Boyce/DiPrima/Meade 11th ed, Ch 6.1: Definition of Laplace...

90
Boyce/DiPrima/Meade 11 th ed, Ch 6.1: Definition of Laplace Transform Elementary Differential Equations and Boundary Value Problems, 11 th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc. Many practical engineering problems involve mechanical or electrical systems acted upon by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are difficult to apply. In this chapter we use the Laplace transform to convert a problem for an unknown function f into a simpler problem for F, solve for F, and then recover f from its transform F. Given a known function K(s,t), an integral transform of a function f is a relation of the form F ( s ) = K ( s, t ) f ( t ) dt , -∞≤ α β α < β ≤∞

Transcript of Boyce/DiPrima/Meade 11th ed, Ch 6.1: Definition of Laplace...

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Boyce/DiPrima/Meade 11th ed, Ch 6.1: Definition of

Laplace Transform

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John

Wiley & Sons, Inc.

• Many practical engineering problems involve mechanical or

electrical systems acted upon by discontinuous or impulsive

forcing terms.

• For such problems the methods described in Chapter 3 are

difficult to apply.

• In this chapter we use the Laplace transform to convert a

problem for an unknown function f into a simpler problem

for F, solve for F, and then recover f from its transform F.

• Given a known function K(s,t), an integral transform of a

function f is a relation of the form

F(s) = K(s,t) f (t)dt, - ¥ £a

b

ò a <b £ ¥

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Improper Integrals

• The Laplace transform will involve an integral from zero

to infinity. Such an integral is a type of improper integral.

• An improper integral over an unbounded interval is

defined as the limit of an integral over a finite interval

where A is a positive real number.

• If the integral from a to A exists for each A > a and if the

limit as A → ∞ exists, then the improper integral is said to

converge to that limiting value. Otherwise, the integral is

said to diverge or fail to exist.

A

aAadttfdttf )(lim)(

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Example 1

• Consider the following improper integral.

• We can evaluate this integral as follows:

• Therefore, the improper integral diverges.

dt

t1

¥

ò

dt

t1

¥

ò = limA®¥

dt

t1

A

ò = limA®¥

lnA( )®¥

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Example 2

• Consider the following improper integral.

• We can evaluate this integral as follows:

• Note that if c = 0, then ect = 1. Thus the following two cases

hold:

11limlim

00

cA

A

Act

A

ct ec

dtedte

0dtect

0. if,diverges

and0; if,1

0

0

cdte

cc

dte

ct

ct

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Example 3

• Consider the following improper integral.

• From Example 1, this integral diverges at p = 1

• We can evaluate this integral for p ≠ 1 as follows:

• The improper integral diverges at p = 1 and

1dtt p

11

1limlim 1

11

p

A

Ap

A

p Ap

dttdtt

11

1lim,1If

1

11

1

1lim,1If

1

1

p

A

p

A

Ap

p

pA

pp

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Piecewise Continuous Functions

• A function f is piecewise continuous on an interval [a, b] if

this interval can be partitioned by a finite number of points

a = t0 < t1 < … < tn = b such that

(1) f is continuous on each (tk, tk+1)

• In other words, f is piecewise continuous on [a, b] if it is

continuous there except for a finite number of jump

discontinuities.

nktf

nktf

k

k

tt

tt

,,1,)(lim)3(

1,,0,)(lim)2(

1

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Theorem 6.1.1

• If f is piecewise continuous for t ≥ a, if | f(t) | ≤ g(t) when

t ≥ M for some positive M and if converges, then

also converges.

• On the other hand, if f(t) ≥ g(t) ≥ 0 for t ≥ M, and if

diverges, then also diverges.

g(t)dtM

¥

òf (t)dt

a

¥

ò

g(t)dtM

¥

òf (t)dt

a

¥

ò

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The Laplace Transform

• Let f be a function defined for t > 0, and satisfies certain conditions to be named later.

• The Laplace Transform of f is defined as an integral transform:

• The kernel function is K(s,t) = e–st.

• Since solutions of linear differential equations with constant coefficients are based on the exponential function, the Laplace transform is particularly useful for such equations.

• Note that the Laplace Transform is defined by an improper integral, and thus must be checked for convergence.

• On the next few slides, we review examples of improper integrals and piecewise continuous functions.

0

)()()( dttfesFtfL st

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Theorem 6.1.2

• Suppose that f is a function for which the following hold:

(1) f is piecewise continuous on [0, b] for all b > 0.

(2) | f(t) | ≤ Keat when t ≥ M, for constants a, K, M, with K, M > 0.

• Then the Laplace Transform of f exists for s > a.

• Note: A function f that satisfies the conditions specified above

is said to to have exponential order as t .

finite )()()(0

dttfesFtfL st

®¥

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Example 4

• Let f (t) = 1 for t ≥ 0. Then the Laplace transform F(s) of f is:

0,1

lim

lim

1

0

0

0

ss

s

e

dte

dteL

bst

b

bst

b

st

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Example 5

• Let f (t) = eat for t ≥ 0. Then the Laplace transform F(s) of f is:

asas

as

e

dte

dteeeL

btas

b

btas

b

atstat

,1

lim

lim

0

)(

0

)(

0

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Example 6

• Consider the following piecewise-defined function f

where k is a constant. This represents a unit impulse.

• Noting that f(t) is piecewise continuous, we can compute

its Laplace transform

• Observe that this result does not depend on k, the function

value at the point of discontinuity.

0,1

)()}({1

00

ss

edtedttfetfL

sstst

10

1,

10,1

)(

t

tk

t

tf

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Example 7

• Let f (t) = sin(at) for t ≥ 0. Using integration by parts twice, the

Laplace transform F(s) of f is found as follows:

0,)()(1

sin/)sin(lim1

coslim1

cos/)cos(lim

sinlimsin)sin()(

222

2

00

0

00

00

sas

asFsF

a

s

a

atea

saate

a

s

a

atea

s

a

atea

saate

atdteatdteatLsF

bst

bst

b

bst

b

bst

bst

b

bst

b

st

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Linearity of the Laplace Transform

• Suppose f and g are functions whose Laplace transforms exist

for s > a1 and s > a2, respectively.

• Then, for s greater than the maximum of a1 and a2, the Laplace

transform of c1 f (t) + c2g(t) exists. That is,

with

finite is )()()()(0

2121

dttgctfcetgctfcL st

)( )(

)( )()()(

21

02

0121

tgLctfLc

dttgecdttfectgctfcL stst

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Example 8

• Let f (t) = 5e-2t – 3sin(4t) for t ≥ 0.

• Then by linearity of the Laplace transform, and using results

of previous examples, the Laplace transform F(s) of f is:

0,16

12

2

5

)4sin(35

)4sin(35

)}({)(

2

2

2

sss

tLeL

teL

tfLsF

t

t

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Boyce/DiPrima/Meade 11th ed, Ch 6.2:

Solution of Initial Value Problems

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley

& Sons, Inc.

• The Laplace transform is named for the French mathematician

Laplace, who studied this transform in 1782.

• The techniques described in this chapter were developed

primarily by Oliver Heaviside (1850 - 1925), an English

electrical engineer.

• In this section we see how the Laplace transform can be used

to solve initial value problems for linear differential equations

with constant coefficients.

• The Laplace transform is useful in solving these differential

equations because the transform of f ' is related in a simple way

to the transform of f, as stated in Theorem 6.2.1.

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Theorem 6.2.1

• Suppose that f is a function for which the following hold:

(1) f is continuous and f ' is piecewise continuous on [0, b] for all b > 0.

(2) | f(t) | ≤ Keat when t ≥ M, for constants a, K, M, with K, M > 0.

• Then the Laplace Transform of f ' exists for s > a, with

• Proof (outline): For f and f ' continuous on [0, b], we have

• Similarly for f ' piecewise continuous on [0, b], see text.

)0()()( ftfsLtfL

bstsb

b

bst

bst

b

bst

b

dttfesfbfe

dttfestfedttfe

0

000

)()0()(lim

)()()(lim)(lim

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The Laplace Transform of f '

• Thus if f and f ' satisfy the hypotheses of Theorem 6.2.1, then

• Now suppose f ' and f '' satisfy the conditions specified for f

and f ' of Theorem 6.2.1. We then obtain

• Similarly, we can derive an expression for L{f (n)}, provided f

and its derivatives satisfy suitable conditions. This result is

given in Corollary 6.2.2

)0()0()(

)0()0()(

)0()()(

2 fsftfLs

fftfsLs

ftfsLtfL

)0()()( ftfsLtfL

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Corollary 6.2.2

• Suppose that f is a function for which the following hold:

(1) f , f ', f '' ,…, f (n-1) are continuous, and f (n) piecewise continuous,

on [0, b] for all b > 0.

(2) | f(t) | ≤ Keat, | f '(t) | ≤ Keat , …, | f (n–1)(t) | ≤ Keat for t ≥ M, for

constants a, K, M, with K, M > 0.

Then the Laplace Transform of f (n) exists for s > a, with

)0()0()0()0()()( )1()2(21)( nnnnnn fsffsfstfLstfL

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Example 1: Chapter 3 Method (1 of 4)

• Consider the initial value problem

• Recall from Section 3.1:

• Thus r1 = –2 and r2 = –3, and general solution has the form

• Using initial conditions:

• Thus

• We now solve this problem using Laplace Transforms.

00,10,02 yyyyy

01202)( 2 rrrrety rt

tt ececty 2

21)(

3/1,3/202

121

21

21

cc

cc

cc

tt eety 23/13/2)(

0.0 0.5 1.0 1.5 2.0t

5

10

15

20

y t

tt eety 23/13/2)(

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Example 1: Laplace Transform Method (2 of 4)

• Assume that our IVP has a solution and

satisfy the conditions of Corollary 6.2.2. Then

and hence

• Letting Y(s) = L{y}, we have

• Substituting in the initial conditions, we obtain

• Thus

0}0{}{2}{}{}2{ LyLyLyLyyyL

0}{2)0(}{)0()0(}{2 yLyysLysyyLs

0)0()0(1)(22 yyssYss

01)(22 ssYss

12

1)(}{

ss

ssYyL

00,10,02 yyyyy

f(t) f '(t) and f ''(t)

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Example 1: Partial Fractions (3 of 4)

• Using partial fraction decomposition, Y(s) can be rewritten:

• Thus

3/2,3/1

12,1

)2()(1

211

1212

1

ba

baba

basbas

sbsas

s

b

s

a

ss

s

12)(}{

3/23/1

sssYyL

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Example 1: Solution (4 of 4)

• Recall from Section 6.1:

• Thus

• Recalling Y(s) = L{y}, we have

and hence

2},{3/2}{3/1

12)( 23/23/1

seLeLss

sY tt

asas

dtedteesFeL tasatstat

,1

)(0

)(

0

}3/13/2{}{ 2tt eeLyL

y(t) =1

3e2t +

2

3e-t

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General Laplace Transform Method

• Consider the constant coefficient equation

• Assume that the solution y(t) satisfies the conditions of

Corollary 6.2.2 for n = 2.

• We can take the transform of the above equation:

where F(s) is the transform of f(t).

• Solving for Y(s) gives:

)(tfcyybya

a(s2Y (s)- sy(0)- y '(0))+b(sY (s)- y(0))+ cY (s) = F(s)

Y (s) =(as + b)y(0)+ ay '(0)

as2 + bs + c+

F(s)

as2 + bs + c

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Algebraic Problem

• Thus the differential equation has been transformed into the

the algebraic equation

for which we seek y = such that L{ } = Y(s).

• Note that we do not need to solve the homogeneous and

nonhomogeneous equations separately, nor do we have a

separate step for using the initial conditions to determine the

values of the coefficients in the general solution.

cbsas

sF

cbsas

yaybassY

22

)()0()0()(

f(t) f(t)

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Characteristic Polynomial

• Using the Laplace transform, our initial value problem

becomes

• The polynomial in the denominator is the characteristic

polynomial associated with the differential equation.

• The partial fraction expansion of Y(s) used to determine

requires us to find the roots of the characteristic equation.

• For higher order equations, this may be difficult, especially if

the roots are irrational or complex.

cbsas

sF

cbsas

yaybassY

22

)()0()0()(

00 0,0),( yyyytfcyybya

f(t)

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Inverse Problem

• The main difficulty in using the Laplace transform method is

determining the function y = such that L{ } = Y(s).

• This is an inverse problem, in which we try to find such

that = L–1{Y(s)}.

• There is a general formula for L–1, but it requires knowledge of

the theory of functions of a complex variable, and we do not

consider it here.

• It can be shown that if f is continuous with L{f(t)} = F(s), then f

is the unique continuous function with f (t) = L–1{F(s)}.

• Table 6.2.1 in the text lists many of the functions and their

transforms that are encountered in this chapter.

f(t) f(t)

f(t)f(t)

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Linearity of the Inverse Transform

• Frequently a Laplace transform F(s) can be expressed as

• Let

• Then the function

has the Laplace transform F(s), since L is linear.

• By the uniqueness result of the previous slide, no other

continuous function f has the same transform F(s).

• Thus L–1 is a linear operator with

)()()()( 21 sFsFsFsF n

)()(,,)()( 1

1

1

1 sFLtfsFLtf nn

)()()()( 21 tftftftf n

)()()()( 1

1

11 sFLsFLsFLtf n

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Example 2: Nonhomogeneous Problem (1 of 2)

• Consider the initial value problem

• Taking the Laplace transform of the differential equation, and

assuming the conditions of Corollary 6.2.2 are met, we have

• Letting Y(s) = L{y}, we have

• Substituting in the initial conditions, we obtain

• Thus

10,20,2sin yytyy

)4/(2}{)0()0(}{ 22 syLysyyLs

)4/(2)0()0()(1 22 sysysYs

)4)(1(

682)(

22

23

ss

ssssY

)4/(212)(1 22 sssYs

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• Using partial fractions,

• Then

• Solving, we obtain A = 2, B = 5/3, C = 0, and D = -2/3. Thus

• Hence

Example 2: Solution (2 of 2)

41)4)(1(

682)(

2222

23

s

DCs

s

BAs

ss

ssssY

)4()4()()(

14682

23

2223

DBsCAsDBsCA

sDCssBAssss

4

3/2

1

3/5

1

2)(

222

sss

ssY

tttty 2sin3

1sin

3

5cos2)(

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Example 3: Solving a 4th Order IVP (1 of 2)

• Consider the initial value problem

• Taking the Laplace transform of the differential equation, and

assuming the conditions of Corollary 6.2.2 are met, we have

• Letting Y(s) = L{y} and substituting the initial values, we have

• Using partial fractions

• Thus

0)0(''',00'',10',0)0(,0)4( yyyyyy

222 )1)(()1)(( ssdcssbas

0}{)0(''')0('')0()0(}{ 234 yLysyysysyLs

)1)(1()1()(

22

2

4

2

ss

s

s

ssY

)1()1()1)(1()(

2222

2

s

dcs

s

bas

ss

ssY

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Example 3: Solving a 4th Order IVP (2 of 2)

• In the expression:

• Setting s = 1 and s = –1 enables us to solve for a and b:

• Setting s = 0, b – d = 0, so d = 1/2

• Equating the coefficients of in the first expression gives

a + c = 0, so c = 0

• Thus

• Using Table 6.2.1, the solution is

0)0(''',00'',10',0)0(,0)4( yyyyyy

2/1,01)(2and1)(2 bababa

222 )1)(()1)(( ssdcssbas

3s

)1()1()(

22

2/12/1

sssY

2

sinsinh)(

ttty

0 1 2 3 4 5 6 7t

50

100

150

200

y t

2

sinsinh)(

ttty

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Boyce/DiPrima/Meade 11th ed, Ch 6.3:

Step Functions

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John

Wiley & Sons, Inc.

• Some of the most interesting elementary applications of the

Laplace Transform method occur in the solution of linear

equations with discontinuous or impulsive forcing functions.

• In this section, we will assume that all functions considered are

piecewise continuous and of exponential order, so that their

Laplace Transforms all exist, for s large enough.

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Step Function definition

• Let c > 0. The unit step function, or Heaviside function, is

defined by

• A negative step can be represented by

ct

cttuc

,1

,0)(

ct

cttuty c

,0

,1)(1)(

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Example 1

• Sketch the graph of y = h(t), where

• Solution: Recall that uc(t) is defined by

• Thus

and hence the graph of h(t) is a rectangular pulse.

0),()()( 2 ttututh

ct

cttuc

,1

,0)(

t

t

t

th

20

2,1

0,0

)(

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Example 2

• For the function

whose graph is shown

• To write h(t) in terms of uc(t), we will need

u4(t), u7(t), and u9(t). We begin with the 2,

then add 3 to get 5, then subtract 6 to get –1,

and finally add 2 to get 1 – each quantity is

multiplied by the appropriate uc(t)

9,1

97,1

74,5

40,2

)(

t

t

t

t

th

0),(2)(6)(32)( 974 ttutututh

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Laplace Transform of Step Function

• The Laplace Transform of uc(t) is

s

e

s

e

s

e

es

dte

dtedttuetuL

cs

csbs

b

b

c

st

b

b

c

st

b

c

st

c

st

c

lim

1limlim

)()(0

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Translated Functions

• Given a function f (t) defined for t ≥ 0, we will often want to

consider the related function g(t) = uc(t) f (t - c):

• Thus g represents a translation of f a distance c in the

positive t direction.

• In the figure below, the graph of f is given on the left, and the

graph of g on the right.

ctctf

cttg

),(

,0)(

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Theorem 6.3.1

• If F(s) = L{f (t)} exists for s > a ≥ 0, and if c > 0, then

• Conversely, if f (t) = L–1{F(s)}, then

• Thus the translation of f (t) a distance c in the positive t

direction corresponds to a multiplication of F(s) by e–cs.

)()()()( sFetfLectftuL cscs

c

)()()( 1 sFeLctftu cs

c

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Theorem 6.3.1: Proof Outline

• We need to show

• Using the definition of the Laplace Transform, we have

)(

)(

)(

)(

)()()()(

0

0

)(

0

sFe

duufee

duufe

dtctfe

dtctftuectftuL

cs

sucs

cusctu

c

st

c

st

c

)()()( sFectftuL cs

c

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Example 3

• Find L{ f (t)}, where f is defined by

• Note that f (t) = sin(t) + u /4(t) cos(t – /4), and

f (t) =

sin t, 0 £ t <p

4

sin t + cos(t -p

4), t ³

p

4

ì

í

ïï

î

ïï

1

1

11

1

cossin

)4/cos()(sin)(

2

4/

2

4/

2

4/

4/

s

se

s

se

s

tLetL

ttuLtLtfL

s

s

s

p p

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Example 4

• Find L–1{F(s)}, where

• Solution:

• The function may also be written as

2

21)(

s

esF

s

2)(

1)(

2

2

21

2

1

ttut

s

eL

sLtf

s

2,2

20,)(

t

tttf

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Theorem 6.3.2

• If F(s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant,

then

• Conversely, if f (t) = L-1{F(s)}, then

• Thus multiplication f (t) by ect results in translating F(s) a

distance c in the positive t direction, and conversely.

• Proof Outline:

cascsFtfeL ct ),()(

)()( 1 csFLtfect

)()()()(0

)(

0csFdttfedttfeetfeL tcsctstct

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Example 5

• To find the inverse transform of

• We first complete the square:

• Since

it follows that

54

1)(

2

sssG

)2(

12

1

144

1

54

1)(

222

sF

ssssssG

tesGLtg t cos)()( 21

)()2(andcos1

1)( 21

2

11 tfesFLts

LsFL t

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Boyce/DiPrima/Meade 11th ed, Ch 6.4: Differential Equations

with Discontinuous Forcing Functions

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John

Wiley & Sons, Inc.

• In this section focus on examples of nonhomogeneous initial

value problems in which the forcing function is discontinuous.

00 0,0),( yyyytgcyybya

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Example 1: Initial Value Problem (1 of 12)

• Find the solution to the initial value problem

• Such an initial value problem might model the response of a

damped oscillator subject to g(t), or current in a circuit for a

unit voltage pulse.

20 and 50,0

205,1)()()(

where

0)0(,0)0(),(22

205tt

ttututg

yytgyyy

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Example 1: Laplace Transform (2 of 12)

• Assume the conditions of Corollary 6.2.2 are met. Then

or

• Letting Y(s) = L{y},

• Substituting in the initial conditions, we obtain

• Thus

)}({)}({}{2}{}{2 205 tuLtuLyLyLyL

s

eeyLyysLysyyLs

ss 2052 }{2)0(}{)0(2)0(2}{2

seeyyssYss ss 2052 )0(2)0(12)(22

seesYss ss 2052 )(22

22

)(2

205

sss

eesY

ss

0)0(,0)0(),()(22 205 yytutuyyy

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Example 1: Factoring Y(s) (3 of 12)

• We have

where

• If we let h(t) = L-1{H(s)}, then

by Theorem 6.3.1.

)20()()5()()( 205 thtuthtuty

)(22

)( 205

2

205

sHeesss

eesY ss

ss

22

1)(

2

ssssH

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Example 1: Partial Fractions (4 of 12)

• Thus we examine H(s), as follows.

• This partial fraction expansion yields the equations

• Thus

2222

1)(

22

ss

CBs

s

A

ssssH

2/1,1,2/1

12)()2( 2

CBA

AsCAsBA

22

2/12/1)(

2

ss

s

ssH

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Example 1: Completing the Square (5 of 12)

• Completing the square,

16/154/1

4/14/1

2

12/1

16/154/1

2/1

2

12/1

16/1516/12/

2/1

2

12/1

12/

2/1

2

12/1

22

2/12/1)(

2

2

2

2

2

s

s

s

s

s

s

ss

s

s

ss

s

s

ss

s

ssH

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Example 1: Solution (6 of 12)

• Thus

and hence

• For h(t) as given above, and recalling our previous results,

the solution to the initial value problem is then

16/154/1

4/15

152

1

16/154/1

4/1

2

12/1

16/154/1

4/14/1

2

12/1)(

22

2

ss

s

s

s

s

ssH

tetesHLth tt

4

15sin

152

1

4

15cos

2

1

2

1)}({)( 4/4/1

)20()()5()()( 205 thtuthtut

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Example 1: Solution Graph (7 of 12)

• Thus the solution to the initial value problem is

• The graph of this solution is given below.

415sin152

1415cos

2

1

2

1)(

where),20()()5()()(

4/4/

205

teteth

thtuthtut

tt

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Example 1: Composite IVPs (8 of 12)

• The solution to original IVP can be viewed as a composite of

three separate solutions to three separate IVPs:

)20()20(),20()20(,022:20

0)5(,0)5(,122:205

0)0(,0)0(,022:50

2323333

22222

11111

yyyyyyyt

yyyyyt

yyyyyt

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Example 1: First IVP (9 of 12)

• Consider the first initial value problem

• From a physical point of view, the system is initially at rest,

and since there is no external forcing, it remains at rest.

• Thus the solution over [0, 5) is y1 = 0, and this can be verified

analytically as well. See graphs below.

50;0)0(,0)0(,022 11111 tyyyyy

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Example 1: Second IVP (10 of 12)

• Consider the second initial value problem

• Using methods of Chapter 3, the solution has the form

• Physically, the system responds with the sum of a constant

(the response to the constant forcing function) and a damped

oscillation, over the time interval (5, 20). See graphs below.

205;0)5(,0)5(,122 22222 tyyyyy

2/14/15sin4/15cos 4/

2

4/

12 tectecy tt

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Example 1: Third IVP (11 of 12)

• Consider the third initial value problem

• Using methods of Chapter 3, the solution has the form

• Physically, since there is no external forcing, the response is a

damped oscillation about y = 0, for t > 20. See graphs below.

4/15sin4/15cos 4/

2

4/

13 tectecy tt

20);20()20(),20()20(,022 2323333 tyyyyyyy

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Example 1: Solution Smoothness (12 of 12)

• Our solution is

• It can be shown that and are continuous at t = 5 and

t = 20, and has a jump of 1/2 at t = 5 and a jump of –1/2

at t = 20:

• Thus jump in forcing term g(t) at these points is balanced by a

corresponding jump in highest order term 2y'' in ODE.

)20()()5()()( 205 thtuthtut

limt®5-

¢¢j (t) = 0, limt®5+

¢¢j (t) = 1/ 2

limt®20

-¢¢j (t) @ –0.0072, lim

t®20+¢¢j (t) @ –0.5072

f f 'f ''

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Smoothness of Solution in General

• Consider a general second order linear equation

where p and q are continuous on some interval (a, b) but g is

only piecewise continuous there.

• If y = is a solution, then and are continuous

on (a, b) but has jump discontinuities at the same

points as g.

• Similarly for higher order equations, where the highest

derivative of the solution has jump discontinuities at the same

points as the forcing function, but the solution itself and its

lower derivatives are continuous over (a, b).

)()()( tgytqytpy

f(t) f f '

f ''

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Example 2: Initial Value Problem (1 of 12)

• Find the solution to the initial value problem

• The graph of forcing function

g(t) is given on right, and is

known as ramp loading.

¢¢y + 4y = g(t), y(0) = 0, ¢y (0) = 0

where

g(t) = u5 (t)t - 5

5- u10 (t)

t -10

5=

0, 0 £ t < 5

1

5(t - 5) 5 £ t < 10

1, t ³10

ì

í

ïï

î

ïï

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Example 2: Laplace Transform (2 of 12)

• Assume that this ODE has a solution y = and that

and satisfy the conditions of Corollary 6.2.2.

Then

or

• Letting Y(s) = L{y}, and substituting in initial conditions,

• Thus

5}10)({5}5)({}{4}{ 105 ttuLttuLyLyL

2

1052

5}{4)0()0(}{

s

eeyLysyyLs

ss

21052 5)(4 seesYs ss

45

)(22

105

ss

eesY

ss

0)0(,0)0(,5

10)(

5

5)(4 105

yy

ttu

ttuyy

f(t)f '(t) f ''(t)

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Example 2: Factoring Y(s) (3 of 12)

• We have

where

• If we let h(t) = L-1{H(s)}, then

by Theorem 6.3.1.

)10()()5()(5

1)( 105 thtuthtuty

)(545

)(105

22

105

sHee

ss

eesY

ssss

4

1)(

22

sssH

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Example 2: Partial Fractions (4 of 12)

• Thus we examine H(s), as follows.

• This partial fraction expansion yields the equations

• Thus

44

1)(

2222

s

DCs

s

B

s

A

sssH

4/1,0,4/1,0

144)()( 23

DCBA

BAssDBsCA

4

4/14/1)(

22

sssH

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Example 2: Solution (5 of 12)

• Thus

and hence

• For h(t) as given above, and recalling our previous results,

the solution to the initial value problem is then

ttsHLth 2sin8

1

4

1)}({)( 1

4

2

8

11

4

1

4

4/14/1)(

22

22

ss

sssH

)10()()5()(5

1)( 105 thtuthtuty

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Example 2: Graph of Solution (6 of 12)

• Thus the solution to the initial value problem is

• The graph of this solution is given below.

ttth

thtuthtut

2sin8

1

4

1)(

where,)10()()5()(5

1)( 105

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Example 2: Composite IVPs (7 of 12)

• The solution to original IVP can be viewed as a composite of

three separate solutions to three separate IVPs (discuss):

)10()10(),10()10(,14:10

0)5(,0)5(,5/)5(4:105

0)0(,0)0(,04:50

232333

2222

1111

yyyyyyt

yytyyt

yyyyt

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Example 2: First IVP (8 of 12)

• Consider the first initial value problem

• From a physical point of view, the system is initially at rest,

and since there is no external forcing, it remains at rest.

• Thus the solution over [0, 5) is y1 = 0, and this can be verified

analytically as well. See graphs below.

50;0)0(,0)0(,04 1111 tyyyy

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Example 2: Second IVP (9 of 12)

• Consider the second initial value problem

• Using methods of Chapter 3, the solution has the form

• Thus the solution is an oscillation about the line (t – 5)/20,

over the time interval (5, 10). See graphs below.

105;0)5(,0)5(,5/)5(4 2222 tyytyy

4/120/2sin2cos 212 ttctcy

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Example 2: Third IVP (10 of 12)

• Consider the third initial value problem

• Using methods of Chapter 3, the solution has the form

• Thus the solution is an oscillation about y = 1/4, for t > 10.

See graphs below.

10);10()10(),10()10(,14 232333 tyyyyyy

4/12sin2cos 213 tctcy

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Example 2: Amplitude (11 of 12)

• Recall that the solution to the initial value problem is

• To find the amplitude of the eventual steady oscillation, we

locate one of the maximum or minimum points for t > 10.

• Solving y' = 0, the first maximum is (10.642, 0.2979).

• Thus the amplitude of the oscillation is about 0.0479.

ttththtuthtuty 2sin8

1

4

1)( ,)10()()5()(

5

1)( 105

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Example 2: Solution Smoothness (12 of 12)

• Our solution is

• In this example, the forcing function g is continuous but g' is

discontinuous at t = 5 and t = 10.

• It follows that and its first two derivatives are continuous

everywhere, but has discontinuities at t = 5 and t = 10 that

match the discontinuities of g' at t = 5 and t = 10.

ttththtuthtuty 2sin8

1

4

1)( ,)10()()5()(

5

1)( 105

ff '''

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Boyce/DiPrima/Meade 11th ed, Ch 6.5:

Impulse Functions

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley

& Sons, Inc.

• In some applications, it is necessary to deal with phenomena of

an impulsive nature.

• For example, an electrical circuit or mechanical system subject

to a sudden voltage or force g(t) of large magnitude that acts

over a short time interval about t0. The differential equation

will then have the form

small. is 0 and

otherwise,0

,big)(

where

),(

00

ttttg

tgcyybya

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Measuring Impulse

• In a mechanical system, where g(t) is a force, the total impulse

of this force is measured by the integral

• Note that if g(t) has the form

then

• In particular, if c = 1/(2 ), then I( ) = 1 (independent of ).

0

0

)()()(t

tdttgdttgI

otherwise,0

,)(

00 tttctg

0,2)()()(0

0

cdttgdttgI

t

t

t t t

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Unit Impulse Function

• Suppose the forcing function has the form

• Then as we have seen, I( ) = 1.

• We are interested acting over

shorter and shorter time intervals

(i.e., ). See graph on right.

• Note that gets taller and narrower

as . Thus for t ≠ 0, we have

dt (t) =

1

2t, -t < t < t

0, otherwise

ì

íï

îï

1)(limand,0)(lim00

Itd

t

dt (t)

dt (t)

dt (t)

t ® 0

t ® 0

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Dirac Delta Function

• Thus for t ≠ 0, we have

• The unit impulse function is defined to have the properties

• The unit impulse function is an example of a generalized

function and is usually called the Dirac delta function.

• In general, for a unit impulse at an arbitrary point t0,

1)(limand,0)(lim00

Itd

1)(and,0for 0)(

dtttt

1)(and,for 0)( 000

dttttttt

d

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Laplace Transform of (1 of 2)

• The Laplace Transform of is defined by

and thus

0,)(lim)( 000

0

tttdLttL

00

0

0

00

0

0

0

0

)cosh(lim

)sinh(lim

2lim

2

1lim

2lim

2

1lim)(lim)(

0

00

00

000

00

stst

stssst

tsts

t

t

st

t

t

stst

es

sse

s

se

ee

s

e

eess

e

dtedtttdettL

d

d

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Laplace Transform of (2 of 2)

• Thus the Laplace Transform of is

• For Laplace Transform of at t0= 0, take limit as follows:

• For example, when t0 = 10, we have L{ (t –10)} = e–10s.

0,)( 000

tettL

st

1lim)(lim)( 0

00 00

0

st

tettdLtL

d

d

d

d

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Product of Continuous Functions and

• The product of the delta function and a continuous function f

can be integrated, using the mean value theorem for integrals:

• Thus

)(

*)(lim

)* where( *)(22

1lim

)(2

1lim

)()(lim)()(

0

0

000

0

00

0

0

0

tf

tf

ttttf

dttf

dttfttddttftt

t

t

)()()( 00 tfdttftt

d

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Example 1: Initial Value Problem (1 of 3)

• Consider the solution to the initial value problem

• Then

• Letting Y(s) = L{y},

• Substituting in the initial conditions, we obtain

or

)}5({}{2}{}{2 tLyLyLyL

sesYyssYysysYs 52 )(2)0()()0(2)0(2)(2

sesYss 52 )(22

22)(

2

5

ss

esY

s

0)0(,0)0(),5(22 yytyyy

)5( t

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Example 1: Solution (2 of 3)

• We have

• The partial fraction expansion of Y(s) yields

and hence

22)(

2

5

ss

esY

s

16/154/1

4/15

152)(

2

5

s

esY

s

5

4

15sin)(

15

2)( 4/5

5 tetuty t

5 10 15 20t

0.2

0.1

0.1

0.2

0.3

0.4

0.5

y t

Plot of the Solution

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Example 1: Solution Behavior (3 of 3)

• With homogeneous initial conditions at t = 0 and no external

excitation until t = 5, there is no response on (0, 5).

• The impulse at t = 5 produces a decaying oscillation that

persists indefinitely.

• Response is continuous at t = 5 despite singularity in forcing

function. Since y' has a jump discontinuity at t = 5, y'' has an

infinite discontinuity there. Thus a singularity in the forcing

function is balanced by a corresponding singularity in y''.

5 10 15 20t

0.2

0.1

0.1

0.2

0.3

0.4

0.5

y t

Plot of the Solution )5( t

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Boyce/DiPrima/Meade 11th ed, Ch 6.6:

The Convolution Integral

Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John

Wiley & Sons, Inc.

• Sometimes it is possible to write a Laplace transform H(s) as

H(s) = F(s)G(s), where F(s) and G(s) are the transforms of

known functions f and g, respectively.

• In this case we might expect H(s) to be the transform of the

product of f and g. That is, does

H(s) = F(s)G(s) = L{f }L{g} = L{f g}?

• On the next slide we give an example that shows that this

equality does not hold, and hence the Laplace transform

cannot in general be commuted with ordinary multiplication.

• In this section we examine the convolution of f and g, which

can be viewed as a generalized product, and one for which the

Laplace transform does commute.

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Observation

• Let f (t) = 1 and g(t) = sin(t). Recall that the Laplace

Transforms of f and g are

• Thus

and

• Therefore for these functions it follows that

1

1sin)()(

2

stLtgtfL

1

1sin)(,

11)(

2

stLtgL

sLtfL

)()()()( tgLtfLtgtfL

1

1)()(

2

sstgLtfL

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Theorem 6.6.1

• Suppose F(s) = L{f (t)} and G(s) = L{g(t)} both exist for

s > a ≥ 0. Then H(s) = F(s)G(s) = L{h(t)} for s > a, where

• The function h(t) is known as the convolution of f and g and

the integrals above are known as convolution integrals.

• Note that the equality of the two convolution integrals can be

seen by making the substitution u = t – .

• The convolution integral defines a “generalized product” and

can be written as h(t) = ( f *g)(t). See text for more details.

tt

dtgtfdgtfth00

)()()()()(

x

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Theorem 6.6.1 Proof Outline

)(

)()(

)()(

)()(

)()()(

)()(

)()()()(

00

0 0

0

0

0 0

)(

0 0

thL

dtdgtfe

dtdgtfe

ddttfge

utdttfedg

duufedg

dgeduufesGsF

tst

tst

st

st

us

ssu

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Example 1: Find Inverse Transform (1 of 2)

• Find the inverse Laplace Transform of H(s), given below.

• Solution: Let F(s) = 1/s2 and G(s) = a/(s2 + a2), with

• Thus by Theorem 6.6.1,

)sin()()(

)()(

1

1

atsGLtg

tsFLtf

)()(

222 ass

asH

t

datthsHL0

1 )sin()()()(

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Example 1: Solution h(t) (2 of 2)

• We can integrate to simplify h(t), as follows.

2

2

2

000

000

)sin(

)sin(11

)sin(1

)][cos(1

1)cos(1

)cos(1

)cos(1

)cos(1

)sin()sin()sin()()(

a

atat

ata

ta

ata

atta

atta

daa

aa

ata

dadatdatth

ttt

ttt

t

datthsHL0

1 )sin()()()(

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Example 2: Initial Value Problem (1 of 4)

• Find the solution to the initial value problem

• Solution:

• or

• Letting Y(s) = L{y}, and substituting in initial conditions,

• Thus

)}({}{4}{ tgLyLyL

)(}{4)0()0(}{2 sGyLysyyLs

)(13)(42 sGssYs

4

)(

4

13)(

22

s

sG

s

ssY

1)0(,3)0(),(4 yytgyy

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Example 2: Solution (2 of 4)

• We have

• Thus

• Note that if g(t) is given, then the convolution integral can be

evaluated.

dgttttyt

)()(2sin2

12sin

2

12cos3)(

0

)(4

2

2

1

4

2

2

1

43

4

)(

4

13)(

222

22

sGsss

s

s

sG

s

ssY

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Example 2:

Laplace Transform of Solution (3 of 4)

• Recall that the Laplace Transform of the solution y is

• Note depends only on system coefficients and initial

conditions, while depends only on system coefficients

and forcing function g(t).

• Further, = L–1[ ] solves the homogeneous IVP

while = L–1{ } solves the nonhomogeneous IVP

)()(4

)(

4

13)(

22sΨsΦ

s

sG

s

ssY

1)0(,3)0(),(4 yytgyy

1)0(,3)0(,04 yyyy

0)0(,0)0(),(4 yytgyy

F(s)

Y(s)

f(t) F(s)

y (t) Y(s)

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Example 2: Transfer Function (4 of 4)

• Examining more closely,

• The function H(s) is known as the transfer function, and

depends only on system coefficients.

• The function G(s) depends only on external excitation g(t)

applied to system.

• If G(s) = 1, then g(t) = and hence h(t) = L–1{H(s)} solves

the nonhomogeneous initial value problem

• Thus h(t) is response of system to unit impulse applied at t = 0,

and hence h(t) is called the impulse response of system.

4

1)( where),()(

4

)()(

22

ssHsGsH

s

sGsΨ

0)0(,0)0(),(4 yytyy

Y(s)

d(t)