Fourier Laplace Transforms

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________________________________________________________ MEP 201 Advanced Engineering Mathematics

Fourier Transforms/Laplace Transforms

1

-p +

f x

FOURIER INTEGRAL AND FOURIER TRANSFORM

Fourier integral representation of ( ) x f

The function ( ) x f shown below is defined only for p x p ≤≤− .

The function can be represented by the Fourier series

( )

( ) ( )

( ) ( ) ( )∑ ∫ ∫

∑ ∫ ∫

∑

∞

=

+−

−π++−

=

∞

=

+−

ππ+ππ++−

=

∞

=

π+π+=

1n

p

p' dx x ' x

pncos' x f

p1 p

p' dx ' x f

p2 1

1n

p

p' dx

p' x nsin

p x nsin

p' x ncos

p x ncos' x f

p1 p

p' dx ' x f

p2

1

1np x nsinnb

p x ncosna

2 oa x f

Above is the Fourier series for ( ) x f no matter how large p is.

Let p approach infinity. Provided ( )

∫

+∞

∞−' dx ' x f exists, the first term vanishes as p approaches infinity.

Let pn

nπ=ω and

( ) p p

n p

1nn1nn

π=π−π+

=ω−+ω=ω∂ . Thus,

( ) ( ) ( )[ ]

( ) ( )[ ] ( )∑∑ ∫

∑ ∫

∞

=ω∂ω

π=ω∂

∞

=

+

−−ω

π=

∞

=

+−

−ωπω∂

=

1nn x ,nF 1

n1n

p

p' dx x ' x ncos' x f 1

1n

p

p' dx x ' x ncos' x f x f n

As ∞→ p , ω→ω∂ d n so that ( ) ( )∫ ∞

ωωπ

=0

d x ,nF 1 x f .

Fourier integral representation of ( ) x f : ( ) ( ) ( )[ ]∫ ∫ ∞

ω

∞+

∞−−ω

π=

0 d ' dx x ' x cos' x f 1 x f [1]

Equation [1] is a valid representation of ( ) x f provided

(a) in every finite interval, ( ) x f satisfies the Dirichlet conditions, and

(b) ( )∫ +∞∞−

dx x f exists.





2

EXAMPLE 1: Obtain the Fourier integral representation of the function ( )

>

+<<−=

1 x 0

1 x 11 x f .

SOLUTION:

( ) ( )∫ ∫ ∫ ∫ ∞

ω

∞+

++

+−

−ω+−

∞−π=

0 d

1' dx .0

1

1' dx x ' x cos.1

1' dx .0 1 x f

( ) ( ) ( ) ( )∫ ∫

∞ω

ωω−ω−−ω−ω

π=

∞ω

+

−ω−ω

π=

0 d

x sin x sin10

d 1

1

x ' x sin1 x f

( ) ( )∫ ∞

ωωω+ωω+ωω−ωωπ

=0

d x sincos x cossin x sincos x cossin1 x f

( ) ∫ ∞

ωω

ωωπ

=0

d x cossin2 x f

Question: What is the value of ∫ ∞

ωω

ωω0

d x cossin ?

Fourier sine and cosine transform of ( ) x f

Given that

( ) ( )[ ]

( ) ( )[ ]∫

∫ ∫ ∞

ωωω+ωωπ

=

∞ω

∞+

∞−ωω+ωω

π=

0 d x sinBcos A1

0 d ' dx x sin' x sin x cos' x cos' x f 1 x f

where ( ) ( )∫ +∞

∞−ω=ω ' dx ' x cos' x f A and ( ) ( )∫

+∞∞−

ω=ω ' dx ' x sin' x f B :

If ( )' x f is even, ( ) 0 B =ω and ( ) ( )∫ ∞

ω=ω0

' dx ' x cos' x f 2 A . Therefore,

( ) ( )∫ ∫ ∞ ∞

ωωωπ= 0 0 d ' xdx cos' x cos' x f 2 x f [2]

If ( )' x f is odd, ( ) 0 A =ω and ( ) ( )∫ ∞

ω=ω0

' dx ' x sin' x f 2 B . Therefore,

( ) ( )∫ ∫ ∞ ∞

ωωωπ

=0 0

d ' xdx sin' x sin' x f 2 x f [3]

Equation [2] can be written in the form

( ) ( )∫ ∞

ωπ

=ω0

xdx cos x f 2 g Fourier cosine transform of ( ) x f ; spectrum of ( ) x f .

( ) ( )∫ ∞

ωωωπ= 0 xd cosg 2

x f Inverse cosine transform of ( )ωg .

Equation [3] can be written in the form

( ) ( )∫ ∞

ωπ

=ω0

xdx sin x f 2 g Fourier sine transform of ( ) x f ; spectrum of ( ) x f .

( ) ( )∫ ∞

ωωωπ

=0

xd sing 2 x f Inverse sine transform of ( )ωg .





3

EXAMPLE 2: Given ( )

>

<<

<<

=

b x 0

b x a1

a x 0 0

x f . Obtain the Fourier cosine transform and the sine transform of ( ) x f .

Obtain also the Fourier cosine and sine integral representations of ( ) x f .

SOLUTION:

(a) To get the Fourier cosine transform of the function, we create an even function whose definition in ∞<< x 0

is identical to ( ) x f . Then,

( ) ( )∫ ∞

ωωωπ

=0

xd cosg 2 x f

where ( ) ( ) ( )∫ ∫ ωπ

=∞

ωπ

=ωb

axdx cos12

0 xdx cos x f 2 g

( )

ωω−ω

π=ω asinbsin2 g The Fourier cosine transform of ( ) x f .

The Fourier cosine integral representation of ( ) x f is then

( )( )

∫ ∞

ωω

ω−ωωπ

=0

d asinbsin x cos2 x f

(b) If an odd function is created by extending ( ) x f through negative x , then

( ) ( )∫ ∞

ωωωπ

=0

xd sing 2 x f

where ( ) ( ) ( )∫ ∫ ωπ

=∞

ωπ

=ωb

axdx sin12

0 xdx sin x f 2 g

( )

ωω−ω

π=ω bcosacos2 g The Fourier sine transform of ( ) x f .

The Fourier sine integral representation of ( ) x f is then

( ) ( )∫

∞ω

ωω−ωω

π=

0 d

bcosacos x sin2 x f

Exponential form of the Fourier integral

Refer back to equation [1]. Since ( )( )[ ] ( )[ ]

2

x ' x i exp x ' x i exp x ' x cos

−ω−+−ω=−ω , then equation [1] can be written in

the form ( ) ( ) ( )[ ] ( ) ( )[ ]∫ ∫ ∫ ∫ ∞

ω

∞+

∞−−ω−

π+

∞ω

∞+

∞−−ω

π=

0 d ' dx x ' x i exp' x f

2 1

0 d ' dx x ' x i exp' x f

2 1 x f .

If ω is replaced by ω− , the first integral becomes

( ) ( )[ ] ( ) ( )[ ]∫ ∫ ∫ ∫ ∞− ω

∞+

∞− −ω−π=

∞−

ω

∞+

∞− −ω−π−

0

d ' dx x ' x i exp' x f 2

1

0 d ' dx x ' x i exp' x f 2

1

.

Therefore

( ) ( ) ( )[ ]∫ ∫ ∞+∞−

ω

∞+

∞−−ω−

π= d ' dx x ' x i exp' x f

2 1 x f

( ) ( )∫ ∫ ∞+∞−

ω

∞+

∞−ω−ω

π= d ' dx ' x i e' x f x i e

2 1 x f

The above expression for ( ) x f can be written in the symmetric form





4

( ) ( )∫ +∞∞−

ωωωπ

= d x i eg 2

1 x f Complex inverse transform of ( )ωg .

( ) ( )∫ +∞

∞−ω−

π=ω dx x i e x f

2

1g Complex Fourier transform of ( ) x f ; spectrum of ( ) x f .

We shall also use the symbol ( ){ } x f F to denote the complex Fourier transform of ( ) x f and the symbol ( ){ }ω− g 1F

to denote the inverse transform of ( )ωg .

EXAMPLE 3. Obtain the Fourier integral representation of ( )

>>−

<=

0 awhere0 x ax e

0 x 0 x f

SOLUTION:

( ) ( ) ( )

( ) ( )[ ]

( )( )ω+π

=∞

=

ω−ω−

ω+π−=

∞ω+−ω+−

ω+π

=

∞ ω+−π

=+∞∞−

ω−π

=ω

∫

∫ ∫

i a2

1

0 x x sini x cosax e

i a1

2

1

0 x i ad x i ae

i a1

2

1

0 dx x i ae

2

1dx x i e x f 2

1g

( ) ( )

∫ ∫

∫

∫ ∫

∞+∞−

ωω+

ωω−ωπ

+∞+∞−

ωω+

ωω+ωπ

=

∞+∞−

ωω+ω−•

ω+ω+ω

π=

∞+∞−

ωω+π

=∞+∞−

ωωωπ

=ω

d a

x cos x sina2 i d

a

x sin x cosa2 1

d i ai a

i a x sini x cos

2 1

d i a

e2 1d x i eg

2

1 x f

2 2 2 2

x i

Since ( ) x f is real, then the second integral is zero.

( ) ∫ +∞

∞− ωω+ ωω+ωπ= d a

x sin x cosa2 1 x f 2 2

Seatwork: What is the value of ∫ ∞

+

β+β=

0 dx

x a

x sin x x cosaI

2 2 if 0 >β and 0 a > ?

EXAMPLE 4: Determine the Fourier cosine transform pair for ( ) ∞<<∞−

−= x

2 x exp x f

2 .

SOLUTION:

The Fourier cosine transform of ( ) x f is ( ) ∫ ∞

ω

−

π=ω

0 xdx cos

2 x exp2 g

2

From the table of integrals, ( )∫ ∞

−π

=−0 a2

bexpa2 bxdx cos

2 x

2 aexp 2

2

.

Thus, ( )

ω−=ω2

expg 2

The inverse transform of ( )ωg is ( )

−=

∞ωω

ω−π

= ∫ 2 x exp

0 xd cos

2 exp2 x f

2 2 .





5

Partial Differential Equations

EXAMPLE 5: A slender infinite rod has its lateral surface insulated. The initial temperature is ( ) ( ) x f 0 , x u = ,

where ( ) x f is bounded. Find ( )t , x u .

SOLUTION:

Let ( ) ( ) ( )t T x X t , x u = . Substitute into the one-dimensional heat equation to get

2 T

' T

a

1 X

' ' X 2

ω−==

Thus, ( )

ω−= t 2 2 aexpt T and ( ) x sin2 c x cos1c x X ω+ω=

( ) [ ] x sin2 c x cos1c t 2 2 aexpt , x u ω+ω

ω−=ω

( ) ( ) ( )[ ] ω∞

ωω+ωω

ω−

π= ∫ d

0 x sinB x cos At 2 2 aexp1t , x u

Apply initial condition:

( ) ( ) ( ) ( )[ ] ( ) ( )[ ] ω∞

∞+

∞−−ω

π=ω

∞ωω+ωω

π== ∫ ∫ ∫ d

0 ' dx x ' x cos' x f 1d

0 x sinB x cos A1 x f 0 , x u

( ) ( ) ( )∫ ∫ ∞+∞−

∞+

ω−ω

ω−

π= ' dx

0 d x ' x cost 2 2 aexp' x f 1t , x u

( )( )

∫ ∞+∞−

−−π

π= ' dx

t a4

x ' x exp

t a2 ' x f 1

2

2

from the integration formula in EXAMPLE 4 above where ω→ x , x ' x b −→ , t 2 a2 a → .

Suppose ( ) <<

= x other all 0

20 x 0 100 x f . Then ( )

( )∫

−−

π=

20

0 ' dx

t a4

x ' x exp

t a2

100 t , x u 2

2

EXAMPLE 6: A semi-infinite string lies on the positive half of the x-axis and is fixed at 0 x = . If the string is

given an initial displacement ( ) ( ) x f 0 , x y = and an initial velocity of ( ) x g , find the displacement ( )t , x y . Assume

that the displacement is bounded everywhere.

SOLUTION:

Let ( ) ( ) ( )t T x X t , x y = . Substitute into the one-dimensional wave equation to get 2 T

' ' T

a

1 X

' ' X 2

ω−== .

Thus, ( ) ( )( )at sin4c at cos3c x sin2 c x cos1c t , x y ω+ωω+ω=

( ) ( )t T 1c 0 t ,0 y == Therefore, 0 1c = .

( ) ( )at sin4c at cos3c x sint , x y ω+ωω=ω

( ) ( ) ( )[ ] ω∞

ωω+ωωωπ

= ∫ d 0

at sinBat cos A x sin1t , x y

( ) ( ) ( )( )

∫ ∫ ∞

ωωπ

ωπ

=∞

ωωωπ

==0

xd sin2

A2 0

xd sin A1 x f 0 , x y





6

Refer to the formula above for the inverse sine transform of a function and observe that the right

hand side of the above equation is the inverse sine transform of ( )π

ω2

A . Therefore,

( )( )∫

∞ω

π=

π

ω0

xdx sin x f 2

2

A. Hence, ( ) ( )∫

∞ω=ω

0 xdx sin x f 2 A .

( ) ( )( )

∫ ∫ ∞

π

ωωωωπ

=∞

ωωωωπ

===∂

∂0 2

xd sinaB2 0

xd sinaB1 x g 0 t t

y

The quantity on the right-hand side is the inverse sine transform of ( )

( )∫ ∞

ωπ

=π

ωω0

xdx sin x g 2

2

aB.

Thus, ( ) ( )∫ ∞

ωω

=ω0

xdx sin x g a

2 B .

Therefore,

( ) ( ) ( )∫ ∫ ∫ ∞

ω

ω

∞ω

ω+ω

∞ωω

π=

0 d at sin

0 ' dx ' x sin' x g

a2 at cos

0 ' dx ' x sin' x f 2 x sin1t , x y

EXAMPLE 7: Find the steady-state temperature in a thin semi-infinite plate bounded by the x-axis and the

positive halves of the line 0 x = and 1 x = . The left and the bottom edges are maintained at zero temperature

while the temperature at the right edge is kept at ( )y f .

SOLUTION: Let ( ) ( ) ( )y Y x X y , x u = and substitute into the two-dimensional steady-state heat equation to

get 2 Y

' ' Y X

' ' X ω=−= .

( ) ( )( ) x sinh4c x cosh3c y sin2 c y cos1c y , x u ω+ωω+ω=

BC 1: ( ) 0 0 , x u = Therefore, 0 1c = .

BC 2: ( ) 0 y ,0 u = Therefore, 0 3c = .

( ) y sin x sinhy , x u ωω=ω

( ) ( )∫ ∞

ωωωω=0

yd sin x sinhC y , x u

BC 3: ( ) ( ) ( )∫ ∞

ωωωω==0

yd sinsinhC y f y ,1u

( ) ( )∫ ∞

ωωωωπ

=π 0

yd sinsinhC 2 y f 2

The right hand side is the inverse sine transform of ( ) ωω sinhC . Therefore,

( ) ( )∫ ∞

ωππ

=ωω0

ydy siny f 2 2 sinhC

( ) ( )

∫

∞ω

ωπ=ω

0 ydy siny f

sinh

2 C

Suppose ( )

>

<<=

ay 0

ay 0 100 y f . Then,

( ) [ ]acos1sinh200 a

0 ydy sin100

sinh2 C ω−

ωπω=ω

ωπ=ω ∫





7

EXAMPLE 8: Solve EXAMPLE 7 by getting the Fourier sine transform of both sides of the PDE.

SOLUTION:

0 0

ydy siny

u 0

ydy sin x

u 2

2

2

2 =

∞ω

∂∂+

∞ω

∂∂ ∫ ∫

( ) 0 0

dy y u

y y sin

0 ydy siny , x u

x 2 2 =∞

∂∂

∂∂ω+∞ ω

∂∂ ∫ ∫

Integrate by parts. In the second integral, let y sins ω= , dy y u

y dt

∂∂

∂∂= ;

y u t

∂∂= .

Hence,

∫ ∫ ∞

ωω∂∂−

∞

∂∂ω=

∞∂∂

∂∂ω

0 ydy cos

y u

0 y u y sin

0 dy

y u

y y sin

∞ωω+∞ωω−=

∞∂∂ωω−= ∫ ∫ 0

ydy sinu 0

y cosu 0

dy y u y cos0

∫ ∞

ωω−=

0

ydy sinu 2

Hence,

( ) ( ) 0 0

ydy siny , x u 2 0

ydy siny , x u dx

d 2

2 =

∞ωω−

∞ω ∫ ∫

Denoting the Fourier sine transform of ( )y , x u by ( ) x ,sU ω , we get the following ordinary differential from the

above equation:

( )( ) 0 x ,sU 2

dx

x ,U d

2

s2

=ωω−ω

Thus, the effect of the Fourier sine transform is to convert the PDE to the ordinary differential equation shown

above. The general solution of this differential equation is( ) x sinh2 c x cosh1c x ,sU ω+ω=ω

BC 1: ( ) ( ) ( ) ( )0 sinh2 c 0 cosh1c 0 0

ydy siny ,0 u 2 0 ,sU ω+ω==∞

ωπ

=ω ∫ Therefore, 0 1c = .

BC 3: ( ) ( ) ω=∞

ωπ

=ω ∫ sinh2 c 0

ydy siny f 2 1,sU Therefore, ( )∫ ∞

ωωπ

=0

ydy siny f sinh

12 2 c

( ) ( ) x sinh0

ydy siny f sinh

12 x ,sU ω

∞ω

ωπ=ω ∫

( )( )

∫ ∞

ωω

ωω∫ ω

π=

∞

0 d

sinh

y sin x sinhydy siny f 2 y , x u 0





8

LAPLACE TRANSFORM

Definition

Let ( )t f be defined for all 0 t > . The Laplace transform of ( )t f is defined by the equation

L ( ){ } ( ) ( )∫ ∞

−== 0 dt t f st esF t f

over the range of values of s for which the integral exists.

EXAMPLE 1:

a) ( ) 1t f =

L { } ( )s1

0 s

e0

dt 1st e1st

=∞

−=∞ −=

−

∫

b) ( ) at et f −=

L

{ }( )

as

1

0 as

e

0 dt

at

e

st

e

at

e

t as

+=

∞

+−=

∞ −−

=

− +−

∫

c) ( ) t t f =

L { }2 2

st st st

s

1

0 s

e0

dt s

e

0 s

te0

tdt st et =∞

−=∞

+∞

−=∞ −=

−−−

∫ ∫

Theorem: L ( ) ( ){ } 1c t 2 f 2 c t 1f 1c =+ L ( ){ } 2 c t 1f + L ( ){ }t 2 f

Definition ( )t f is said to be of exponential order if there exists a number 0 M > and an α such that

( ) M t f t e

t

lim <α−

∞→

for all t greater than some finite number T .

EXAMPLE 2: ( ) 2 t t f =

1e

2

t lim

e

t 2

t lim2 t t e

t lim

t t <

∞→=

∞→=−

∞→

Hence, 1=α and 1M = .

Conclusion: 2 t is of exponential order.

EXAMPLE 3: ( ) 1,t t f −>νν=

L { } ∫ ∞ ν−=ν0

dt t st et Letsz t =

sdz dt =

For 0 s >

L { } ( )11 s

1

0 dz z z e

s

10 s

dz z esz t

+ν+ν

+νΓ=

∞ ν−=∞ −ν

=ν ∫ ∫

For n=ν (an integer), L 1ns

! nt +

=ν





9

1

a

u(t-a)

t

Theorem: Let ( )t f be sectionally continuous and of exponential order. Then, L ( ){ }t f exists when α>s .

Theorem: Let ( )t f be continuous and ( )t ' f be sectionally continuous in every finite interval T t 0 ≤≤ . Let

( ){ }t f be of exponential order. Then, when α>s ,

L ( ){ } st ' f = L ( ){ } ( )0 f t f −

Similarly, L ( ){ } st ' ' f = L ( ){ } ( )0 ' f t ' f −

[ss= L ( ){ } ( )] ( ) 2 s0 ' f t f =−0 f - L ( ){ } ( ) ( )0 ' f 0 sf t f −−

L ( ) nst nf = L ( ){ } ( ) ( ) ( )( )0 1nf ...0 ' f t f −−−−− 2-n s0 f 1-n s provided ( ) ( ),...t ' ' f ,t ' f satisfy the same conditions as

( )t f .

EXAMPLE 4:

a) ( ) ( ) ( ) bt cos2 bt ' ' f bt sinbt ' f bt cost f −=−==

L ( ) 2 st ' ' f = L ( ){ } ( ) ( )0 ' f t f −− 0 sf

L 2 sbt cos2 b =− L { } 0 bt cos −− s Therefore, L { }2 2 bs

sbt cos+

=

b) ( ) bt sint f = L ( ){ } st ' f = L ( ){ } ( )0 f t f −

L { } sbt cosb = L { } 0 bt sin − Therefore, L { }2 2 bs

bbt sin+

=

Theorem: L ( ) ( )asF t f at e −= where ( ) =sF L ( ){ }t f .

EXAMPLE 5:

a) L ( )31s

! 2 2 t t e−

= b) L ( ) 2 2

32 s

3t 3sint 2 e++

=−

Definition: Unit step function ( )

>

<=−

at 1

at 0 at u

A function that is zero for all at < and which is equal to ( )t f

when at > can be written simply as ( ) ( )at u t f − .

Theorem: L ( ) ( ){ } ( )sF aseat u at f −=−−

where ( ) =sF L ( ){ }t f .





10

EXAMPLE 6: Find the Laplace transform of the function on the

right.

For ( )

−=<<

32 t

5 3t f ,4t 1 . Therefore,

( ) ( ) ( )[ ]4t u 1t u 32 t

5 3t f −−−

−=

( ) ( )

−

+−−−

+−= 4t u

310 4t 1t u

311t

5 3

( ) ( ) ( ) ( ) ( ) ( )

−+−−−−+−−= 4t u

310 4t u 4t 1t u

311t u 1t

5 3

L ( ){ }

+−+=

−−−−

s3e10

s

es3

e

s

e5 3t f

s4

2

s4s

2

s

Partial Fractions

Let ( )( )( )sB

s Asf = where ( )sB is a polynomial of higher degree than the polynomial ( )s A . ( )sf is said to have a pole of

order k at as = if there exists a positive integer k such that ( )k as − is defined and not zero at as = . When 1k = ,

as = is called a simple pole.

Case 1:

( )sf has simple poles only, i.e., ( ) 0 sB = has distinct roots na,...,3a,2 a,1a so that ( ) ( )( ) ( )nas...2 as1assB −−−=

Then,

( )

n

n

k

k

2

2

1

1

as

c ...

as

c ...

as

c

as

c sf

−++

−++

−+

−= [a]

To determine k c , multiply both sides of [a] by ( )k as − and take the limit as s approaches a , i.e.,

( ) ( )sf k asas

limk c −→

=

Case 2:

( )sf has higher order poles, say, a pole of order k at 1as = :

( ) ( ) ( )( )...3as2 ask 1assB −−−= .

( )( ) ( ) ( ) ( ) ( ) ( )

...as

c

as

c

as

c ...

as

c

as

c

as

c sf

3

3

2

2

1

k 1

2 k 1

13

1k 1

12

k 1

11 +−

+−

+−

++−

+−

+−

=−−

[b]

Multiply both sides of [b] by ( )k 1as − :

( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )...

asc as

asc ask 1c 1k 1as...13c 2 1as12 c 1as11c sf k 1as

3

3k

1

2

2 k

1 +−

−+−

−+−−++−+−+=− [c]

11c is obtained by taking the limit as s approaches 1a . 12 c is obtained by differentiating [c] with respect to s

and taking the limit as s approaches 1a .

( ) ( )1as

sf k 1as

dsd

12 c =

−=

0.2

1

f(t)

t

2.0

4





11

( )( ) ( )

1

1k

1k

as

sf k 1as

ds

d ! 1k

1k 1c

=

−

−=

−

−

2 c , 3c ,…, k c are obtained using the procedure for case 1.

Inverse Transform

The inverse transform of a function ( )sF is that function of t whose Laplace transform is ( )sF . The inverse

transform of ( )sF is denoted by L -1 ( ){ }sF . It can be shown that (see Sec. 7.9 of text)

( ) =t f L -1 ( ){ } ( )∫

∞+∞−π

=i a

i adsst esF

i 2 1sF

The inverse transform of many functions can be found without using the above complex integral. Some

examples will illustrate how this is done.

EXAMPLE 7: Find the inverse transform of

a) ( )( )( )2 s2 ss

2 s2 ssF 2

−+−+=

Solution:

( )( )( ) 2 s

c

2 s

c

s

c

2 s2 ss2 s2 ssF 32 1

2

−+

++=

−+−+=

( )( ) 2 1

0 s2 s2 s2 s2 s

1c 2

==−+

−+=

( ) 41

2 s2 ss

2 s2 s2 c

2 −=

−=+−+=

( ) 432 s

2 ss 2 s2 s3c

2

==− −+=

( )( ) ( )2 s4

32 s4

1s2

1sF −

++

+=

L -1 ( ){ } t 2 e

43t 2 e

41

2 1sF +−−=

b) ( )( ) ( ) ( ) 3s

c

1s

c

1s

c

3s1s

2 ssF 2 12

2

11

2 ++

++

+=

++

+=

( ) ( ) 2 1

1s

3s1s

2 s11c

2 =

−=++

+=

41

1s3s2 s

dsd

12 c =−=+

+=

( ) 41

3s1s

2 s2 c

2 −=

−=+

+=

( )( ) ( ) ( )3s4

11s4

1

1s2

1sF 2 +

−+

++

=





12

L -1 ( ){ } t 3e

41t e

41t te

2 1sF −−−+−=

Convolution Integral

L -1 ( ) ( ){ } ( ) ( ) ( ) ( )∫ ∫ τττ−=ττ−τ= t 0

d g t f t 0

d t g f sGsF

EXAMPLE 8:

L -1

( )=

+ 2 2 3

4

bss

b2 L

-1

( )=

+•

2 2 3 bs

b

s

2 3b L -1 3b L

2 t L { }}bt sin

( ) ( ) ( )

( ) ( ) ( )2 t t g bt sinf

t bsint g 2 f

τ−=τ−=τ

τ−=τ−τ=τ

L -1

( )( ) ( ) [ ]∫ ∫ τ−τ−=τττ−=

+

t

0 bcosd 2 t 2 b

t

0 d bsin2 t 3b

bss

b2 2 2 3

4

( ) ( )[ ]

∞ττ−τ+

∞ττ−−= ∫ 0

d t 2 bcos0

bcos2 t 2 b

( ) [ ]

∞ττ−+−−= ∫ 0

bsind t b2 2 t 2 b

2 bt cost 2 2 t 2 b −+=

Linear Differential Equations with Constant Coefficients

Consider the second-order equation ( ) ( ) ( ) ( )t g t cy t ' by t ' ' ay =++ .

Take the Laplace transform of both sides of the equation (assuming that the Laplace transforms of both sides

exist):

aL ( ){ } bt ' ' y + L ( ){ } c t ' y + L ( ){ } =t y L ( ){ }t g

Assuming that ' ' y and ' y satisfy the conditions of the theorem on the transform of derivatives, we have

( ) ( ) ( ) ( ) ( )[ ] ( ) ( )sGscY 0 y ssY b0 ' y 0 sy sY 2 sa =+−+

−−

where ( ) =sY L ( ){ }t y and ( ) =sG L ( ){ }t g .

Solve for ( )sY to get ( )( ) ( ) ( ) ( )

c bsas

0 ' ay 0 y bassGsY

2 ++

+++=

OBSERVATION: The Laplace transform converts the ordinary differential equation in the unknown ( )t y into

the algebraic equation in the unknown ( )sY . The same thing will happen (i.e., a differential equation in y is





13

transformed into an algebraic equation in Y ) if higher ordered linear equations with constant coefficients are

considered.

If ( )0 y and ( )0 ' y are known in the problem under consideration, then the quantity on the right hand side is

completely known and the solution of the initial value problem is ( ) =t y L -1 ( ){ }sY .

EXAMPLE 9: Solve the equation ( ) ( ) t e2 y 3t ' y 4t ' ' y −=++ if the initial conditions are ( ) 0 0 y = and ( ) 10 ' y −= .

SOLUTION: Take the Laplace transform of both sides of the differential equation

( ) ( ) ( ) ( ) ( )[ ] ( ) =+−+

−− sY 30 y ssY 40 ' y 0 sy sY 2 s L

t e2 −

( ) ( ) ( )1s

2 sY 3ssY 41sY 2 s+

=+++

( )( )

( )( ) ( )( ) ( ) ( )1s

c

1s

c

3s

c

1s3s

1s

1s3s4s

1ssY 12

2

111

2 2 ++

++

+=

++

−−=+++

−−=

112

c 111

c 11

c −===

( ) =t y L -1 ( ){ } t et tet 3esY −−−+−=

PARTIAL DIFFERENTIAL EQUATIONS: Boundary-Value Problems

EXAMPLE 10: Find the temperature distribution in a slender semi-infinite rod whose lateral surface is

insulated and whose axis coincides with the positive x axis if the temperature is initially 0 and the left end of the

rod is subsequently maintained at the temperature ( )t f .

SOLUTION: The governing partial differential equation is

t u

a

1

x

u 2 2

2

∂∂=

∂

∂ or 2

2

x

u 2 at u

∂

∂=∂∂

Boundary Conditions: (1) ( ) 0 0 , x u = (2) ( ) ( )t f t ,0 u =

Take the Laplace transform with respect to t of the PDE:

L t2 a

t u =

∂∂

L t

∂

∂2

2

x

u

s L t ( ){ } ( )2

2

dx

d 2 a0 , x u t , x u =− L t ( ){ }t , x u

( )( )2

2

dx

s, x U d 2 a0 s, x sU =−

0 U a

s

dx

U d 2 2

2 =− The roots of the characteristic equation are

a

sm ±=





14

Note that the effect of the Laplace transformation is the conversion of the PDE in ( )t , x u into an ordinary

differential equation in ( )s, x U .

General solution of the ODE:

( )

−+

= x

a

sexp2 c x

a

sexp1c s, x U

0 1c = for ( )s, x U to be bounded at ∞= x

L ( ){ } =t ,0 U L ( ){ } ( ) 2 c s,0 U t f == from the general solution above.

Therefore,

( ) ( )

−= x

a

sexpsF s, x U where ( ) =sF L ( ){ }t f

From table of L ( ){ }t f ,a

2 πL

-1

−

−=

−

t 4aexpt ase 2

3

L -1 ( )t g

t a4

x exp

t a2

x x

a

sexp

2

2

3

=

−

π

=

−

( ) =s, x U L ( ){ }t f L ( ){ }t g Use convolution to get inverse.

( )( )

( )

( )∫ ττ

τ−

−

π=

τ−t

0 d f

t

exp

a2

x t , x u

2 3

t 2 a4

2 x

EXAMPLE 11: Find the displacement of a string of length l that is fixed at both ends if the initial displacement

is l

x m

sin

π

and the initial velocity is l

x n

sin

π

. Use Laplace transform method.

SOLUTION: The governing partial differential equation is

2

2

2

2

x

y 2 at

y

∂

∂=

∂

∂BC 1: ( ) 0 t ,0 y = BC 2: ( ) 0 t ,1y =

Take the Laplace transform of both sides of the PDE:

2 s L t ( ){ } ( ) ( )2

2

dx

d 2 a0 , x ' y 0 , x sy t , x y =−− L t ( ){ }t , x y

( ) ( )s, x Y dx

d 2 al x nsin

l x msinss, x Y 2 s

2

2 =π−π−

( ) ( )l x nsin

a

1l x msin

a

ss, x Y a

s

dx

s, x Y d 2 2 2

2

2

2 π−π−=− [a]

Complimentary function is

( )

+

−=

asx exp2 c

asx exp1c s, x c Y

Particular solution:

( )l x ncosD

l x mcosC

l x nsinB

l x msin As, x P Y π+π+π+π=





15

Substitute into [a] and solve for A, B, C and D:

( )

+

=π 2

l am2 s

s A ( )

+

=π 2

l an2 s

1B 0 DC ==

General solution:

( )( ) ( )

+

+

+

+

+

−=π

π

π

π

2

l an2

l

x n

2

l am2

l

x m

s

sin

s

sinsasx exp2 c

asx exp1c s, x Y

L ( ){ } =t ,0 y L ( ){ } 0 t ,1y =

( ) 0 0 2 c 1c 0 s,0 Y +++== Therefore, 2 c 1c −=

( ) 0 0 asl exp2 c

asl exp2 c 0 s,1Y ++

+

−−==

0 asl exp

asl exp2 c =

−−

Therefore, 0 2 c 1c ==

( )

( ) ( )

l x nsin

s

1l x msin

s

ss, x Y 2

l

an2 2

l

am2

π

++π

+=

ππ

Answer: ( )l x nsin

sin

l x msin

l at mcost , x y

l an

l at n

π+ππ=π

π

If ( ) ( ) ∑∞

=

π==1m

l x msinma x f 0 , x y and ( ) ( ) ∑

∞

=

π==1n

l x nsinnb x g 0 , x ' y , then

( ) ∑∑∞

=

π+∞

=

ππ=π

π

1nl x nsin

sinnb

1ml x msin

l at mcosmat , x y

l an

l at n

EXAMPLE 12: A slender rod of length l has its lateral surface insulated. Initially, the temperature in the bar

isl x nsinb π . Subsequently, the two ends of the bar are maintained at temperature 0 . Determine the temperature

( )t , x v in the bar.

SOLUTION:

2

2

2 x

v t v

a

1

∂

∂=∂∂ Let ( ) =s, x V L ( ){ }t , x v and take the Laplace transform of both sides of the PDE.

( ) ( )[ ]( )2

2

2 dx

s, x V d 0 , x v s, x sV

a

1 =−

l x nsin

a

bV a

s

dx

V d 2 2 2

2 π−=−

Characteristic equation: 0 a

s2 m2

=− a

sm ±=

Complementary function:

−+

= x

a

sexp2 c x

a

sexp1c c V





16

Particular solution:l x nsinB

l x ncos A pV π+π= Substitute into the differential equation and solve

for A and B to get: 0 A = and( )2

l ans

bBπ++

= .

Therefore, ( )( ) l

x nsin

s

b x asexp2 c x

asexp1c s, x V

2

l an

π

++

−+

=

π

L ( ){ } ( ) 2 c 1c s,0 V t ,0 v +== Therefore, 1c 2 c −= .

L ( ){ } ( )

−+

=== l

a

sexp2 c l

a

sexp1c s,1V 0 t ,l v

Since 1c 2 c −= , then 0 l a

sexpl

a

sexp1c =

−−

. Therefore, 0 2 c 1c == .

( )( ) l

x nsin

s

bs, x V 2

l an

π

+=

π

Answer: ( ) =t , x v L -1 ( ){ }

l x nsint

2

l anexpbs, x V π

π−=

If the initial temperature is ( ) x f , expand ( ) x f into a Fourier sine series, i.e.,

( ) ∑∞

=

π=1n

l x nsinnb x f

Then, the temperature distribution in the rod is

( ) ∑∞

=

π

π−=

1nl x nsint

2

l anexpnbt , x v





17

FORMULA SHEET

FOURIER INTEGRAL, FOURIER TRANSFORM, AND LAPLACE TRANSFORM

Fourier integral representation of ( ) x f :

( ) ( ) ( ) ( ) ( )[ ]

∫ ∫ ∫

∞ωωω+ωω

π

=∞ ∞

∞−

ω−ω

π

=0

d x sinB x cos A1

0

d ' dx x ' x cos' x f 1

x f

where ( ) ∫ ∞

∞−ω=ω ' dx ' x cos' x f A and ( ) ∫

∞∞−

ω=ω ' dx ' x sin' x f B

( ) ( ) ( )∫ ∫ ∞

∞−ω

∞∞−

−ω−π

= d ' dx x ' x x i e' x f 2

1 x f

(NOTE: There is no uniformity among authors on the definition of Fourier transform pairs, particularly on what

constant to use before the integral sign.)

Fourier Cosine Transform Pair Fourier Sine Transform Pair Exponential Fourier Transform Pair

( ) ( )∫ ∞

ωπ

=ω0

xdx cos x f 2 g ( ) ( )∫ ∞

ωπ

=ω0

xdx sin x f 2 g ( ) ( )∫ ∞

∞−ω−

π=ω dx x i e x f

2

1g

( ) ( )∫ ∞ ωωωπ

=0

xd cosg 2 x f ( ) ( )∫ ∞ ωωωπ

=0

xd sing 2 x f ( ) ( )∫ ∞∞−ωω−ω

π= d x i eg

2

1 x f

Short Table of Laplace Transforms( )t f ( )sF ( )t f ( )sF

1 s1 bt cos

2 2 bs

s

+

nt 1ns

! n

+

bt sin 2 2 bs

b

+

νt ( )

1s

1

+ν

+νΓ

at e as

1−

( )t f at e ( )asF − where

( ) =sF L ( ){ }t f

( ) ( )at u at f −− ( )sF ase−

( )t δ 1 t 4

a2 3

et −−

ase

a2 −π

( )at −δ ase− ( )

( )( )

( )2 2 2 2 bab

bt sin

aba

at sin

−+

−

+

+ 2 b2 s2 a2 s

1

( )t nf ( ) ( ) ( ) ( )( )0 1nf ...0 ' f 2 ns0 f 1nssF ns −−−−−−

L

( ){ }t f

L ( ){ }t g ( ) ( ) == sGsF L ( ) ( ) =

ττ−τ∫ t

0 d t g f L ( ) ( )

τττ−∫ t

0 d g t f

OR L -1 ( ) ( ){ } ( ) ( ) ( ) ( )∫ ∫ τττ−=ττ−τ=

t

0 d g t f

t

0 d t g f sGsF

Definition and properties of the unit impulse or delta function ( )0 t t −δ :

( )

=∞

≠=−δ

0 t t

0 t t 0 0 t t ( ) 1dt 0 t t =

∞∞−

−δ∫ ( ) ( ) ( )t f dt 0 t t t f =∞

∞−−δ∫

Documents

Fourier Laplace Transforms