HAPTER 6. LAPLACE TRANSFORMSocw.snu.ac.kr/sites/default/files/NOTE/08 Laplace... · 2018-09-13 · Seoul National Univ. 3 Engineering Math, 6. Laplace Transforms Theorem 1 Linearity

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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.

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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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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.


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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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


00

)( )}({)]([)()( tfedttfeedttfeasF atatsttas L

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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


(a = -1, = 20)

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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.


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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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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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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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


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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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


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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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


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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


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Ex. 5 Solve

Step 1

Step 2

Step 3


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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37 12 21 , (0) 3.5, (0) 10ty y y e y y

Q? Solve

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[ 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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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


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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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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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6.3 Unit Step Function (Heaviside Function).Second Shifting Theorem (t-Shifting)

Unit Step Function

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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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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


𝜋

2

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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


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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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


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


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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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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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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Ex. 1 Mass Spring System Under a Square Wave

Step 1

Step 2

Step 3


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


Time shifting

ate f t F s a L

S-shifting


Time shifting

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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


21 ( )= , ( )=at t te f t e g t e

s a

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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9 ( / 2), (0) 2, (0) 0y y t y y

Q? Solve

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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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Convolution:

Theorem 1 Convolution Theorem

If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order,

then


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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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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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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Property of convolution

Commutative Law (교환법칙):

Distributive Law (분배법칙):

Associative Law (결합법칙):

Unusual Properties of Convolution:


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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Application to Nonhomogeneous Linear ODEs by convolution


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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Theorem 1 Convolution Theorem

If f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order, then


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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Ex. 5 Response of a Damped Vibrating System to a Single Square Wave


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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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

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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

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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

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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

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2 2sin t

s

L

By differentiation

2

2 2sin

2

t st

s

L

2 2cos

st

s

L

By differentiation


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

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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

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1 s s

f t f tF s ds F s ds

t t

L L Integration of Transforms:


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.


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


)()}({

),()}({

1 tfsF

sFttf

- L

L

Because the lower limit of the integral is “s”

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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


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

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Ex.3 Laguerre’s Equation, Laguerre Polynomials

Laguerre’s ODE: '' 1 ' 0 0 1 2 , (0) '(0) 0ty t y ny n , , , y y


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

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Q? Solve 2 sin 3 ?t t L

Q? Solve 2 2, ?

( 16)

sf t f t

s

L

Seoul NationalUniv.


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.

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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.


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

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HAPTER 6. LAPLACE TRANSFORMSocw.snu.ac.kr/sites/default/files/NOTE/08 Laplace... · 2018-09-13 · Seoul National Univ. 3 Engineering Math, 6. Laplace Transforms Theorem 1 Linearity