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Ch 6.1: Definition of Laplace TransformMany 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
∞≤<≤∞= ∫ βαβ
α,)(),()( dttftsKsF
The Laplace TransformLet f be a function defined for t ≥ 0, and satisfies certain conditions to be named later. The Laplace Transform of f is defined as
Thus 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
Example 1Consider the following improper integral.
We can evaluate this integral as follows:
Note that if s = 0, then est = 1. Thus the following two cases hold:
( )1lim1limlim0
00−===
∞→∞→∞→
∞
∫∫ sb
b
bst
b
b st
b
st ess
edtedte
∫∞
0dtest
0. if,diverges
and0; if,1
0
0
≥
<−=
∫
∫∞
∞
sdte
ss
dte
st
st
Example 2Consider the following improper integral.
We can evaluate this integral using integration by parts:
Since this limit diverges, so does the original integral.
[ ]( )[ ]1cossinlim
cossinlim
sinsinlim
coslimcos
00
00
00
−+=
+=
⎥⎦⎤
⎢⎣⎡ −=
=
∞→
∞→
∞→
∞→
∞
∫
∫∫
bsbsb
tstst
tdtstst
tdtsttdtst
b
bb
b
bb
b
b
b
∫∞
0cos tdtst
Piecewise Continuous FunctionsA function f is piecewise continuous on an interval [a, b] if this interval can be partitioned by a finite number of pointsa = 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
K
K
=∞<
−=∞<
−+
+
→
→
Example 3Consider the following piecewise-defined function f.
From this definition of f, and from the graph of f below, we see that f is piecewise continuous on [0, 3].
⎪⎩
⎪⎨
⎧
≤<+≤<−≤≤
=32121,310,
)(
2
tttttt
tf
Example 4Consider the following piecewise-defined function f.
From this definition of f, and from the graph of f below, we see that f is not piecewise continuous on [0, 3].
( )⎪⎩
⎪⎨
⎧
≤<≤<−≤≤+
= −
32,421,210,1
)( 1
2
ttttt
tf
Theorem 6.1.2Suppose 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 5Let f (t) = 1 for t ≥ 0. Then the Laplace transform F(s) of f is:
{ }
0,1
lim
lim
1
0
0
0
>=
−=
=
=
−
∞→
−
∞→
∞ −
∫
∫
ss
se
dte
dteL
bst
b
b st
b
st
Example 6Let f (t) = eat for t ≥ 0. Then the Laplace transform F(s) of f is:
{ }
asas
ase
dte
dteeeL
btas
b
b tas
b
atstat
>−
=
−−=
=
=
−−
∞→
−−
∞→
∞ −
∫
∫
,1
lim
lim
0
)(
0
)(
0
Example 7Let 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
asFsFas
a
ateasaate
as
a
ateas
a
ateasaate
atdteatdteatLsF
b stbst
b
b st
b
b stbst
b
b st
b
st
Linearity of the Laplace TransformSuppose 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
020121
tgLctfLc
dttgecdttfectgctfcL stst
+=
+=+ ∫∫∞ −∞ −
Example 8Let 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
122
5)4sin(35
)4sin(35)}({)(
2
2
2
>+
−+
=
−=
−=
=
−
−
sss
tLeLteL
tfLsF
t
t
Ch 6.2: Solution of Initial Value Problems 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 6.2.1Suppose 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 −=′
⎥⎦⎤
⎢⎣⎡ +−=
⎥⎦⎤
⎢⎣⎡ −−=′
∫
∫∫−−
∞→
−−
∞→
−
∞→
b stsb
b
b stbst
b
b st
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 fand f ' of Theorem 6.2.1. We then obtain
Similarly, we can derive an expression for L{f (n)}, provided fand its derivatives satisfy suitable conditions. This result is given in Corollary 6.2.2
{ } { }{ }[ ]{ } )0()0()(
)0()0()()0()()(
2 fsftfLsfftfsLs
ftfsLtfL
′−−=
′−−=
′−′=′′
{ } { } )0()()( ftfsLtfL −=′
Corollary 6.2.2Suppose 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 L
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.
( ) ( ) 30,20,065 =′==+′+′′ yyyyy
( )( ) 032065)( 2 =++⇔=++⇒= rrrrety rt
tt ececty 32
21)( −− +=
7,93322
2121
21 −==⇒⎭⎬⎫
=−−=+
cccccc
tt eety 32 79)( −− −=
Example 1: Laplace Tranform Method (2 of 4)Assume that our IVP has a solution φ and that φ'(t) and φ''(t) 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
( ) ( ) 30,20,065 =′==+′+′′ yyyyy
0}0{}{6}{5}{}65{ ==+′+′′=+′+′′ LyLyLyLyyyL
[ ] [ ] 0}{6)0(}{5)0()0(}{2 =+−+′−− yLyysLysyyLs
( ) ( ) 0)0()0(5)(652 =′−+−++ yyssYss
( ) ( ) 0352)(652 =−+−++ ssYss
( )( )23132)(}{++
+==
ssssYyL
Using partial fraction decomposition, Y(s) can be rewritten:
Thus
( )( ) ( ) ( )( ) ( )
9,71332,2
)32()(13232132
2323132
=−==+=+
+++=++++=+
++
+=
+++
BABABA
BAsBAssBsAs
sB
sA
sss
( ) ( )29
37)(}{
++
+−==
sssYyL
Example 1: Partial Fractions (3 of 4)
Recall from Section 6.1:
Thus
Recalling Y(s) = L{y}, we have
and hence
( ) ( ) ,2},{9}{72
93
7)( 23 −>+−=+
++
−= −− seLeLss
sY tt
{ } asas
dtedteesFeL tasatstat >−
==== ∫∫∞ −−∞ − ,1)(
0
)(
0
}97{}{ 23 tt eeLyL −− +−=
tt eety 23 97)( −− +−=
Example 1: Solution (4 of 4)
General Laplace Transform MethodConsider the constant coefficient equation
Assume that this equation has a solution y = φ(t), and that φ'(t) and φ''(t) satisfy the conditions of Corollary 6.2.2. Then
If we let Y(s) = L{y} and F(s) = L{ f }, then
)(tfcyybya =+′+′′
)}({}{}{}{}{ tfLycLybLyaLcyybyaL =+′+′′=+′+′′
[ ] [ ]( ) ( )
( )cbsas
sFcbsasyaybassY
sFyaybassYcbsassFycLyysLbysyyLsa
+++
++′++
=
=′−+−++
=+−+′−−
22
2
2
)()0()0()(
)()0()0()()(}{)0(}{)0()0(}{
Algebraic ProblemThus the differential equation has been transformed into the the algebraic equation
for which we seek y = φ(t) such that L{φ(t)} = 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
sFcbsasyaybassY
+++
++′++
= 22
)()0()0()(
Characteristic PolynomialUsing 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 φrequires 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
sFcbsasyaybassY
+++
++′++
= 22
)()0()0()(
( ) ( ) 00 0,0),( yyyytfcyybya ′=′==+′+′′
Inverse ProblemThe main difficulty in using the Laplace transform method is determining the function y = φ(t) such that L{φ(t)} = Y(s). This is an inverse problem, in which we try to find φ such that φ(t) = 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.
Linearity of the Inverse TransformFrequently 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+++= L
{ } { })()(,,)()( 11
11 sFLtfsFLtf nn
−− == K
)()()()( 21 tftftftf n+++= L
{ } { } { })()()()( 11
11 sFLsFLsFLtf n−−− ++== L
Example 2
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
ssY 2)( =
⎟⎠⎞
⎜⎝⎛==
sssY 122)(
{ } ( ) 212122)( 111 ==⎭⎬⎫
⎩⎨⎧=
⎭⎬⎫
⎩⎨⎧= −−−
sL
sLsYL
2)( =ty
Example 3
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
53)(−
=s
sY
⎟⎠⎞
⎜⎝⎛
−=
−=
513
53)(
sssY
{ } tes
Ls
LsYL 5111 35
135
3)( =⎭⎬⎫
⎩⎨⎧
−=
⎭⎬⎫
⎩⎨⎧
−= −−−
tety 53)( =
Example 4
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
4
6)(s
sY =
44
!36)(ss
sY ==
{ } 34
11 !3)( ts
LsYL =⎭⎬⎫
⎩⎨⎧= −−
3)( tty =
Example 5
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛== 333
!24!2!2
88)(sss
sY
{ } 23
13
11 4!24!24)( ts
Ls
LsYL =⎭⎬⎫
⎩⎨⎧=
⎭⎬⎫
⎩⎨⎧
⎟⎠⎞
⎜⎝⎛= −−−
24)( tty =
3
8)(s
sY =
Example 6
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
⎥⎦⎤
⎢⎣⎡
++⎥⎦
⎤⎢⎣⎡
+=
++
=9
331
94
914)( 222 ss
ss
ssY
{ } tts
Ls
sLsYL 3sin313cos4
93
31
94)( 2
12
11 +=⎭⎬⎫
⎩⎨⎧
++
⎭⎬⎫
⎩⎨⎧
+= −−−
ttty 3sin313cos4)( +=
914)( 2 +
+=
sssY
Example 7
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
⎥⎦⎤
⎢⎣⎡
−+⎥⎦
⎤⎢⎣⎡
−=
−+
=9
331
94
914)( 222 ss
ss
ssY
{ } tts
Ls
sLsYL 3sinh313cosh4
93
31
94)( 2
12
11 +=⎭⎬⎫
⎩⎨⎧
−+
⎭⎬⎫
⎩⎨⎧
−= −−−
ttty 3sinh313cosh4)( +=
914)( 2 −
+=
sssY
Example 8
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
{ }( )
tets
LsYL −−− −=⎭⎬⎫
⎩⎨⎧
+−= 2
311 5
1!25)(
tetty −−= 25)(
( )3110)(+
−=s
sY
( ) ( ) ( ) ⎥⎦⎤
⎢⎣
⎡
+−=⎥
⎦
⎤⎢⎣
⎡
+−=
+−= 333 1
!251!2
!210
110)(
ssssY
Example 9
For the function Y(s) below, we find y(t) = L-1{Y(s)} by using a partial fraction expansion, as follows.
tt eetyss
sY
BAABBA
ABsBAssBsAs
sB
sA
sss
ssssY
34
2
710
711)(
31
710
41
711)(
7/10,7/11134,3
)34()(13)4()3(13
34)3)(4(13
1213)(
+=⇒⎥⎦⎤
⎢⎣⎡−
+⎥⎦⎤
⎢⎣⎡+
=
===−=+
−++=+++−=+
−+
+=
−++
=−++
=
−
Example 10
For the function Y(s) below, we find y(t) = L-1{Y(s)} by completing the square in the denominator and rearranging the numerator, as follows.
Using Table 6.1, we obtain
( ) ( )( )( ) ( ) ( ) ⎥
⎦
⎤⎢⎣
⎡
+−+⎥
⎦
⎤⎢⎣
⎡
+−−
=+−+−
=
+−+−
=++−
−=
+−−
=
1312
1334
13234
132124
196104
106104)(
222
222
sss
ss
ss
sss
ssssY
tetety tt sin2cos4)( 33 +=
Example 11: Initial Value 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
( ) ( ) 60,00,0258 =′==+′−′′ yyyyy
[ ] [ ] 0}{25)0(}{8)0()0(}{2 =+−−′−− yLyysLysyyLs
( ) ( ) 0)0()0(8)(2582 =′−−−+− yyssYss
2586)(}{ 2 +−
==ss
sYyL
( ) 06)(2582 =−+− sYss
Completing the square, we obtain
Thus
Using Table 6.2.1, we have
Therefore our solution to the initial value problem is
( ) 91686
2586)( 22 ++−
=+−
=ssss
sY
( ) ⎥⎦
⎤⎢⎣
⎡
+−=
9432)( 2s
sY
{ } tesYL t 3sin2)( 41 =−
tety t 3sin2)( 4=
Example 11: Solution (2 of 2)
Example 12: 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 12: Solution (2 of 2)
41)4)(1(682)( 2222
23
++
+++
=+++++
=s
DCss
BAsss
ssssY
( )( ) ( )( ))4()4()()(
1468223
2223
DBsCAsDBsCAsDCssBAssss
+++++++=
+++++=+++
43/2
13/5
12)( 222 +
−+
++
=sss
ssY
tttty 2sin31sin
35cos2)( −+=
Ch 6.3: Step Functions 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 definitionLet c ≥ 0. The unit step function, or Heaviside function, is defined by
A negative step can be represented by
⎩⎨⎧
≥<
=ctct
tuc ,1,0
)(
⎩⎨⎧
≥<
=−=ctct
tuty c ,0,1
)(1)(
Example 1Sketch the graph of
Solution: Recall that uc(t) is defined by
Thus
and hence the graph of h(t) is a rectangular pulse.
0),()()( 2 ≥−= ttututh ππ
⎩⎨⎧
≥<
=ctct
tuc ,1,0
)(
⎪⎩
⎪⎨
⎧
∞<≤<≤<≤
=t
tt
thπ
πππ
202,1
0,0)(
Laplace Transform of Step FunctionThe Laplace Transform of uc(t) is
{ }
se
se
se
es
dte
dtedttuetuL
cs
csbs
b
b
c
st
b
b
c
st
b
c
stc
stc
−
−−
∞→
−
∞→
−
∞→
∞ −∞ −
=
⎥⎦
⎤⎢⎣
⎡+−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡−==
==
∫
∫∫
lim
1limlim
)()(0
Translated FunctionsGiven 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.
⎩⎨⎧
≥−<
=ctctfct
tg),(
,0)(
Example 2Sketch the graph of
Solution: Recall that uc(t) is defined by
Thus
and hence the graph of g(t) is a shifted parabola.
.0,)( where),()1()( 21 ≥=−= tttftutftg
⎩⎨⎧
≥<
=ctct
tuc ,1,0
)(
⎩⎨⎧
≥−<≤
=1,)1(
10,0)( 2 tt
ttg
Theorem 6.3.1If 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 tdirection corresponds to a multiplication of F(s) by e-cs.
{ } { } )()()()( sFetfLectftuL cscsc
−− ==−
{ })()()( 1 sFeLctftu csc
−−=−
Theorem 6.3.1: Proof OutlineWe 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
cst
c
−
∞ −−
∞ +−−=
∞ −
∞ −
=
=
=
−=
−=−
∫∫
∫∫
{ } )()()( sFectftuL csc
−=−
Example 3Find the Laplace transform of
Solution: Note that
Thus
)()1()( 12 tuttf −=
⎩⎨⎧
≥−<≤
=1,)1(
10,0)( 2 tt
ttf
{ } { } { } 322
12)1)(()(setLettuLtfL
ss
−− ==−=
Example 4Find L{ f (t)}, where f is defined by
Note that f (t) = sin(t) + uπ/4(t) cos(t - π/4), and
⎩⎨⎧
≥−+<≤
=4/),4/cos(sin
4/0,sin)(
πππ
ttttt
tf
{ } { } { }{ } { }
11
111
cossin
)4/cos()(sin)(
2
4/
24/
2
4/4/
++
=
++
+=
+=
−+=
−
−
−
sse
sse
s
tLetL
ttuLtLtfL
s
s
s
π
π
π
π π
Example 5Find L-1{F(s)}, where
Solution: 4
73)(sesF
s−+=
( )373
471
41
4
71
41
7)(61
21
!361!3
21
3)(
−+=
⎭⎬⎫
⎩⎨⎧ ⋅+
⎭⎬⎫
⎩⎨⎧=
⎭⎬⎫
⎩⎨⎧
+⎭⎬⎫
⎩⎨⎧=
−−−
−−−
ttut
seL
sL
seL
sLtf
s
s
Theorem 6.3.2If 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 4Find the inverse transform of
To solve, we first complete the square:
Since
it follows that
521)( 2 ++
+=
ssssG
( )( )
( ) 411
4121
521)( 222 ++
+=
++++
=++
+=
ss
sss
ssssG
{ } { } ( )tetfesFLsGL tt 2cos)()1()( 11 −−−− ==+=
{ } ( )ts
sLsFLtf 2cos4
)()( 211 =
⎭⎬⎫
⎩⎨⎧
+== −−
Ch 6.4: Differential Equations with Discontinuous Forcing Functions 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 50,0205,1
)()()(
where0)0(,0)0(),(22
205 ttt
tututg
yytgyyy
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 −=+′+′′
[ ] [ ]seeyLyysLysyyLs
ss 2052 }{2)0(}{)0(2)0(2}{2
−− −=+−+′−−
( ) ( ) ( ) seeyyssYss ss 2052 )0(2)0(12)(22 −− −=′−+−++
Example 1: Laplace Transform (2 of 12)
( ) ( ) seesYss ss 2052 )(22 −− −=++
( )( )22
)( 2
205
++−
=−−
ssseesY
ss
0)0(,0)0(),()(22 205 =′=−=+′+′′ yytutuyyy
We have
where
If we let h(t) = L-1{H(s)}, then
by Theorem 6.3.1.
)20()()5()()( 205 −−−== thtuthtuty φ
Example 1: Factoring Y(s) (3 of 12)
( )( ) ( ) )(
22)( 205
2
205
sHeessseesY ss
ss−−
−−
−=++
−=
( )221)( 2 ++
=sss
sH
Thus we examine H(s), as follows.
This partial fraction expansion yields the equations
Thus
Example 1: Partial Fractions (4 of 12)
( ) 22221)( 22 ++
++=
++=
ssCBs
sA
ssssH
2/1,1,2/112)()2( 2
−=−==⇒=++++
CBAAsCAsBA
222/12/1)( 2 ++
+−=
sss
ssH
Completing the square,
Example 1: Completing the Square (5 of 12)
( )( )
( ) ⎥⎦
⎤⎢⎣
⎡
++++
−=
⎥⎦
⎤⎢⎣
⎡
+++
−=
⎥⎦⎤
⎢⎣⎡
++++
−=
⎥⎦⎤
⎢⎣⎡
+++
−=
+++
−=
16/154/14/14/1
212/1
16/154/12/1
212/1
16/1516/12/2/1
212/1
12/2/1
212/1
222/12/1)(
2
2
2
2
2
ss
s
ss
s
sss
s
sss
s
sss
ssH
Thus
and hence
For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then
Example 1: Solution (6 of 12)
( )( )
( )( ) ( ) ⎥
⎦
⎤⎢⎣
⎡
++−⎥
⎦
⎤⎢⎣
⎡
+++
−=
⎥⎦
⎤⎢⎣
⎡
++++
−=
16/154/14/15
1521
16/154/14/1
212/1
16/154/14/14/1
212/1)(
22
2
sss
s
ss
ssH
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−== −−− tetesHLth tt
415sin
1521
415cos
21
21)}({)( 4/4/1
)20()()5()()( 205 −−−= thtuthtutφ
Thus the solution to the initial value problem is
The graph of this solution is given below.
Example 1: Solution Graph (7 of 12)
( ) ( )415sin152
1415cos21
21)(
where),20()()5()()(
4/4/
205
teteth
thtuthtut
tt −− −−=
−−−=φ
The solution to original IVP can be viewed as a composite of three separate solutions to three separate IVPs:
Example 1: Composite IVPs (8 of 12)
)20()20(),20()20(,022:200)5(,0)5(,122:205
0)0(,0)0(,022:50
2323333
22222
11111
yyyyyyytyyyyytyyyyyt
′=′==+′+′′>=′==+′+′′<<=′==+′+′′<≤
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.
Example 1: First IVP (9 of 12)
50;0)0(,0)0(,022 11111 <≤=′==+′+′′ tyyyyy
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.
Example 1: Second IVP (10 of 12)
205;0)5(,0)5(,122 22222 <<=′==+′+′′ tyyyyy
( ) ( ) 2/14/15sin4/15cos 4/2
4/12 ++= −− tectecy tt
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.
Example 1: Third IVP (11 of 12)
( ) ( )4/15sin4/15cos 4/2
4/13 tectecy tt −− +=
20);20()20(),20()20(,022 2323333 >′=′==+′+′′ tyyyyyyy
Our solution is
It can be shown that φ and φ' are continuous at t = 5 and t = 20, and φ'' 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.
Example 1: Solution Smoothness (12 of 12)
)20()()5()()( 205 −−−= thtuthtutφ
-.5072)(lim,-.0072)(lim
2/1)(lim,0)(lim
2020
55
≅′′≅′′
=′′=′′
+−
+−
→→
→→
tt
tt
tt
tt
φφ
φφ
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 = ψ(t) is a solution, then ψ and ψ ' are continuous on (a, b) but ψ '' 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).
Smoothness of Solution in General
)()()( tgytqytpy =+′+′′
Find the solution to the initial value problem
The graph of forcing function g(t) is given on right, and isknown as ramp loading.
( )⎪⎩
⎪⎨
⎧
≥<≤−<≤
=−
−−
=
=′==+′′
10,11055550,0
510)(
55)()(
where0)0(,0)0(),(4
105
tttt
ttuttutg
yytgyy
Example 2: Initial Value Problem (1 of 12)
Assume that this ODE has a solution y = φ(t) and that φ'(t) and φ''(t) 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(}{
seeyLysyyLs
ss −− −=+′−−
Example 2: Laplace Transform (2 of 12)
( ) ( ) 21052 5)(4 seesYs ss −− −=+
( )( )45
)( 22
105
+−
=−−
sseesY
ss
0)0(,0)0(,510)(
55)(4 105 =′=
−−
−=+′′ yyttuttuyy
We have
where
If we let h(t) = L-1{H(s)}, then
by Theorem 6.3.1.
[ ])10()()5()(51)( 105 −−−== thtuthtuty φ
Example 2: Factoring Y(s) (3 of 12)
( )( ) )(
545)(
105
22
105
sHeess
eesYssss −−−− −
=+
−=
( )41)( 22 +
=ss
sH
Thus we examine H(s), as follows.
This partial fraction expansion yields the equations
Thus
Example 2: Partial Fractions (4 of 12)
( ) 441)( 2222 +
+++=
+=
sDCs
sB
sA
sssH
4/1,0,4/1,0144)()( 23
−====⇒=+++++
DCBABAssDBsCA
44/14/1)( 22 +
−=ss
sH
Thus
and hence
For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then
Example 2: Solution (5 of 12)
( )ttsHLth 2sin81
41)}({)( 1 −== −
⎥⎦⎤
⎢⎣⎡
+−⎥⎦
⎤⎢⎣⎡=
+−=
42
811
41
44/14/1)(
22
22
ss
sssH
[ ])10()()5()(51)( 105 −−−== thtuthtuty φ
Thus the solution to the initial value problem is
The graph of this solution is given below.
Example 2: Graph of Solution (6 of 12)
[ ]
( )ttth
thtuthtut
2sin81
41)(
where,)10()()5()(51)( 105
−=
−−−=φ
The solution to original IVP can be viewed as a composite of three separate solutions to three separate IVPs (discuss):
Example 2: Composite IVPs (7 of 12)
)10()10(),10()10(,14:100)5(,0)5(,5/)5(4:105
0)0(,0)0(,04:50
232333
2222
1111
yyyyyytyytyytyyyyt
′=′==+′′>=′=−=+′′<<=′==+′′<≤
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.
Example 2: First IVP (8 of 12)
50;0)0(,0)0(,04 1111 <≤=′==+′′ tyyyy
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.
Example 2: Second IVP (9 of 12)
105;0)5(,0)5(,5/)5(4 2222 <<=′=−=+′′ tyytyy
( ) ( ) 4/120/2sin2cos 212 −++= ttctcy
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.
Example 2: Third IVP (10 of 12)
10);10()10(),10()10(,14 232333 >′=′==+′′ tyyyyyy
( ) ( ) 4/12sin2cos 213 ++= tctcy
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.
Example 2: Amplitude (11 of 12)
[ ] ( )ttththtuthtuty 2sin81
41)( ,)10()()5()(
51)( 105 −=−−−==φ
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 φ and its first two derivatives are continuous everywhere, but φ''' has discontinuities at t = 5 and t = 10 that match the discontinuities of g' at t = 5 and t = 10.
Example 2: Solution Smoothness (12 of 12)
[ ] ( )ttththtuthtuty 2sin81
41)( ,)10()()5()(
51)( 105 −=−−−== φ
Ch 6.5: Impulse Functions 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
>⎩⎨⎧ +<<−
=
=+′+′′
τ
ττ ttttg
tgcyybya
Measuring ImpulseIn a mechanical system, where g(t) is a force, the total impulseof this force is measured by the integral
Note that if g(t) has the form
then
In particular, if c = 1/(2τ), then I(τ) = 1 (independent of τ ).
∫∫+
−
∞
∞−==
τ
ττ 0
0
)()()(t
tdttgdttgI
⎩⎨⎧ +<<−
= otherwise,0
,)( 00 ττ tttc
tg
0,2)()()( 0
0
>=== ∫∫+
−
∞
∞−τττ
τ
τcdttgdttgI
t
t
Unit Impulse FunctionSuppose the forcing function dτ(t) has the form
Then as we have seen, I(τ) = 1.
We are interested dτ(t) acting over shorter and shorter time intervals (i.e., τ → 0). See graph on right.
Note that dτ(t) gets taller and narrower as τ → 0. Thus for t ≠ 0, we have
⎩⎨⎧ <<−
= otherwise,0
,21)(
ττττ
ttd
1)(limand,0)(lim00
==→→
ττττ
Itd
Dirac Delta FunctionThus for t ≠ 0, we have
The unit impulse function δ 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
==→→
ττττ
Itd
1)(and,0for 0)( =≠= ∫∞
∞−dtttt δδ
1)(and,for 0)( 000 =−≠=− ∫∞
∞−dttttttt δδ
Laplace Transform of δ (1 of 2)
The Laplace Transform of δ is defined by
and thus { } { } 0,)(lim)( 0000 >−=−
→tttdLttL ττ
δ
{ }
( ) ( )[ ]
00
00
00
0
0
0
0
)cosh(lim
)sinh(lim2
lim
21lim
2lim
21lim)(lim)(
0
00
00
00 000
stst
stssst
tstst
t
st
t
t
stst
es
sse
sseee
se
eess
e
dtedtttdettL
−
→
−
→
−−−
→
−−+−
→
+
−
−
→
+
−
−
→
∞ −
→
=⎥⎦⎤
⎢⎣⎡=
⎥⎦⎤
⎢⎣⎡=⎥
⎦
⎤⎢⎣
⎡ −=
+−=−
=
=−=− ∫∫
τ
ττ
τ
ττ
τδ
τ
τ
ττ
τ
ττ
τ
τ
ττ
τ
ττττ
Laplace Transform of δ (2 of 2)
Thus the Laplace Transform of δ is
For Laplace Transform of δ at t0= 0, take limit as follows:
For example, when t0= 10, we have L{δ(t -10)} = e-10s.
{ } 0,)( 000 >=− − tettL stδ
{ } { } 1lim)(lim)( 0
00 000==−= −
→→
st
tettdLtL
ττδ
Product of Continuous Functions and δThe product of the delta function and a continuous function fcan be integrated, using the mean value theorem for integrals:
Thus
[ ]
)(
*)(lim
)* where( *)(221lim
)(21lim
)()(lim)()(
0
0
000
0
000
0
0
tf
tf
ttttf
dttf
dttfttddttftt
t
t
=
=
+<<−=
=
−=−
→
→
+
−→
∞
∞−→
∞
∞−
∫
∫∫
τ
τ
τ
ττ
ττ
ττττ
τ
δ
)()()( 00 tfdttftt =−∫∞
∞−δ
Consider the solution to the initial value problem
Then
Letting Y(s) = L{y},
Substituting in the initial conditions, we obtain
or
)}7({}{2}{}{2 −=+′+′′ tLyLyLyL δ
[ ] [ ] sesYyssYysysYs 72 )(2)0()()0(2)0(2)(2 −=+−+′−−
Example 1: Initial Value Problem (1 of 3)
( ) sesYss 72 )(22 −=++
22)( 2
7
++=
−
ssesY
s
0)0(,0)0(),7(22 =′=−=+′+′′ yytyyy δ
We have
The partial fraction expansion of Y(s) yields
and hence
Example 1: Solution (2 of 3)
22)( 2
7
++=
−
ssesY
s
( ) ⎥⎦
⎤⎢⎣
⎡
++=
−
16/154/14/15
152)( 2
7
sesY
s
( ) ( )7415sin)(
1521)( 4/7
7 −= −− tetuty t
With homogeneous initial conditions at t = 0 and no external excitation until t = 7, there is no response on (0, 7). The impulse at t = 7 produces a decaying oscillation that persists indefinitely. Response is continuous at t = 7 despite singularity in forcing function. Since y' has a jump discontinuity at t = 7, y'' has an infinite discontinuity there. Thus singularity in forcing function is balanced by a corresponding singularity in y''.
Example 1: Solution Behavior (3 of 3)
Ch 6.6: The Convolution Integral 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.
Example 1Let 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 ≠
{ } { } ( )11)()( 2 +
=ss
tgLtfL
Theorem 6.6.1Suppose 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 - τ. The convolution integral defines a “generalized product” and can be written as h(t) = ( f *g)(t). See text for more details.
∫∫ −=−=tt
dtgtfdgtfth00
)()()()()( τττττ
Theorem 6.6.1 Proof Outline
{ })(
)()(
)()(
)()(
)()()(
)()(
)()()()(
00
0 0
0
0
0 0
)(
0 0
thL
dtdgtfe
dtdgtfe
ddttfge
utdttfedg
duufedg
dgeduufesGsF
tst
t st
st
st
us
ssu
=⎥⎦⎤
⎢⎣⎡ −=
−=
−=
+=−=
=
=
∫∫
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∞ −
∞ −
∞ ∞ −
∞ ∞ −
∞ ∞ +−
∞ ∞ −−
τττ
τττ
τττ
ττττ
ττ
ττ
τ
τ
τ
τ
Example 2Find the Laplace Transform of the function h given below.
Solution: Note that f (t) = t and g(t) = sin2t, with
Thus by Theorem 6.6.1,
42}2{sin)}({)(
1}{)}({)(
2
2
+===
===
stLtgLsG
stLtfLsF
∫ −=t
tdtth0
2sin)()( ττ
{ } ( )42)()()()( 22 +
===ss
sGsFsHthL
Example 3: Find Inverse Transform (1 of 2)
Find the inverse Laplace Transform of H(s), given below.
Solution: Let F(s) = 2/s2 and G(s) = 1/(s - 2), with
Thus by Theorem 6.6.1,
{ }{ } tesGLtg
tsFLtf21
1
)()(2)()(
==
==−
−
)2(2)( 2 −
=ss
sH
{ } ∫ −==− tdetthsHL
0
21 )(2)()( ττ τ
Example 3: Solution h(t) (2 of 2)
We can integrate to simplify h(t), as follows.
[ ] [ ]
21
21
21
21
1211
22)(2)(
2
222
222
0
2
0
2
0
2
0
2
0
2
0
2
−−=
−+−−=
⎥⎦⎤
⎢⎣⎡ −−−−=
⎥⎦⎤
⎢⎣⎡ −−=
−=−=
∫
∫∫∫
te
eettte
eetet
deete
dedetdetth
t
ttt
ttt
ttt
ttt
ττ
τττττ
τττ
τττ
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 =+′−−
Example 4: Initial Value Problem (1 of 4)
( ) )(13)(42 sGssYs +−=+
4)(
413)( 22 ++
+−
=s
sGs
ssY
1)0(,3)0(),(4 −=′==+′′ yytgyy
We have
Thus
Note that if g(t) is given, then the convolution integral can be evaluated.
τττ dgttttyt
)()(2sin212sin
212cos3)(
0∫ −+−=
Example 4: Solution (2 of 4)
)(4
221
42
21
43
4)(
413)(
222
22
sGsss
ss
sGs
ssY
⎥⎦⎤
⎢⎣⎡
++⎥⎦
⎤⎢⎣⎡
+−⎥⎦
⎤⎢⎣⎡
+=
++
+−
=
Recall that the Laplace Transform of the solution y is
Note Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t).
Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP
while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP
Example 4: Laplace Transform of Solution (3 of 4)
)()(4)(
413)( 22 sΨsΦ
ssG
sssY +=
++
+−
=
1)0(,3)0(),(4 −=′==+′′ yytgyy
1)0(,3)0(,04 −=′==+′′ yyyy
0)0(,0)0(),(4 =′==+′′ yytgyy
Examining Ψ (s) 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) = δ(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.
Example 4: Transfer Function (4 of 4)
41)( where),()(
4)()( 22 +
==+
=s
sHsGsHs
sGsΨ
0)0(,0)0(),(4 =′==+′′ yytyy δ
Consider the general initial value problem
This IVP is often called an input-output problem. The coefficients a, b, c describe properties of physical system, and g(t) is the input to system. The values y0 and y0' describe initial state, and solution y is the output at time t. Using the Laplace transform, we obtain
or
Input-Output Problem (1 of 3)
00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′
[ ] [ ] )()()0()()0()0()(2 sGscYyssYbysysYsa =+−+′−−
)()()()()( 2200 sΨsΦ
cbsassG
cbsasyaybassY +=
+++
++′++
=
We have
As before, Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t).
Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP
while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP
Laplace Transform of Solution (2 of 3)
00 )0(,)0(,0 yyyycyybya ′=′==+′+′′
0)0(,0)0(),( =′==+′+′′ yytgcyybya
00 )0(,)0(),( yyyytgcyybya ′=′==+′+′′
)()()()()( 2200 sΨsΦ
cbsassG
cbsasyaybassY +=
+++
++′++
=
Examining Ψ (s) more closely,
As before, H(s) is the transfer function, and depends only on system coefficients, while G(s) depends only on external excitation g(t) applied to system.Thus if G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous IVP
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, with
Transfer Function (3 of 3)
0)0(,0)0(),( =′==+′+′′ yytcyybya δ
cbsassHsGsH
cbsassGsΨ
++==
++= 22
1)( where),()()()(
{ } τττψ dgthsGsHLtt
)()( )()()(0
1 ∫ −== −