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1Engineering Math, 6. Laplace Transforms
CHAPTER 6. LAPLACE TRANSFORMS
2018.6서울대학교
조선해양공학과
서 유 택
※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다.
Seoul NationalUniv.
2Engineering Math, 6. Laplace Transforms
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
Laplace Transform (라플라스 변환):
- 적분 과정을 통해 주어진 함수를 새로운 함수로 변환하는 것
Inverse Transform (역 변환):
0
stF s f e f t dt
L
1 F f t L
Ex.1 Let f (t) = 1 when t ≥ 0. Find F(s).
Ex.2 Let f(t) = eat when t ≥ 0, where a is a constant. Find L( f ).
00
1 11 0st stf e dt e s
s s
L L
00
1 1
s a tat st ate e e dt ea s s a
L
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3Engineering Math, 6. Laplace Transforms
Theorem 1 Linearity of the Laplace Transform
The Laplace transform is a linear operation;
that is, for any functions f(t) and g(t) whose transforms exist and any constants a
and b, the transform of af(t) + bg(t) exists, and
af t bg t a f t b g t L L L
Ex.3 Find the transform of cosh at and sinh at.
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
2 2
1 1 1 1cosh , inh
2 2
1 1 1 1 cosh
2 2
1 sinh
2
at at at at at at
at at
at
at e e at e e e , es a s a
sat e e
s a s a s a
at e
L L
L
L
L L
L L 2 2
1 1 1
2
at ae
s a s a s a
s
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4Engineering Math, 6. Laplace Transforms
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
Seoul NationalUniv.
5Engineering Math, 6. Laplace Transforms
Theorem 2 First Shifting Theorem (제 1 이동정리), s-shifting (S-이동)
If f (t) has the transform F(s) (where s > k for some k),
then eatf (t) has the transform F(s ‒ a) (where s – a > k). In formulas,
or, if we take the inverse on both sides, . 1at -e f t F s a L
Proof) We obtain F(s‒a) by replacing s with s ‒ a in the integral, so that
ate f t F s a L
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
00
)( )}({)]([)()( tfedttfeedttfeasF atatsttas L
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6Engineering Math, 6. Laplace Transforms
1at -e f t F s a L Ex. 5 Formulas 11 and 12 in Table 6.1,
For instance, use these formulas to find the inverse of the transform.
2 22 2cos , sinat ats a
e t e ts a s a
L L
2
3 137
2 401
sf
s s
L
1 1 1
2 2 2
3 1 140 1 203 7 3cos 20 7sin 20
1 400 1 400 1 400
ts s
f e t ts s s
L L L
ate f t F s a L
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
(a = -1, = 20)
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7Engineering Math, 6. Laplace Transforms
Theorem 3 Existence Theorem for Laplace Transform
If f (t) is defined and piecewise continuous on every finite interval on the semi-
axis t ≥ 0
and satisfies | f(t) | ≤ Mekt for all t ≥ 0 and some constants M and k,
that is, f (t) does not grow too fast (“growth restriction (증가제한)”),
then the Laplace transform L ( f ) exists for all s > k.
6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting)
0 0
0
| ( ) | ( ) ( )st st
kt st
f e f t dt f t e dt
MMe e dt
s k
L
Proof) Piecewise continuous → e-st f (t) is
integrable over any finite interval on t-axis.
2
(1) ( ) cosh , (2) ( ) , (3) ( )n t f t t f t t f t e
2 3
0
1 !! 2! 3! !
n nt t n t
n
t x x te x e t n e
n n
1 1
cosh2 2
t t t t tt e e e e e
| f(t) | ≤ Mekt for all t ≥ 0 ?2
cosh , ! , ?t n t tt e t n e e
2t kte Me (not satisfied)
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8Engineering Math, 6. Laplace Transforms
Theorem 1 Laplace Transform of Derivatives (도함수의 라플라스 변환)
6.2 Transform of Derivatives and Integrals. ODEs
2
' 0
'' 0 ' 0 .
f s f f ,
f s f sf f
L L
L L
00 0{ ( )} ( ) ( ) ( )
(0) { ( )}
st st stf t e f t dt e f t s e f t dt
f s f t
L
L
{ ( )} (0)f t s f f L L
00 0{ ( )} ( ) ( ) ( )
(0) { ( )}
[ (0)] (0)
st st stf t e f t dt e f t s e f t dt
f s f t
s s f f f
L
L
L
2{ ( )} (0) (0)f t s f sf f L L
3 2{ ( )} (0) (0) (0)f t s f s f sf f L L
Proof)
000
000
)()()(
)()()(
)()()(
dttfedttfsetfe
dttfedttfsetfe
tfetfsetfe
ststst
ststst
ststst
dttftgtftgdttftg )()()()()()(
(integration by parts)
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9Engineering Math, 6. Laplace Transforms
6.2 Transform of Derivatives and Integrals. ODEs
Ex. 1 Let f (t) = tsinωt. Find the transform of f (t).
2
2 2
2 2
22 2
0 0,
' sin cos , ' 0 0,
'' 2 cos sin
'' 2 (0) (0)
2 sin
f
f t t t t f
f t t t t
sf f s f sf f
s
sf t t
s
L L L
L L
Theorem 2 Laplace Transform of Derivatives
Let be continuous for all t ≥ 0 and satisfy the growth restriction
| f(t) | ≤ Mekt.
Furthermore, let f (n) be continuous on every finite interval on the semi-axis t
≥ 0. 11 20 ' 0 0n nn n nf s f s f s f f
L L
)1(,....,, nfff
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10Engineering Math, 6. Laplace Transforms
Theorem 3 Laplace Transform of Integral (적분의 라플라스 변환)
Let F(s) denote the transform of a function f(t) which is piecewise
continuous for all t ≥ 0
and satisfies a growth restriction | f(t) | ≤ Mekt. Then, for s > 0, s > k, and t >
0,
1
0 0
1 1
t t
-f t F s f d F s , f d F ss s
L L L
6.2 Transform of Derivatives and Integrals. ODEs
t
dftg0
)()( Proof)
→ g(t) also satisfies a growth restriction.
g'(t) = f (t), except at points at which f(t) is discontinuous.
g'(t) is piecewise continuous on each finite interval.
g(0) = 0.
)}({)0()}({)}('{)}({ tgsgtgstgtf LLLL
0 0 0| ( ) | ( ) ( ) ( 1) ( 0)
t t tk kt ktM M
g t f d f d M e d e e kk k
1{ ( )} { ( )}g t f t
s L L
From Theorem 1
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11Engineering Math, 6. Laplace Transforms
Theorem 3 Laplace Transform of Integral
Ex. 3 Find the inverse of and . 2 2 2 2 2
1 1
s s s s
1
2 2
1 1sin t
s
L
1
0 0
1 1
t t
-f t F s f d F s , f d F ss s
L L L
6.2 Transform of Derivatives and Integrals. ODEs
22}{sin
stL
1
22 2
0
1 sin 1 1 cos
t
d ts s
L
1
2 2 32 2 2
0
1 1 sin 1 cos
tt t
ds s
L
2 2{cos }
st
s
L
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12Engineering Math, 6. Laplace Transforms
Differential Equations, Initial Value Problems:
Step 1 Setting up the subsidiary equation (보조방정식).
Step 2 Solution of the subsidiary equation by algebra.
Transfer Function (전달함수):
Solution of the transfer function:
Step 3 Inversion of Y to obtain y.
y(t) = L -1 (Y).
0 1'' ' , 0 , ' 0y ay by r t y K y K
, Y y R r L L
2 20 ' 0 0 0 ' 0s Y sy y a sY y bY R s s as b Y s a y y R s
22
2
1 1
1 1
2 4
Q ss as b
s a b a
0 ' 0Y s s a y y Q s R s Q s
6.2 Transform of Derivatives and Integrals. ODEs
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13Engineering Math, 6. Laplace Transforms
Ex. 4 Solve
Step 1
Step 2
Step 3
'' , 0 1, ' 0 1y y t y y
2 2
2 2
1 10 ' 0 1 1s Y sy y Y s Y s
s s
2 2 2 2 22 2
1 1 1 1 1 1 1 1
1 1 1 11
sQ Y s Q Q
s s s s s ss s
1 1 1 1
2 2
1 1 1sinh
1 1
ty t Y e t ts s s
L L L L
6.2 Transform of Derivatives and Integrals. ODEs
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14Engineering Math, 6. Laplace Transforms
Ex. 5 Solve
Step 1
Step 2
Step 3
6.2 Transform of Derivatives and Integrals. ODEs
0)0(16.0)0(.09 yyyyy
4
35)2/1(
08.0)2/1(16.0
9
)1(16.0
22
s
s
ss
sY
/2
/2
35 0.08 35( ) ( ) 0.16cos sin
2 235 / 2
0.16cos 2.96 0.027sin 2.96
t
t
y t Y e t t
e t t
L -1
)1(16.0)9(.0916.016.0 22 sYssYsYsYs
(a = -1/2, = 35
4)
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15Engineering Math, 6. Laplace Transforms
6.2 Transform of Derivatives and Integrals. ODEs
37 12 21 , (0) 3.5, (0) 10ty y y e y y
Q? Solve
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16Engineering Math, 6. Laplace Transforms
6.2 Transform of Derivatives and Integrals. ODEs
[ Relation of Time domain and s-Domain ]
Time domain analysis
(differential equation)
Problem Answer
Laplace Transform
S-domain analysis
(algebraic equation)
Inverse Transform
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17Engineering Math, 6. Laplace Transforms
The process of solution consists of three steps.
Step 1 The given ODE is transformed into an algebraic equation (“subsidiary
equation”).
Step 2 The subsidiary equation is solved by purely algebraic manipulations.
Step 3 The solution in Step 2 is transformed back, resulting in the solution of
the given problem.
IVP
Initial Value
Problem
(초기값문제)
AP
Algebraic
Problem
(보조방정식)
①Solving
AP
by Algebra
②Solution
of the
IVP
③
Solving an IVP by Laplace transforms
6.2 Transform of Derivatives and Integrals. ODEs
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18Engineering Math, 6. Laplace Transforms
6.2 Transform of Derivatives and Integrals. ODEs
Advantages of the Laplace Method
1. Solving a nonhomogeneous ODE does not require first solving the
homogeneous ODE.
2. Initial values are automatically taken care of.
3. Complicated inputs can be handled very efficiently.
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19Engineering Math, 6. Laplace Transforms
6.3 Unit Step Function (Heaviside Function).Second Shifting Theorem (t-Shifting)
Unit Step Function (단위 계단 함수)
• Unit Step Function (Heaviside function):
• Laplace transform of unit step function:
0
1
t au t a
t a
s
e
s
edtedtatueatu
as
at
st
a
stst
1)()(0
L
“on off function”
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20Engineering Math, 6. Laplace Transforms
6.3 Unit Step Function (Heaviside Function).Second Shifting Theorem (t-Shifting)
Unit Step Function
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21Engineering Math, 6. Laplace Transforms
6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting)
Theorem 1 Second Shifting Theorem (제 2이동 정리); Time Shifting
1 , as - asf t F s f t a u t a e F s f t a u t a e F s L L L
dtdatta ,thus,
at
at
atfatuatftf
if
if
)(
0)()()(ˆ
0
)(
0)()()( dfedfeesFe assasas
0 0
( ) ( )
ˆ( ) ( ) ( )
as st
a
st st
e F s e f t a dt
e f t a u t a dt e f t dt
Proof)
s
e
s
edtedtatueatu
as
at
st
a
stst
1)()(0
L
0
stF s f e f t dt
L
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22Engineering Math, 6. Laplace Transforms
Ex. 1 Write the following function using unit step functions and find its
transform.
Step 1 Using unit step functions:
2
2 0 1
12 2
cos2
t
tf t t
t t
21 1 12 1 1 1 cos
2 2 2f t u t t u t u t t u t
6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting)
𝜋
2
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23Engineering Math, 6. Laplace Transforms
Ex. 1 Write the following function using unit step functions and find its
transform.
Step 1 Using unit step functions:
Step 2
21 1 12 1 1 1 cos
2 2 2f t u t t u t u t t u t
6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting)
22
3 2
1 1 1 1 1 11 1 1 1
2 2 2 2
st u t t t u t es s s
L L
2 2 22 2
3 2
1 1 1 1 1 1 1
2 2 2 2 2 2 8 2 2 8
s
t u t t t u t es s s
L L
22
1 1 1 1cos sin
2 2 2 1
s
t u t t u t es
L L
2
2 23 2 3 2 2
2 2 1 1 1 1 1
2 2 8 1
s ss sf e e e e
s s s s s s s s s
L
asf t a u t a e F s L
1
!n
n
nt
s L
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24Engineering Math, 6. Laplace Transforms
Ex. 2 Find the inverse transform f(t) of
2 3
22 2 2 22
s s se e eF s
s s s
2 31 1 sin 1 1 sin 2 2 3 3
tf t t u t t u t t e u t
1
2 2
1 sin t
s
L
1 1 2
22
1 1
2
tt tes s
L L
6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting)
2 3
0 0 t 1
sin 1 t 2
0 2 t 3
3 t 3t
t
t e
1
2 2
21
2 2
1 sin 1 1
1 sin 2 2
s
s
et u t
s
et u t
s
L
L
asf t a u t a e F s L
22}{sin
stL
Time shifting
ate f t F s a L
S-shifting
where, sin 1 sin sin ,sin 2 sin 2 sint t t t t t
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25Engineering Math, 6. Laplace Transforms
6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting)
3 2 1 if 0 1 and 0 1
(0) 0, (0) 0
y y y t if t
y y
Q? Solve using Laplace transform.
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26Engineering Math, 6. Laplace Transforms
6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions
Dirac delta function or the unit impulse function:
Definition of the Dirac delta function
Laplace transform of the Dirac delta function
0 otherwise
t at a
0
1 lim
0 otherwisek k
k
a t a kkf t a t a f t a
0 0
11 1
a k
k
a
f t a dt dt t a dtk
1
kf t a u t a u t a kk
1 1
ks
a k sas as as
k
ef t a e e e t a e
ks ks
L L
0
0 00
1 1 1 1lim = lim = = ( ) 1
0
ks ks s ks
k kk
e e e des
ks s k s dk s
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27Engineering Math, 6. Laplace Transforms
Ex. 1 Mass Spring System Under a Square Wave
Step 1
Step 2
Step 3
6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions
0)0(0)0().2()1()(23 yytututryyy
tt eeFtf 2
2
1
2
1)()( 1-L
)()23(
1)()(
123 2
2
22 ssss eesss
sYees
YsYYs
2
2/1
)1(
12/1
)2)(1(
1
)23(
1)(
2
ssssssssssF
)2(
)21(
)10(
2/12/1
2/12/1
0
)2(2)1(2)2()1(
)1(2)1(
t
t
t
eeee
eetttt
tt
)2()2()1()1(
))()(( 2
tutftutf
esFesFy ss-1L
asf t a u t a e F s L
Time shifting
ate f t F s a L
S-shifting
asf t a u t a e F s L
Time shifting
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28Engineering Math, 6. Laplace Transforms
Ex. 2 Find the response of the system in Example 1 with the square wave
replaced by a unit impulse at time t = 1.
Initial value problem:
Subsidiary equation:
'' 3 ' 2 1 , 0 0, ' 0 0y y y t y y
2 3 2 ss Y sY Y e
1 1
1 2 1 2
sse
Y s es s s s
6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions
21 ( )= , ( )=at t te f t e g t e
s a
L
asf t a u t a e F s L
1
1 2 1
0 0 t 1
t 1t t
y t Ye e
L
( 1) ( 1) ( 1)( 1)f t u t g t u
( 1) 2( 1)( 1) ( 1)t te u t e u
ast a e L
Dirac’s Delta
Time shifting
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29Engineering Math, 6. Laplace Transforms
6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions
9 ( / 2), (0) 2, (0) 0y y t y y
Q? Solve
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30Engineering Math, 6. Laplace Transforms
Transform of a product is generally different from the product
of the transforms of the factors.
Convolution (합성곱):
6.5 Convolution. Integral Equations
0
t
f g t f g t d
)()()( gfgf LLL )()()( gffg LLL
Ex. Transform of a Convolution
Evaluate
Solution) ttgetf t sin)(,)(
0
2 2
sin( ) ( ) ( ) { } {sin }
1 1 1
1 1 ( 1)( 1)
tτ te t dτ F s G s e t
s s s s
L L L
1
2 2
2 2
!{ }
1{ }
{sin }
{cos }
n
n
at
nt
s
es a
ts
st
s
L
L
L
L
0
sin( ) *sint
te t d e t L L
f g f g L L L
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31Engineering Math, 6. Laplace Transforms
Convolution:
Theorem 1 Convolution Theorem
If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order,
then
6.5 Convolution. Integral Equations
0
t
f g t f g t d
f g f g L L L
Ex. 1 Let . Find h(t).
1H s
s a s
1 1
0
1 1 1, 1 1 1 1
t
at at a ate h t e e d es a s a
L L
( ) ( )h H sLs
gas
fsas
gfgfh1
)(,1
)(,11
)()()*()(
LLLLLL
1
! 1{ } , { }n at
n
nt e
s s a
L L
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32Engineering Math, 6. Laplace Transforms
6.5 Convolution. Integral Equations
Ex. 2 Convolution
Evaluate1
2 2 2
1
( )
-
s
L
Solution))(
1)()(Let
22
ssGsF
1
2 2
1 1( ) ( ) sinf t g t t
s
L
1
2 2 2 2 0
1 1sin sin ( )
( )
t- t d
s
L
t
t
dtt
dtt
02
02
)]cos()2[cos(2
1
)]cos()[cos(2
11
tt
t
cos
sin
2
12
2
0
1 1sin (2 ) cos
2 2
t
t t
)(find,)(
1)(
222th
ssH
1
2 2
2 2
!{ }
1{ }
{sin }
{cos }
n
n
at
nt
s
es a
ts
st
s
L
L
L
L
0
t
f g t f g t d
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33Engineering Math, 6. Laplace Transforms
6.5 Convolution. Integral Equations
Ex. 4 Repeated Complex Factors. Resonance – Undamped mass-spring system
tt
tKty 0
0
0
2
0
0 cossin
2)(
0)0(0)0(sin 0
2
0 yytKyy
2 2 0 00 2 2 2 2 2
0 0
( )( )
K Ks Y Y Y s
s s
The subsidiary equation
222 )(
1)(
ssH
tt
tth
cos
sin
2
1)(
2
In Ex. 2, we found
The solution
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34Engineering Math, 6. Laplace Transforms
Property of convolution
Commutative Law (교환법칙):
Distributive Law (분배법칙):
Associative Law (결합법칙):
Unusual Properties of Convolution:
6.5 Convolution. Integral Equations
f g g f
1 2 1 2f g g f g f g
f g v f g v
000 ff
1f f
Ex. 3 Unusual Properties of Convolution
ttdtt
2
0 2
111*
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35Engineering Math, 6. Laplace Transforms
Application to Nonhomogeneous Linear ODEs by convolution
6.5 Convolution. Integral Equations
If
0 1'' ' , 0 , ' 0y ay by r t y K y K
2
2
0 ' 0 0
0 ' 0
s Y sy y a sY y bY R s
s as b Y s a y y R s
, Y y R r L L
22
2
1 1
1 1
2 4
Q ss as b
s a b a
0)0(0)0( yy RQY
t
drtqty0
)()()(
0
t
f g t f g t d
f g g f
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36Engineering Math, 6. Laplace Transforms
Theorem 1 Convolution Theorem
If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order, then
6.5 Convolution. Integral Equations
f g f g L L L
Proof)
),[,, ttppt
0 0( ) ( ) ( ) ( )s spF s e f d and G s e g p dp
dttgeedttgesG ststs )()()( )(
ddttgefddttgeefesGsF ststss
)()()()()()(00
)()()()()()()(000
sHhdtthedtdtgfesGsF stt
st
L
0)(
ddttf
0 0)(
t
dtdf
( ) ( )f g h H s L L
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37Engineering Math, 6. Laplace Transforms
Ex. 5 Response of a Damped Vibrating System to a Single Square Wave
6.5 Convolution. Integral Equations
0)0(0)0().2()1()(23 yytututryyy
tt eetq 2)(
2
1
1
1
)23(
1)()(
123
2
22
sssssQee
sYsYYs ss
0,1For yt
( ) 2( )
1 1
( ) 2( ) ( 1) 2( 1)
1
For 1 2, ( ) 1
1 1 1
2 2 2
t tt t
t
t t t t
t y q t d e e d
e e e e
2 2( ) 2( )
0 1 1For 2 , ( ) 1 ( ) 1
tt tt y q t d q t d e e d
( ) 1, only for 1 2r t t
2
( ) 2( ) ( 2) 2( 2) ( 1) 2( 1)
1
1 1 1
2 2 2
t t t t t te e e e e e
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38Engineering Math, 6. Laplace Transforms
Integral Equation: Equation in which the unknown function appears in an integral.
Ex. 6 Solve the Volterra integral equation of the second kind.
Equation written as a convolution:
Apply the convolution theorem:
0
sin
t
y t y t d t
siny y t t
2 2
1 1
1Y s Y s
s s
2 3
4 2 4
1 1 1
6
s tY s y t t
s s s
6.5 Convolution. Integral Equations (적분방정식)
)()( tYy L
Unknown
t
dthftgtf0
)()()()( : Volterra integral equation for f(t)
1
!{ }n
n
nt
s L
Seoul NationalUniv.
39Engineering Math, 6. Laplace Transforms
6.5 Convolution. Integral Equations
Solve
ttt tffordefettf
0
2 )()(3)(
Solution)
1 1 1 1
3 4
2! 3! 1 1( ) 3 2
1f t
s s s s
L L L L
tett 213 32
Example An Integral Equation
2
0( ) 3 ( )
tt tf t t e f e d L L L L
1
1)(
1
1!23)(
3
ssF
sssF
1
2166)(
43
sssssF
0
1
2 2
2 2
!{ }
1{ }
{sin }
{cos }
t
n
n
at
f g t f g t d
f g f g
nt
s
es a
ts
st
s
L L L
L
L
L
L
Seoul NationalUniv.
40Engineering Math, 6. Laplace Transforms
6.5 Convolution. Integral Equations
0( ) 2 ( )
tt ty t e y e d te Q? Solve
Q? find if equals: f t f tL
9
, ?( 3)
f t f ts s
L
Seoul NationalUniv.
41Engineering Math, 6. Laplace Transforms
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
Differentiation of Transforms
L (f ) f (t)
Ex.1 We shall derive the following three formulas.
2
2 2
1
s 3
1sin cos
2t t t
222 s
s
tt
sin2
222
2
s
s ttt
cossin
2
1
0
stF s f e f t dt
L 0
stdFF s e tf t dt tf
ds
L
1 , tf t F s F s tf t L L
Seoul NationalUniv.
42Engineering Math, 6. Laplace Transforms
2 2sin t
s
L
By differentiation
2
2 2sin
2
t st
s
L
2 2cos
st
s
L
By differentiation
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
222
22
222
222
222
2
22
)()(
2
)(
21)()cos(
s
s
s
ss
s
s
ssFttL
222 )(
2)()sin(
s
ssFttL
L (f ) f (t)
2
2 2
1
s 3
1sin cos
2t t t
222 s
st
t
sin
2
222
2
s
s ttt
cossin
2
1
)()}({
),()}({
1 tfsF
sFttf
- L
L
2 2
2 2 2( cos )
( )
st t
s
L
Seoul NationalUniv.
43Engineering Math, 6. Laplace Transforms
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
L (f ) f (t)
2
2 2
1
s 3
1sin cos
2t t t
222 s
s
tt
sin2
222
2
s
s ttt
cossin
2
1
)()}({
),()}({
1 tfsF
sFttf
-L
L
2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2
1 1 1 1 1sin cos
2 2 ( ) ( ) 2 ( ) ( )
s s s st t t
s s s s
L
2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2
1 1 1 1 1sin cos
2 2 ( ) ( ) 2 ( )
s s st t t
s s s
L
2 2sin t
s
L
2 2
2 2 2( cos )
( )
st t
s
L
2
2 2 2 2 2 2 2
1 2 1
2 ( ) ( )s s
Seoul NationalUniv.
44Engineering Math, 6. Laplace Transforms
1 s s
f t f tF s ds F s ds
t t
L L Integration of Transforms:
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
Proof)
sddttfesdsFs
ts
s
~)(~)~(0
~
dtsdetfdtsdtfesdsFs
ts
s
ts
s
0
~
0
~ ~)(~)(~)~(
t
e
t
esde
st
s
ts
s
ts
~~ ~
t
tfdt
t
tfesdsF st
s
)()(~)~(0
L
Seoul NationalUniv.
45Engineering Math, 6. Laplace Transforms
1 , s s
f t f tF s ds F s ds
t t
L L
Ex. 2 Find the inverse transform of
Case 1 Differentiation of transform
Case 2 Integration of transform
2 2 2
2 2ln 1 ln
s
s s
21 1
2 2 2 2
2 2ln 1 2cos 2
2cos 2 2 1 cos
s sF s f F s t tf t
s s s
tf t t
t t
L L L
2 2 2
2 2 2
2 2ln ln
d s ss s
ds s s
2 2 2
2 2ln 1 ln
s
s s
Derivative
1
2 2
21 1
2
2 2 2 cos 1
2ln 1 1 cos
s
sG s g t G t
s s
g tG s ds t
s t t
L
L L
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
)()}({
),()}({
1 tfsF
sFttf
- L
L
Because the lower limit of the integral is “s”
Seoul NationalUniv.
46Engineering Math, 6. Laplace Transforms
Special Linear ODEs with Variable Coefficients
2 2
' 0
'' 0 ' 0 2 0
d dYty sY y Y s
ds ds
d dYty s Y sy y sY s y
ds ds
L
L
Ex. 3 Laguerre’s Equation, Laguerre Polynomials
Laguerre’s ODE: '' 1 ' 0 0 1 2 , (0) '(0) 0ty t y ny n , , , y y
22 0 0 0 dY dY
sY s y sY y Y s nYds ds
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
2
' 0
'' 0 ' 0 .
f s f f ,
f s f sf f
L L
L L
)()}({ sFttf L
1
!{ }
( 1)
n t
n
nt e
s
L
2 1 0dY
s s n s Yds
2
1 1
1
dY n s n nds ds
Y s s s s
1
1
n
n
sY
s
ate f t F s a L
S-shiftingIn the mean time,
1
!{ }n
n
nt
s L
1( ) ?nl Y L
Rodrigues’s formula
1
( 1) ln ln( 1) ( 1) ln ln ln
n
n
sY n s n s Y
s
Seoul NationalUniv.
47Engineering Math, 6. Laplace Transforms
Ex.3 Laguerre’s Equation, Laguerre Polynomials
Laguerre’s ODE: '' 1 ' 0 0 1 2 , (0) '(0) 0ty t y ny n , , , y y
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
1
!( )
( 1)
n nn t
n n
d n st e
dt s
L
Ys
s
s
sn
net
dt
d
n
el
n
n
n
ntn
n
nt
n
11
)1(
)11(
)1(!
!
1)(
!}{ LL ate f t F s a L
S-shifting
1
1 0
, 1, 2, !
t nn n t
n
, n
l t Y e dt e n
n dt
L
11 20 ' 0 0n nn n n nf s f s f s f f s f
L L L
( 1)( ) (0) (0) ,..., (0) 0n t nf t t e f f f 1
!{ }
( 1)
n t
n
nt e
s
L
1
1n
n
sY
s
Back to 1( ) ?nl Y L
Seoul NationalUniv.
48Engineering Math, 6. Laplace Transforms
6.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients
Q? Solve 2 sin 3 ?t t L
Q? Solve 2 2, ?
( 16)
sf t f t
s
L
Seoul NationalUniv.
49Engineering Math, 6. Laplace Transforms
6.7 Systems of ODEs
1 11 1 12 2 1
2 21 1 22 2 2
( )
( )
y a y a y g t
y a y a y g t
)()0(
)()0(
222212122
121211111
sGYaYaysY
sGYaYaysY
)()0()(
)()0()(
22222121
11212111
sGyYsaYa
sGyYaYsa
)(),(
),(),(
2211
2211
gGgG
yYyY
L L
L L
)(),(, 2
1
21
1
121 YyYyYY -- L L
The Laplace transform method may also be used for solving systems of
ODE (연립상미분방정식),
a first-order linear system with constant coefficients.
Seoul NationalUniv.
50Engineering Math, 6. Laplace Transforms
6.7 Systems of ODEs
Ex. 3 Model of Two Masses on Springs
The mechanical system consists of two bodies of mass 1 on three springs of
the same spring constant k. Damping is assumed to be practically zero.
2
1 1 2 1
2
2 2 1 2
3
3
s Y s k kY k Y Y
s Y s k k Y Y kY
1 1 2 1
2 2 1 2
''
''
y ky k y y
y k y y ky
1 2
1 2
0 1, 0 1,
' 0 3 , 0 3
y y
y k y k
Laplace
transform
Model of the
physical
system:
Initial
conditions:
2
' 0
'' 0 ' 0 .
f s f f ,
f s f sf f
L L
L L
Seoul NationalUniv.
51Engineering Math, 6. Laplace Transforms
6.7 Systems of ODEs
2
1 1 2 1
2
2 2 1 2
3
3
s Y s k kY k Y Y
s Y s k k Y Y kY
2
1 2 2 22 2
2
2 2 2 22 2
3 2 3 3
32
3 2 3 3
32
s k s k k s k s kY
s k s ks k k
s k s k k s k s kY
s k s ks k k
Cramer’s rule or Elimination
1
1 1
1
2 2
cos sin 3
cos sin 3
y t Y kt kt
y t Y kt kt
L
L
Inverse transform
Ex. 3 Model of Two Masses on Springs
Recommended