality:

()

If C is a simple contour and f z is a complex function

Proposition 11.3-6, Deformation Theorem: Let C1 and C2 be two simple closed contours within a domain

continuous over C , then:

∫

!

()

f z dz ≤

C

∫

C

()

f z

D , and let C2 be interior to C1 (see Figure 11.3-1). If f ( z ) is a complex function holomorphic at all points within or on C1 , and

dz

outside or on C2 , then:

Proposition 11.3-4, ML-Inequality:

!

()

()

()

f z dz

C2

()

Let f z be a holomorphic complex function in a simply

()

f z ≤M

connected domain D . If C is a simple closed contour that lies within D , and if z0 is a fixed point inside the contour C , then:

then: !

!∫

Proposition 11.3-7, Cauchy’s Integral Formula:

number M along C : !

()

f z dz =

C1

Let C be a simple contour of length L , and let f z be a continuous complex function on C . If f z is bounded by a real

!∫

∫

C

()

f z dz ≤ M L

!

( )

f z0 =

1 2π i

!∫

C

( ) dz

f z

z − z0

247

Proposition 11.3-8, Cauchy’s Integral Formula for Derivatives:

!∫

!

C

()

Let f z be a holomorphic complex function in a simply

1

( z − z0 )n

⎧⎪ 2 π i n = 1 dz = ⎨ n ≠1 ⎪⎩ 0

where n is an integer.

connected domain D . If C is a simple closed contour within the domain D , and if z0 is any given point inside the contour C ,

Proposition 11.4-1, Term-by-Term Integration:

()

then f z has derivatives of all orders at the point z0 given by: ! !

n! n f ( ) ( z0 ) = 2π i

()

( ) dz !∫C ( z − z0 )n+1

converges

uniformly to the sum f z in a domain D , then for any simple

f z

contour C that lies within D the integral of the sum along C can be determined using term-by-term integration of the series:

where n is a nonnegative integer.

Proposition 11.3-9, Higher Order Derivatives of a

()

If a series of continuous complex functions f k z

n

!

lim

n→ ∞

n

∫ ∑ f ( z ) dz = lim ∑ ∫ C

k

k=0

n→ ∞

k=0

C

()

fk z dz

Holomorphic Function Exist and are Holomorphic:

Proposition 11.4-2:

{

holomorphic at the point z0 . Proposition 11.3-10: If C is a simple closed contour within a domain D , and if z = z0 is a point inside the contour C , then:

}

If fk ( z ) is a sequence of holomorphic functions in a simply

()

If a complex function f z is holomorphic at a point z = z0 , then its derivatives f ( n) ( z ) of all orders n exist and are

connected domain D , and if the series: ∞

!

∑ f (z) k

k= 0

()

()

converges uniformly to f z in D , then f z is holomorphic on D .

248

Proposition 11.4-3, Term-by-Term Differentiation:

()

If a series of continuous complex functions fn z

()

Proposition 11.5-2, Uniform Convergence of a Power Series:

()

Every function f z that can be represented by a power series:

converges

uniformly to the sum f z in a domain D , then the derivative

()

of f z at any point z0 in D can be determined by term-by-

!

()

f z =

!

∑a (z− z ) n

A power series provides a unique representation of a function

()

f z within its circle of convergence z − z0 = r .

0

n=0

!

r = lim

n→ ∞

series will converge absolutely, and outside this circle it will diverge.

Proposition 11.5-4, Power Series is Holomorphic:

()

an an+1

if this limit exists. Within the circle of convergence the power

n

0

Proposition 11.5-3, Uniqueness of Power Series Representation:

n

has a circle of convergence z − z0 = r with radius given by:

n

converges uniformly inside its circle of convergence z − z0 = r .

Proposition 11.5-1, Circle of Convergence of a Power Series:

∞

∑ a (z − z ) n=0

term differentiation of the series.

Every power series:

∞

If a function f z can be represented by a power series: !

()

f z =

∞

∑

(

an z − z0

)n

n=0

()

then f z is holomorphic at every point inside its circle of convergence z − z0 = r

249

Proposition 11.5-5, Derivatives of All Orders:

Proposition 11.6-2, Existence of Taylor Series:

!

()

f z =

∞

∑ a (z − z ) n

()

exist on a domain D if and only if f z is holomorphic on D .

n

0

Proposition 11.9-1:

n=0

()

If an analytic function f z has a simple pole at a point z = z0 ,

at every point inside its circle of convergence so that: !

k f ( )( z) =

∞

∑ ( n − k )! a ( z − z ) n!

n

n−k

0

then: !

n= k

()

If a complex function f z

!

()

f z =

∑ a (z − z ) n

n

0

n=0

inside its circle of convergence z − z0 = r , then the coefficients of the series are given by: !

an =

n f ( ) z0

( )=

n!

1 2π i

(

) ()

!

( ) ds !∫C ( s − z0 )n+1 f s

where n = 0, 1, 2, ! , and the series is a Taylor series.

()

If an analytic function f z has the form:

can be represented by the power

series: ∞

()

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ z → z0

Proposition 11.9-2:

Proposition 11.6-1, Taylor Series:

!

()

A Taylor series representation of a complex function f z will

Derivatives of all orders exist for a power series:

!

()

f z =

()

P z

=

A

+

B

+

C

( z − a ) ( z − b) ( z − c ) ( z − a ) ( z − b) ( z − c ) where P ( z ) is a polynomial of degree ≤ 2 , and where a , b , and c are the poles of f ( z ) , then: P( a) A = Res ⎡⎣ f ( z ) , a ⎤⎦ = ( a − b) ( a − c ) ()

B = Res ⎡⎣ f z , b ⎤⎦ =

() ( b − a )( b − c ) P b

250

!

() C = Res ⎡⎣ f ( z ) , c ⎤⎦ = ( c − a ) ( c − b) P c

!

()

() ()

()

() q ( z ) has a

( )

both analytic at a point z = z0 . If p z0 ≠ 0 and

()

()

If a complex function f z is analytic in a domain D except at an isolated singularity at a point z = z0 , then:

simple zero at z0 , then z0 is a simple pole of f z , and we have:

()

Res ⎡⎣ f z , z0 ⎤⎦ =

( ) q′ ( z0 ) p z0

!

()

z = z0 . If q ( z ) has a simple zero at z0 , then z0 is a simple pole of

Proposition 11.10-2, Cauchy’s Residue Theorem:

()

Let C be a simple closed contour that lies within a simply

f z and we have:

()

()

connected domain D . If a complex function f z is analytic in

1

D except at a finite number of isolated singular points

( )

q′ z0

Proposition 11.9-5:

z1 , z2 , ! , zk that lie inside C , then: !

()

If an analytic function f z has a pole of order k ≥ 2 at a point

z = z0 , then:

()

within D and enclosing z0 .

()

Res ⎡⎣ f z , z0 ⎤⎦ =

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦

where C is any positively oriented simple closed contour lying

Let f z = 1 q z where the function q z is analytic at a point

!

!∫

C

Proposition 11.9-4:

()

)

)n f ( z )⎤⎥⎦

Isolated Singularity:

Let f z = p z q z where the functions p z and q z are

!

(

(

Proposition 11.10-1, Cauchy’s Residue Theorem for an

Proposition 11.9-3:

()

1 d n−1 ⎡ ⎡ ⎤ Res ⎣ f z , z0 ⎦ = lim z − z0 z → z0 n − 1 ! dz n−1 ⎢ ⎣

!∫

C

()

f z dz = 2 π i

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ n

n =1

251

Proposition 12.4-1, Existence of the Inverse Laplace Transform:

Proposition 12.2-1:

If f ( t ) is a piecewise continuous function of exponential order

The inverse Laplace transform of the unit step function is: U (t ) = L

!

−1⎧

1⎫ 1 ⎨ ⎬= ⎩ s ⎭ 2πi

∫

σ 0 +i ∞

σ 0 −i ∞

!

The inverse Laplace transform of the shifted unit step function is:

U (t − t0 ) = L

−1⎧ e

⎫ 1 ⎨ ⎬= ⎩ s ⎭ 2πi − s t0

∫

σ0 + i ∞

σ0 − i ∞

f (t ) = L

{F ( s )} =

−1

∫

σ0 + i ∞

F ( s ) es t ds

σ0 − i ∞

!If f ( t ) is a piecewise continuous function of exponential order, over [ 0, ∞ ) with L

{ f (t )} = F ( s ) , then the inverse Laplace

transform: !

f (t ) = L

{F ( s )} =

−1

The inverse Laplace transform for the shifted unit impulse

1 2πi

∫

σ0 + i ∞

F ( s ) es t ds

σ0 − i ∞

is given by:

function is: δ (t − t0 ) = L

1 2πi

Proposition 12.5-1:

e− s t0 s t e ds s

Proposition 12.3-1:

!

{ f (t )} = F ( s ) , then the inverse Laplace

transform exists where Re s > σ 0 , and is given by:

Proposition 12.2-2:

!

over [ 0, ∞ ) with L

es t ds s

{e }

−1

− s t0

1 = 2πi

∫

σ0 + i ∞

e

σ0 − i ∞

− s t0

st

e ds

!

f (t ) =

N

∑ Res ⎡⎣F ( s ) e , s st

n =1

n

⎤⎦

where sn are the poles of F ( s ) es t , and where Re s > σ 0 .

252

Proposition 12.5-2: !If f ( t ) is a piecewise continuous function of exponential order

with Laplace transform F ( s ) , and if F ( s ) es t has only N simple poles sn , then the inverse Laplace transform L f (t ) =

!

{F ( s )} is:

−1

N

∑ lim ⎡⎣ ( s − s ) F ( s ) e n =1

n

s → sn

st

⎤⎦

Proposition 13.1-1: If f ( t ) is a piecewise-smooth function of exponential order, and if L

{ f (t )} = F ( s ) , then:

!

L

{ t f ′ (t )} = − s F ′ ( s ) − F ( s )

!

L

{ t f ′′ (t )} = − s 2 F ′( s ) − 2 s F ( s ) + f ( 0 )

Proposition 13.2-1: If f ( t ) and f ( n ) ( t ) are piecewise-smooth functions of exponential order over [ 0, ∞ ) , then: ! !

L

{t

m

}

f ( n )( t ) = ( −1)

m

( )

( )

dm ⎡ n n−1 f 0 − − s n−2 f ′ 0 − m ⎣s F (s) − s ds

( )

( )

− s n−3 f ′′ 0 − − !− f ( n−1) 0 − ⎤⎦ where n and m are positive integers, and L

{ f (t )} = F ( s ) . 253

U (t − t0 )!

!

c!

c s

!

t!

1 s2

!

t2!

2 s3

!

tn!

n! s n+1

!

tα!

!

ea t !

!

t n eb t !

e− s t0

!

sin ( at ) !

1 s

!

cos ( at ) !

Appendix E

Laplace Transform Tables !

In the table of Laplace transforms presented below n is a

positive integer, a , b , c are real constants, U ( t ) is the unit step function, and δ ( t ) is the unit impulse function.

f ( t )! !

δ (t ) !

!

δ (t − t0 ) !

!

1 or U ( t ) !

L

e− s t0 s

!

{ f (t )} 1

α > −1 !

Γ ( α + 1) s α +1

1 s−a n!

( s − b )n+1 a s2 + a2 s s + a2 2

254

!

!

!

!

sinh ( at ) ! cosh ( at ) ! ebt sin (at ) ! ebt cos (at ) !

!

eb t sinh (at ) !

!

e cosh ( at ) !

!

t sin ( at ) !

!

!

bt

t cos ( at ) ! 1− cos ( at ) !

a 2

!

at − sin ( at ) !

s − a2

!

sin ( at ) − at cos ( at ) !

!

sin ( at ) + at cos ( at ) !

s −a 2

(

s2 s2 + a2

s 2

(s

a

( s − b )2 + a 2 s−b

(s

( s − b )2 + a 2 !

a

( s − b)

2

−a

( s − b)

(s

2

−a

2as 2

+a

2

)

2 2

!

!

cos 2 ( at ) !

!

sinh 2 ( t ) !

s2 − a2

(s

(

2

+a

)

2 2

a2

s s2 + a2

)

2 a s2 + a2

2

(s

) )

2

( (s

+a

)

2

2

)

)

2 2

s s 2 + 3a 2

cos ( at ) + at sin ( at ) !

sin 2 ( at ) !

+ a2

2

(

cos ( at ) − at sin ( at ) !

!

2 a3

s s2 − a2

2

s−b

a3

2

(

+a

)

2 a2

s s2 + 4 a2 s2 + 2 a2

(

s s2 + 4 a2

(

2

s s −4 2

)

!

)

2 2

) ) s>2

255

!

cosh 2 ( t ) !

!

t sinh ( at ) !

!

s2 − 2

(

s s2 − 4

(s

)

s>2

!

2as 2

−a

)

2 2

s2 + a2

t cosh ( at ) !

(s

2

+a

e− s t0 s

!

U (t − t0 ) !

!

U ( t − t1 ) − U ( t − t 2 ) !

!

erf

!

ea t erf

!

1− erf

1 ! ( n − 1)!

)

(

(

1 − s t1 − s t 2 e −e s

)

a

at !

s s+a

2 2

(

)

)

a s ( s − a)

at !

sinh ( at ) − sin ( at ) !

2 a3 s4 − a4

!

cosh ( at ) − cos ( at ) !

2 a2 s s4 − a4

!

sinh ( at ) + sin ( at ) !

2 a s2 s4 − a4

!

!

cosh ( at ) + cos ( at ) !

2 s3 s4 − a4

!

a s ( s + a)

!

sin ( ω t + θ ) !

s sinθ + ω cosθ s2 + ω2

!

cos ( ω t + θ ) !

s cosθ − ω sinθ s2 + ω2

!

!

1− e−at !

!

1 e−at − e−bt ! b−a

(

)

a≠b!

1 t

a 2 t

= erfc

cos 2 at !

1

( s + a) ( s + b)

a 2 t

!

e−a s

s

1 sn

π −a e s

s

256

(

)

!

e−a t sin ( ω t + θ ) !

( s + a ) sinθ + ω cosθ ( s + a )2 + ω 2

!

cos ( at ) sinh ( at ) !

a s2 − 2 a2

!

e−a t cos ( ω t + θ ) !

( s + a ) cosθ − ω sinθ ( s + a )2 + ω 2

!

cos ( at ) cosh ( at ) !

s3 s 4 + 4a 4

1 s − ia

!

at cosh ( at ) − sinh ( at ) !

a2 s2 ( s + a)

!

i at

!

e !

!

at − 1+ e−at !

!

1− e−at − at e−at !

!

!

!

!

t! 1 t

!

a2 s ( s + a)

2

π 2 s3

2

b a ⎡ − at − bt ⎤ ⎢⎣1− b − a e + b − a e ⎥⎦ !

!

1 ab

!

e−a t e− b t ! t + (b − a )( c − a ) ( a − b )( c − b ) +

π s 2

sin ( at ) sinh ( at )!

2a s s + 4a 4

sin ( at ) cosh ( at ) !

a s2 + 2 a2

(

s + 4a 4

4

)

sin ( at ) cosh ( at ) − cos ( at ) sinh ( at ) !

!

erf

( t )!

2

− a2

)

2

1 s ( s + a) ( s + b) 1 ( s + a) ( s + b) ( s + c)

e− ct ( a − c )(b − c )

!

4

(s

2 s3

4

a 1+ a s

e− t a !

!

s + 4a 4

4 s3 s4 + 4 a4

1 s s +1

257

!

⎛ a ⎞ erf ⎜ ! ⎝ 2 t ⎟⎠

!

eat erfc

!

!

!

!

!

!

( at ) !

a 2 πt3 1 πt

e

1− e− a s

e− a

2

−a2 4t

4t

!

1 s + as

!

a>0!

a>0!

t

e− a

1

e

s

π

sin t !

cos t

2s

!

t sin ( at ) − at 2 cos at !

( 3 − a t ) sin ( at ) − 3at cos at ! 2 2

s

32

s

e−1

(s

8 a 3s 2

+ a2 8 a5

2

+ a2

) )

(

)

ln s 2 + 1

Ci ( t ) !

!

J0 ( at ) !

2s

1 s2 + a2

(

s2 + a2 − s

!

Jn ( at ) !

!

J0 2 at !

1 −a e s

!

δ′ ( t ) !

s

!

δ( n ) ( t ) !

sn

!

1 bt at e −e ! t

4s

4s

1 −1 1 tan s s

Si ( t ) !

!

−a s

π −1 e s

(s

!

(

)

n

an s2 + a2

)

s

3

(

)

3

!

1 2 πt

3

(e

bt

ln

)

− eat !

s−a s−b

s−a − s−b 258

!

sin ( at ) t

!

tan −1

a s

!

1− cos ( at ) ! t

1 ⎛ s2 + a2 ⎞ ln 2 ⎜⎝ s 2 ⎟⎠

!

1− cosh ( at ) ! t

1 ⎛ s2 − a2 ⎞ ln 2 ⎜⎝ s 2 ⎟⎠

!

sinh 2 ( at ) !

!

cosh 2 ( at ) !

!

t sin ( at ) !

!

t cos ( at ) !

2

2

2 a2

(

s s2 − 4 a2 s2 − 2 a2

(

s s2 − 4 a2

(

) )

2 a 3s 2 − a 2

(s

2

+a

(

)

2 s s 2 − 3a 2

(s

2

+a

)

2 3

)

)

2 3

259

Appendix F

!

f ( at )!

!

f ′ (t )!

!

f ′′ ( t ) !

!

f ( n ) ( t )!

1 ⎛ s⎞ F a ⎜⎝ a ⎟⎠

( )

s F ( s ) − f 0−

( )

( )

s2 F ( s ) − s f 0− − f ′ 0−

s n F(s) −

n−1

∑

s n−1−k  f (k)(0−)

k  =  0

Properties of Laplace Transforms !

In the table of Laplace transforms presented below n is a

positive integer, a and b are constants, and U ( t ) is the unit step function.

functions ! !

L

∞

{ f ( t )} = ∫ f (t ) e− s t dt ! 0−

!

a f (t ) + b g (t ) !

!

ebt f ( t ) !

L {functions }

!

f (t − t0 ) U (t − t0 ) !

!

f (t ) U (t − t0 ) !

!

t f (t ) !

!

t n f (t ) !

!

f (t ) ∗ g (t ) =

a F (s) + b G (s) F ( s − b)

e− s t0 L

!

∫

t

( −1)n F (n ) ( s )

∫

t

f ( τ ) g ( t − τ ) dτ !

f ( τ ) dτ !

0

!

{ f ( t + t0 )}

− F′ ( s)

0

F (s)

e− s t0 F ( s )

f (t ) = f (t + T ) !

F (s) G (s) F (s) s

1 1− e− sT

∫

T

0

−

e− s t f ( t ) dt 260

!

!

!

f (t )

!

t

1 2πi

f (t ) ! t

τ

0

0

∫∫

!

f 0+ !

a−i∞

F (s) s2 1 a

( )

⎛ s⎞ f⎜ ⎟ ⎝ a⎠

lim ( s F ( s ))

s→∞

lim ( s F ( s ))

lim f ( t ) !

s→0

t→∞

f (t ) =

es t F ( s ) ds

f ( u ) du dτ !

f ( at ) !

!

∫

a+i∞

F ( u ) du

s

!

!

∫

∞

∞

∑ n=0

!

t f ′ (t ) !

!

t f ′′ ( t ) !

n

an t !

σ 0n an ≤ M ! n!

∞

∑ n=0

an n! s n+1

− s F′ ( s) − F ( s)

− s 2 F ′( s ) − 2 s F ( s ) + f ( 0 )

261

!

Appendix G

Power Series !

A few of the most common series representations of

z2 z4 cos z = 1− + −! = 2! 4!

∞

∑ (−1)n n=0

!

z 3 2 z 5 17 z 7 tan z = z + + + +! ! 3 15 315

!

z 3 z5 sinh z = z + + +! = 3! 5!

!

∞

∑( n=0

z2 z4 cosh z = 1+ + +! = 2! 4!

∞

( )

z