hapter 6. laplace transforms laplace... · 2019. 9. 6. · seoul national univ. 3 engineering math,...

Seoul NationalUniv.

1Engineering Math, 6. Laplace Transforms

CHAPTER 6. LAPLACE TRANSFORMS

2019.5서울대학교

조선해양공학과

서 유 택

※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다.

Seoul NationalUniv.


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

Seoul NationalUniv.


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 21 1 1

2

at aes a s a s a

s

Seoul NationalUniv.



Seoul NationalUniv.


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

Seoul NationalUniv.


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 at

s ae t e t

s a s a

L L

23 137

2 401

sf

s s

L

1 1 12 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)

Seoul NationalUniv.


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)

Seoul NationalUniv.


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)

Seoul NationalUniv.



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

Seoul NationalUniv.


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,

10 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 Mg t f d f d M e d e e k

k k

1{ ( )} { ( )}g t f t

s L L

From Theorem 1

Seoul NationalUniv.


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

10 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

Seoul NationalUniv.


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

222

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


Seoul NationalUniv.


Ex. 4 Solve

Step 1

Step 2

Step 3

'' , 0 1, ' 0 1y y t y y

2 22 21 1

0 ' 0 1 1s Y sy y Y s Y ss 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 12 21 1 1

sinh1 1

ty t Y e t ts s s

L L L L


Seoul NationalUniv.


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, = 354

)

Seoul NationalUniv.



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

Q? Solve

Seoul NationalUniv.



[ Relation of Time domain and s-Domain ]

Time domain analysis

(differential equation)

Problem Answer

Laplace Transform

S-domain analysis

(algebraic equation)

Inverse Transform

Seoul NationalUniv.


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


Seoul NationalUniv.



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.

Seoul NationalUniv.


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)()( 0L

“on off function”

Seoul NationalUniv.


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

Unit Step Function

Seoul NationalUniv.


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)()( 0L

0

stF s f e f t dt

L

Seoul NationalUniv.


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 1

2 1 1 1 cos2 2 2

f t u t t u t u t t u t


𝜋

2

Seoul NationalUniv.


Ex. 1 Write the following function using unit step functions and find its

transform.

Step 1 Using unit step functions:

Step 2

21 1 1

2 1 1 1 cos2 2 2

f 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

221 1 1 1

cos sin2 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

Seoul NationalUniv.


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 t

f 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

Seoul NationalUniv.



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.

Seoul NationalUniv.


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

Seoul NationalUniv.


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



Time shifting

Seoul NationalUniv.


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

sseY s e

s s s s


21

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

Seoul NationalUniv.



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

Q? Solve

Seoul NationalUniv.


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

Seoul NationalUniv.


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 10

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

Seoul NationalUniv.



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

Seoul NationalUniv.



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 02

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

Seoul NationalUniv.


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*

Seoul NationalUniv.


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

222

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

Seoul NationalUniv.


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

Seoul NationalUniv.


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

Seoul NationalUniv.


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 21 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.



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 ( )t

t 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.



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.


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 ss

tt

sin2

2222

s

s ttt

cossin

2

1

0

stF s f e f t dt

L 0

stdF

F s e tf t dt tfds

L

1 , tf t F s F s tf t L L

Seoul NationalUniv.


2 2sin t s

LBy differentiation

2

2 2sin

2

t st

s

L

2 2coss

ts

LBy 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 ss

tt

sin2

2222

s

s ttt

cossin

2

1

)()}({

),()}({

1 tfsF

sFttf

- L

L

2 2

2 2 2( cos )

( )

st t

s

L

Seoul NationalUniv.



L (f ) f (t)

2

2 2

1

s 3

1sin cos

2t t t

222 ss

tt

sin2

2222

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.


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 22 2

ln lnd s s

s sds 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”

Seoul NationalUniv.


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


In 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.


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

Seoul NationalUniv.



Q? Solve 2 sin 3 ?t t L

Q? Solve 2 2 , ?( 16)s

f t f ts

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

)(),(, 21

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.


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

hapter 6. laplace transforms laplace... · 2019. 9. 6. · seoul national univ. 3 engineering math,...

Documents