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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 £ ¥
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)(
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( )®¥
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
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
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
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
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
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
®¥
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
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
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
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
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
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
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.
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
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
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
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)(
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)
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
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
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
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)
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)
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)
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
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
• 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)(
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
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
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.
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)(
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
)(
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
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
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)(
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 ''
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 ''
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
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)
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
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
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
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
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
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
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
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
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
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 '''
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
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
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
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
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
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
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
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
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
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
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.
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
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
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
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()()()(
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()()()(
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
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
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)
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)
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