Chapter 1

Fourier Transform
1.1. DIRICHLET’S CONDITION
A function f ( x ) is said to be satisfy the dirichlet’s condition in (a, b), if
(i) f ( x ) is defined and single valued function in (a, b)
(ii) f ( x ) and f ( x ) are sectionally (piecewise) continuous in (a, b).
1.2. THE FOURIER TRANSFORM
The Fourier transform of f ( x),  x   is denoted by F f (x) or F(s),
and is defined as F  f ( x )  F ( s )  1 
 f ( x )e dx where f ( x )
isx

2
satisfies the dirichlet’s conditions in the interval (  , ). The function
F(s) is also called Complex Fourier transform of f(x).
1.2. THE INVERSE FOURIER TRANSFORM
The inverse Fourier transform of F(s), is denoted by f(x), and is defined
1  isx
as f ( x )   F ( s )e dx.
2
1.3. FOURIER SINE TRANSFORM
The Fourier Sine transform of f ( x ), 0  x   is denoted by Fs(s), and is
2
defined as Fs{f(x)}  Fs(s)   f(x)sin sx dx where f ( x ) satisfies the
π0
dirichlet’s conditions in the interval ( 0,  ).
1.4. THE INVERSE FOURIER SINE TRANSFORM
The inverse Fourier Sine transform of Fs(s) is denoted by f(x), and is
2
defined as f(x)   Fs(s) sin sx ds.
π0
1.5. FOURIER COSINE TRANSFORM
The Fourier Cosine transform of f ( x ), 0  x   is denoted by Fc(s) is
2
defined as Fc{f(x)}  Fc(s)   f(x)cos sx dx where f ( x ) satisfies the
π0
dirichlet’s conditions in the interval ( 0,  ).
1.6. THE INVERSE FOURIER COSINE TRANSFORM
The inverse Fourier Cosine transform of Fc(s) is denoted by f(x), and is
2
defined as f(x)   Fc(s) cos sx ds.
π0

93
1.7. PROPERTIES OF FOURIER TRANSFORM
1. Linearity property: If F(s) and G(s) are Fourier transform of f(x) and
g(x) then F{C1f(x)+C2g(x)}=C1F(s)+C2G(s), where C1 and C2 are constants.
2. Shifting property: If F{f(x)}=F(s) then we have F{f(x-a) = eias F(s).
3. Change of Scale Property: If F(s) is the Fourier transform of f(x) then
1  s  is the Fourier transform of f(ax).
F 
a a
4. Modulation theorem: If F{f(x)} = F(s) then we have
1
F{f(x)cos ax}  [F(s  a)  F(s  a)]
2
5. Fourier transform of derivatives: If F{f(x)}=F(s) then we have
F{ f (x)}  isF ( s ) where f(x)  0 as x  .
In general, F{f ( n )(x)}  ( is )n F( s ).
6. Fourier integral representation of a function f(x)
(i) Fourier integral formula: Let f(x) be a function defined in   ,  
1
then Fourier integral formula is f(x)    f(t)cos s(t  x)dt ds
π 0 
(ii) Fourier sine integral formula: Let f(x) be a function defined in 0, 
2 
then Fourier sine integral formula is f(x)    f(t)sin st sin sx dt ds
π 00
(iii) Fourier cosine integral: Let f(x) be a function defined in 0,  then
2 
Fourier cosine integral formula is f(x)    f(t)cos st cos sx dt ds
π 00
1.8. CONVOLUTION THEOREM FOR FOURIER TRANSFORM
If F(s) and G(s) are the Fourier transform of f(x) and g(x), then we have
F{f(x)g(x)}=F{f(x)}.F{g(x)}.
1.8. APPLICATIONS OF FOURIER TRANSFORM
To solve partial differential equation through Fourier transform, first take
Fourier transform on both sides of differential equation and convert it in
ordinary differential equation. Then we take inverse Fourier transform on
both sides to obtain solution of the given differential equation.
Transform of Ist and 2nd order partial derivative with respect to x and t:
 u 
(i) F  u   isF{u(x, t)}  isU(s, t) (ii) Fs     sU s(s, t).
 x   x 
u   2u 
(iii) Fc    sU c(s, t) 
2
u( 0, t). (iv) F  2   (  is) 2U(s, t).
 x  π  x 

94
   2u  2
(v) Fs  2 
 s 2U s(s, t)  su( 0, t).
 x  π
 2    u  
(vi) FC   u2   s 2U s(s, t)  2 u x( 0, t)  u x ( 0, t )    
 x  π   x  x 0 
 u  dU(s,t)  u  dU s (s,t)
(vii) F    (viii) f s   
 t  dt  t  dt
u  dU c(s, t)   2u  d 2
(ix) Fc 
  (x) F  2   2 U(s, t)
 t  dt  t  dt
 2    2u  d 2
(xi) Fs   u2   d 2 U s(s, t)
2
(xii) Fc  2   2 U c(s, t)
 t  dt  t  dt
SOLVED PROBLEMS

1  x x  1
2

Pr.1. Find the Fourier transform of f ( x )  


 0 x 1
x cos x  sin x

x
Also evaluate  x 3
cos dx.
2
(RTU 2007, 09)
0

Sol. We know that the Fourier transform of f(x) is

1 
F( s )   f ( x )e dx  1 (1  x 2 )eisx dx
isx 1
[by def. of f(x)]
2  
2 1
1 1
  (1  x )[cos sx  i sin sx]dx
2
2 1
1 1 1

   ( 1  x ) cos sxdx  i  ( 1  x ) sin sxdx
2 2

2 1 1 
1 1 
  ( 1  x ) cos sxdx  0
2
[using property of odd functions]
2  1 
2  
1
2 1  2 sin sx  21
  ( 1  x 2
) cos sxdx   ( 1  x )    x sin sxdx
2  0    s 0 s 0 
2  2  cos sx  2 1 cos sx  2  2  cos s  2  sin sx  
1 1

 0   x    dx     0  2   
  s  s 0 s 0 s    s  s  s  s 0 

2  2 cos s 2 sin s  2  sin s  s cos s 

   3  2
  s2 s    s3 


1 isx
Now we have the inverse Fourier transform is f ( x )   F ( s )e ds
2 

95
 1 
 2 sin s  s cos s  isx 2   sin s  s cos s 
  2.   e ds    (cos sx  i sin sx )ds
2   s 3
    s3 
2    sin s  s cos s  
 sin s  s cos s  
     cos sxds  i    sin sxds
   s 3
   s 3
 
2    sin s  s cos s  
    cos sxds  0 [using property of an odd function]
   s3  
4   sin s  s cos s 
f(x)   cos sxds
 0 s3 

sin s  s cos s   x 1

cos sxds  f ( x )   4 ( 1  x ),
2

s 3
4  x 1
 0,
0


(sin s  s cos s ) s   1  3
Put x 
1
 cos ds  1    s  x 
2 4  4  16
3
2 0 s

 x cos x  sin x   x  3
   cos dx   .
0   
3
x 2 16

x x  a
2

Pr.2. Obtain the Fourier transform of f ( x )   Hence

0 xa


 as  a s  2 sin as  2as cos as
 2 2
 
evaluate:  cos 2  ds (RTU 2007, 10)
0   s3
Sol. We know that the Fourier transform of f(x) is
1  x 2 eisx  
 a
1 1 a isx 2 2 a
F( s )   e f ( x )dx 
isx
 e x dx      x e dx
isx

2  2 a 2  is  a is a 

1  a 2 ias 2  xeisx  

a

 ( e  e ias )    
1
e isx  
a
a 
2  is is  is  a ( is )2
 

1  a 2 ias
 ias 2  a ias ias
 (e e )  (e  e ) 2 e e
1 ias ias
 
2  is is  is s 
1  2a 2
2 2 
  sin as  2 ( 2 cos as )  3 ( 2 sin as )
2  s s s 


1 2
  
 s 3 a s  2 sin as  2as cos as 
2 2

2  

 2  a s  2 sin as  2as cos as
F( s )   
2 2
 
  s3

96
 
1 isx
Now we have the inverse Fourier transform is f ( x )   F ( s )e ds
2 


1 

e
isx
 
 
 2  a s  2 sin as  2as cos as
2 2
ds

2    s3

1 1
  (cos sx  i sin sx )  s 3 [( a s  2 ) sin as  2as cos as ].ds
2 2

 


1 
 cos sx
 ( a s  2 ) sin as  2as cos as s 3 ds
 
2 2


 ( a s  2 ) sin as  2as cos as s 3 ds

1 
sin sx 
 2 2

  


1  2 2
 cos sx 

 ( a s  2 ) sin as  2as cos as s 3 ds  0 [using pro. of odd function]
  



2  cos sx a 2 s 2  2 sin as  2as cos s  ds
 
 0 s3 



 
cos sx a 2 s 2  2 sin as  2as cos as as 
 f(x)

 s3 2
0

xa 
 x x  a
2 2
  x 2 a
   2 Put x 
2
0 xa  0 xa 2


 as  a s  2 sin as  2as cos as as
 2 2

a a 2  2

Hence  cos 2  ds    .
0   s3 2 2 8
Pr.3. Find the Fourier sine and cosine transform of f(x), where
1, 0  x  a
f(x)  . (RTU 2007)
0, xa
Sol. We know that the Fourier sine transform of f(x) is
2 2a
Fs ( s )   f ( x ) sin sx dx 
0
1. sin sx dx [by def. of f(x)]
0
a
2   cos sx  2  1  cos ax 
     .
  s 0  s 
Now using the definition of Fourier cosine transform we have
2 2a 2  sin sx 
a
2  sin as 
FC ( s )   f ( x ) cos sx dx   1. cos sx dx      .
 0  0   s 0  s 
s
Pr.4. Find f(x) if its sine transform is . (RTU 2007)
1 s2

97
 s
Sol. Given that Fs ( s ) 
1  s2
Using the definition of inverse Fourier transform, we have
2 2  s sin sx 2 1 (1  s 2  1 )
f(x)  F ( s ) sin sx ds    1  s 2 ds   sin sx ds
0 s 0  0 s( 1  s 2 )
2  sin sx 2  sin sx
  ds   ds
 0 s  0 s( 1  s 2 )
2   2  sin sx   sin sxds  
    ds    
 2  0 s( 1  s 2 )  0 s 2
 2  sin sx
f(x)   ds ….(1)
2  0 s( 1  s 2 )
df 2  s cos sx
Differentiating with respect to x, we get   ds ….(2)
dx  0 s( 1  s 2 )
Again differentiating with respect to x, we get
d2 f 2  s sin sx d
  ds  f or ( D2 1) f  0 where D  ….(3)
dx 2
 0 (1  s 2 ) dx
df
Solution of (3) is f  Aex  Be  x and  Aex  Be  x …..(4)
dx


and df   2  ds 2   2 tan1 s 0   2 .      
 
At x = 0, 
f 
2 dx  0 1 s   2 2
[from (1) and (2)]
 
At x = 0,  A B and   A B [from (4)]
2 2

Solving above these, we get A = 0 and B    f ( x )   .e  x [from (4)]

2 2
Pr.5. Find the Fourier sine and cosine transform of
x, 0  x 1

f ( x )   2  x , 1  x  2. (RTU 2007)
0, x2

2
Sol. The Fourier sine transform of f(x) is Fs ( s )   f ( x ) sin sx dx
 0

2 1 2
 2 1 2

   f ( x ) sin sx dx   f ( x ) sin sxdx    x sin sx dx   ( 2  x ) sin sx dx
 0 1   0 1 

98
 2   x cos sx sin sx   cos sx sin sx  
1 2

   2    ( 2  x )  2  
  s s 0  s s 1 

2  cos s sin s sin 2s cos s sin s  2  2 sin s  sin 2s 

    2 0 2   2 
 s s s s s    s2 

2 ( 1  cos) sin s
Fs ( s )  2 .
 s2
Now using the definition of Fourier cosine transform, we have
2 2 1 2

FC ( s )   f ( x ) cos sx dx    x cos sx dx   ( 2  1 ) cos sx dx
0 0 1 

2  x sin sx cos sx   sin sx cos sx  

1 2

   2   ( 2  x )  2  
  s s 0  s s 1 

2  sin s cos s 1 cos 2s sin s cos s 

  2  2   2 
  s s s s2 s s 
2  2 cos s  cos 2s  1 2  2 cos s  2 cos 2 s 
   
  s 2 
   s2 
2 ( 1  cos) cos s
FC ( s )  2 .
 s2

Pr.6. (i) Solve the integral equation:  f ( x ) cos x dx  e  (RTU 2011)
0

Sol. Using the definition of Fourier cosine transform pair, we have

2 2 ….(1)
Fc ( s )   f ( x ) cos sx dx and f(x)  Fc ( s ) cos sx ds
 0  0

Replacing s by  , we get Fc ( s ) 
2
 f ( x ) cos  x dx 
2
e  .
 0 
2  2  2  
Now f ( x )   e cos  x d   e cos  x d
 0   0

2  e    
   cos x   sin x  e ax
  e cos bx dx  a 2  b 2 (a cos bx  b sin bx
ax
 1   2
  
2 1  2
 0
  1  2    1  2  .
Pr.6. (ii) Find the Fourier integral representation of f(x) where

99
  0, x0
1 (RTU 2007)
f ( x ) , x0
 2 x x0
e ,
Sol. Using the definition of f(x), we have f ( x )  1    f ( t ) cos( t  x )dt  ds
 

 0  
1    1   1 
    f ( t ) cos s( t  x )dt ds     e cos s( t  x )dt ds
 0 0   0 0 

1   e 1  1
    cos( t  x )  s sin s( t  x ) .ds   
1
 cos sx  s sin sxds
 0 1  s 2
0  0 1  s2
Hence f ( x )  1  (cos sx  s sin sx ds.


 0 1  s2
Pr.7. Find the function f(x) satisfying the integral equation
 1  s , 0  s  1
0 f(x)cos sx dx   0, s 1
(RTU 2010, 12)

Sol. Using definition of Fourier cosine transform we have
2 2 1  s , 0  s 1
FC (s)   f(x)cos sx dx  
π0 π  0, s 1
Using inverse Fourier cosine transform, we have
2 2 1 

f(x)   FC (s) cos sx ds    FC (s) cos sx ds   FC (s) cos sx ds
π0 π 0 1 
2 1 2  2
1
sin sx cos sx 
  ( 1  s) cos sx dx  0  ( 1  s ) 
π  0 π  π  x x 2  0
2  1 cos sx  2 4  x
   ( 1  cos x)  2 sin 2  .
π  x 2 x 2  πx 2 πx 2
Pr.8. Find the Fourier sine transform of f ( x )  e  x , x  0. Also show
 x sin mx 
that  dx  e m , m  0. (RTU 2007, 10)
0 x 1
2
2
Sol. Using the definition of Fourier sine transform we have
2 2  x
Fs ( s )   f ( x ) sin sx dx   e sin sx dx
 0  0

2  e x  2  2 s 
 sin sx  s cos sx    0  1  s 2  s     1  s 2 
1
 
 1  s 2
    
Now using the definition of inverse Fourier sine transform we have
2 2 2 s 
f(x)  F ( s ) sin sx ds     sin sx ds
0 s  0   1  s2 

100
 2  s sin sx  s sin sx  
or f ( x )  
 0 1  s2
ds  ds  f ( x )  e  x x  m
0 1  s2 2 2


 1  s 2 ds  2 e s  x . Hence  x sin mxdx   e  m .
s sin ms m 
or
0 0
2
1 x 2
Pr.9. Prove that (i) F  f ( x )   2 f ( 0 )  s 2 F  f ( x )
C
 C

(ii) F  f ( x )   2 s f ( 0 )  s 2 F  f ( x ) where f ( x )  0 as x   (RTU 2010)

s
 s


Sol. (i) FC  f ( x )  2  f ( x ) cos xdx
 0

2
( x ) cos sx0  s  f ( x ) sin sxdx 

  f
  0



2
[ 0  f ( 0 )]  s
2
 f ( x ) sin sx dx  f ( x )  0 as x  
  0

2 2
 f ( 0 )  sFs { f ( x )}   f ( 0 )  s [  sFC { f ( x )}]
 
Hence FC { f ( x )}   2 f ( 0 )  s 2 FC { f ( x )}.


2
( x ) sin sx0  s  f ( x ) cos sxdx
2 
(ii) Fs { f ( x )}   f ( x ) sin sxds   f
 0   0


2 
 2
  0  s  f ( x ) cos sxdx   s  f ( x ) cos sxds
 0  0
 2  2
 sFC { f ( x )}  s  f ( 0 )  sFs { f ( x )}  sf ( 0 )  s 2 F { f ( x )}.
   s

Pr.10. Use the method of Fourier integral to determine the
displacement u(x, t) of an infinite string, given that the string in
initially at rest and that the initial displacement is f ( x),  x   .
Show that the solution can also be put in the form
1
u( x , t )  [ f ( x  Ct )  f ( x  Ct )]. (RTU 2007, 10)
2
Sol. The displacement u(x, t) is governed by two dimensional wave
 2u 2  u
2
equation, so we have  C ,  x  ,t  0 ….(1)
t 2 x 2
Subject to boundary condition:
(i) u(x, 0) = initial displacement = f(x) …..(2)
u
(ii)  0 at t = 0 [becouse string is initially at rest] ……(3)
t

101
Taking Fourier transform of (1), we have
 2U  2U  2U
  C 2 ( is )2 U ( s , t )   C 2 s 2U   C 2 s 2U  0
t 2
t 2 t 2
Hence solution is U ( s ,t )  A cos Cst  B sin Cst …..(4)
dU ….(5)
   ACs sin Cst  BCs cos Cst
dt
From equation (2), we have u(x, 0) = f(x)  U ( x ,0 )  F { f ( x )}  F ( s ).
From equation (2), we have (t = 0), A = F(s)
From equation (3), we have  u   0 
d u( s ,0 )
0
 t  t 0 dt

From equation (5), we have B = 0

Thus from equation (4), we have U ( s ,t )  F ( s ) cos Cst


Taking inverse Fourier transform u( x ,t )  1  F ( s ) cos Cst e isx ds
2 
1  isx  e iCst  e  iCst 
  e F ( s ) ds
2   2 
1  1  is 1  is 
   e ( x  Ct )F ( s )ds   e ( x  Ct )F ( s )ds
2  2  2  
Hence u( x ,t )  1 [ F( x  Ct )  f ( x  Ct )].
2
1 as
Pr.11. Find f(x), if its Fourier sine transform is e . (RTU 2011, 12)
s
Sol. The Fourier sine transform is f(x)  2  Fs(s) sin sx ds
π 0
2  sin sx.e-as …..(1)
  ds
π 0 s
Differentiating under the sign of integral with respect to ‘a’ by treating ‘a’
as parameter, we get
2   s  sin sx
 as
df 2  as 2 x
  ds    e sin sx ds   π a 2  x 2 ….(2)
da π 0
s π 0
Hence, integrating both sides of (2) with respect to ‘a’ again, we get
2 x 2 x  1 a  …(3)
f(x)  dx   tan C
π a2  x2 π x  x 
From (1), we get f(0)=0 so by (3), we get

tan1    C
2 2
0  C …(4)
π 2 

102
Substituting the value of C in (3), we have
  
1 a 
cot 1   tan 1 x .
a
f(x)   tan 
22 x 2 x 2 a
u  2u
Pr.12. Solve  C 2 2 ,t  0 subject to u( x ,0 )  f ( x ),  x  
t t
(RTU 2007)
Sol. Given that the domain x is  ,  , so we use Fourier transform.
Taking Fourier transform on both sides of u  C 2  u
2

t t 2
 u   u 
2
F    C 2 F  2   dU  C 2 s 2U or
dU
 C 2 s 2U  0

 t  x  dt dt
2s 2t
The solution is U ( s ,t )  Ae C ….(1)
Given also u( x ,0 )  f ( x )  U ( s ,0 )  F ( s ) [using Fourier transform]
Put t = 0 in (1), we get A=F(s). Then by (1), we have
2s 2t
Using this in (i) U ( s ,t )  F ( s )e C

Taking inverse Fourier transform we get u( x ,t ) 

1  C 2s 2t isx
 F( s )e e ds.
2 
Pr.13. Solve u   u2 , given that u  0 at x  0 and u(x, 0 )   x , 0  x  1
2

t x 
x 0 , x 1
u(x, t) is bounded and x > 0, t > 0. (RTU 2007, 11, 12)
u
Sol. Given that   so we use Fourier cosine transform. Taking
 x  x0
u  2 u  u    2u 
Fourier transform on both sides of   FC    FC  2 
t x 2  t   x 

 dU C   2  u   s 2U C U C  U C s, t   FC{u(x,y)} u sin g given conditions

dt π  x  x 0
dUC  dUC
 s 2U C  s 2U C  0
dt dt
2t
The solution is U C ( s ,t )  Ae s ….(1)
x, 0  x  1
Given that u(x,0 )  

0, x 1
Taking its Fourier cosine transform we have
1
2 2 2  x sin x cos sx 
U C (s, 0 )   u(x, 0 ) cos sx dx   x cos sx dx  
π0 π0 π  s s 2  0

103
 2  sin s cos s 1 
  2  2 .…(2)
π  s s s 

Put t = 0 in (i) A  U C ( s ,0 ) 2  sin s cos s-1 

 A   
π s s2 

From (1), becomes U C(s,0 )  2  sin s  cos 2s-1 e s2t

π s s 
Taking inverse Fourier transform, we get
2  2  sin s cos - 1  s t
u( x ,t )  
2

 
π 0 π s s2 
e cos sxds

2   s sin s  cos -1   s 2t
 u(x,t)   e cos sxds.
π 0 s2 
Pr.14. Using Fourier sine transform solve the partial differential
u  2u
equation  , x  0,t  0 . Subject to the conditions u(0, t) = 0.
t x 2
1, 0  x  1 It may be assumed that u(x, t) is bounded;
u(x, 0 )   .
0 , x 1
u
also u and approach zero as x   . (RTU 2010, 11)
x
Sol. Given that u   u
2

t x 2
 u    2u 
Taking Fourier sine transform of both sides, we get Fs    Fs  2 
 t   x 
dU s 2 dU s
su( 0,t )  s 2U s  s U s
2
or  or  s 2U s  0
dt  dt
2t
The solution is U s ( s ,t )  Ae s ….(1)

Now we have U ( s , 0 )  2  u( x ,0 ) sin sx dx  2 1 sin sx dx

s
π 0  0
1
2  cos sx  2  1  cos s 
     
 s 0  s 

At t = 0, becomes U s ( s ,0 )  A  2  1  cos s 
  A
 s 
2  1  cos s   s 2t
From (1) U s ( s , t ) 
 e
 s 
Taking inverse Fourier sine transform, we get

104
 2   1  cos s  s t 2  sin sx s t 
cos s sin sx s t 
u( x ,t ) 
2

  sin sx e dx    e ds  
2 2
 e ds
 0 s   0 s 0 s 
2
u( x ,t )  [ I1  I 2 ] ….(2)


ds dv
I1   e  s t sin sx put s t  v  s t  v  ds 
2 2 2

0 s t

 v  t dv  v2  xv  dv
I1   e v sin x    e sin  or I1   erf  x 
2

0  t  v t 0  t t 2 2 t 

ds 1 
  sin( x  1 )s  sin( x  1 )s e s t
2 ds
and I 2   cos s sin s x e s t
2

0 s 20 s
dv
Put s 2t  v 2  s t  v  ds 
t
1   v  x 1  x  1   t dv
I2    e sin v  e v sin v
2 2

2  0  t   t   v t

1   v   x  1   dv  v  x  1   dv
   e sin  v    v 
2 2

2  0   e sin
 t   v 0  t   v

1   x 1   x  1 
  erf    erf  
2  2  t  2  t 

2   x     x 1  x  1 
From (2), we have u( x ,t )   erf    erf    erf  
  2 2 t  4 2 t   2 t  
 x  1   x 1  x  1 
or u( x ,t )  erf    erf    erf   .
2 t  2  2 t   2 t 
u  2u
Pr.15. Solve the heat equation  C 2 2 ,  x  , t  0
t x
Subject to u(x, 0) = f(x), where f ( x )  u0 , x  a (RTU 2010)
 0, x  a
Sol. Given that the u  C 2  u
2

t x 2
Here the domain of x is  ,  so we use Fourier transform.
 u    2u 
Taking Fourier transform of both sides, we get F    C 2 F  2 
 t   x 

105
dU
dt

 C 2  s 2U 
[where U  U ( s ,t ) and F { u( x ,t )}  U ( s ,t )]

C 2 s 2t
 C 2 s 2U  0 on solving U ( x ,t )  Ae
or dU ….(1)
dt
u0 , x  a
Given that u( x ,0 )  f ( x )  

 0, x  a
Taking Fourier transform, we have

 
a
u0 a isx u0  e isx   u0 e ias  e ias
U ( s, 0 )   0
u e dx   
2 a a 2  is  a is 2
u0 2 ….(2)
U( s, 0 )  .2 sin as  ( u0 sin as )
s 2 

2 sin as
Using (1) and (2), we get A  u0
 s

From (1), we have U ( s ,t )  sin as sC 2s2t isx 2

.e .e dx u0
 s
Now take inverse Fourier transform, we have
 
2 1 sin as C 2s 2t isx u0 sin as C s t
 s .e .e ds   (cos sx  i sin sx )ds
2 2
u( x ,t ) 

u0  .e
2   s
u0   sin as C s t 
 . cos sxds  0 [Using property of odd function]
2 2


  s
.e

2u  e C s t
2 2

u( x ,t )  0  [sin( a  x )s  sin( a  x )s ] ds
 0 2s
u0  e  C s t
2 2

  [sin( a  x )s  sin( a  x )s ] ds
 0 s
Put C 2 s 2t  v 2  Cs t  v  ds  dv
C t
e v  ( a  x )v  a  x   dv
2

u0
u( x ,t )   sin  sin v
 0 v / C t  C t  C t   C t

u0 
 v ( a  x )v  v ax   dv  …(3)
   e sin v 
2 2
  e sin
  0 C t 0 C t   dt 

t b 
 e cos 2btdt  2 e
2 2
Since
0

Integrating both sides w.r.t. ‘b’ from 0 to C , we get

2 x

106
  C/2
C/2 x
 2  sin 2bt  x 2
  e t   dt  b
 e db
0  2b  0 2 0


sin Ct x   2 C/2 
  e t dt  b
2 2
. .  e db
0 2t 2 2  0


sin( Ct / x )   C 
  e t dt  erf   …..(4)
2

0 t 2 2 x 
u0    ax    ax 
Using (4) and (5), we get u( x ,t )   erf    erf  
  2  2C t  2  2C t 

or u0   ax   ax 
u( x , t )  erf    erf   .
2   2C t   2C t 

Pr.16. Using Fourier cosine transform, solve   C 2  2 Subject to

2

t t

the conditions; (i)   0 , when t  0, x  0 (ii)   (constant
t

when x=0 and (t>0). Assume that  ( x, t ) and both tend to zero
x
as x  . (RTU 2009)
  2
Sol.: Given that the  C2 2
t t
Taking Fourier cosine transform of both sides, we get F      2 
   C FC  2 
2

  x 
C
 t

or d  C   2 C 2     s 2C 2  C [where  C   C ( s ,t ) ]
dt   x  x0
C 2
  s 2C 2  C  C 2 This is a linear differential equation
dt 
I .F .  e 
s2C 2dt
 es C t
2 2

 es C t 
2 2
2 2
The solution is  C e
s 2c 2t
 C e2 s 2C 2t
dt  A  C 2  2 2
A
   s C 
2c 2t 2  s2C 2t
  C es  e A …..(1)
 s2
Given also  x , 0  0   C  ( s , 0 )  0 [using Fourier cosine transform]
2  2 
At t=0, by equation (1), we have 0  .  A .
 s2  s 2e

107
 2 
From (1), we have  C  . ( 1  e s C t )
2 2

 s2
Now using inverse Fourier transform we have
2  2  ( 1  e e )  2 
s 2C 2t
s 2C 2t cos sx
 ( x ,t ) 
  cos sxds   ( 1  e e ) 2 ds.
   0 s 2
  0 s

Pr.17. Using Fourier transform, show that the solution of the partial
equation  u  C 2  u ,    x  , t  0 Satisfying the
2 2
differential
t 2
x 2
u
conditions (i) u  f ( x ), 2  0 at t = 0 (ii) u( x ,t ), u , both tend to zero
t x

as x   can be written in the form u( x , t )  1  F ( s ) cos C st e isx ds
2 
where F(s) = Fourier transform of f(x). (RTU 2009)
 2u 2  u
2
Sol. Given that  C
t 2 x 2
Taking Fourier transform of both sides, we get
  2u    2u 
F  2   C2F  2 
 t   x 
d 2U d 2U
 C 2 2
s U or  C 2 s 2U  0
dt 2 dt 2
Solving U ( s ,t ) = Acos Cst + Bsin Cst ….(1)
Given condition u( x ,0 )  f ( x )
Taking Fourier transform, we get. u( s ,0 )  F ( s ) …..(2)
u( x ,0 ) U ( s ,0 )
and 0 0 .…(3)
t t
Using (1) and (2), we get A  F( s )
Now differentiating (1) with respect to t, we get
dU
( s ,t )   ACs sin Cst  BCs sin Cst  0  Bcs  B  0 [using (3)]
dt
Now from (1), we have U ( s ,t )  F ( s ) cos Cst

Taking inverse Laplace u( x ,t )  1  F ( s ) cos Cst e isx ds.
2 
Pr.18. Solve the boundary value problem u  K  u , u( 0,t )  u0 , t  0
2

t x 2
u
given u( x , 0 )  0, x  0 also u and approach to zero x   . (RTU 2007)
x

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Sol. Given that the u  K  u
2

t x 2
Taking Fourier sine transform on both sides, we get
 u    2u 
Fs    KFs  2 
 kdt   x 

 dU s  2 
 K su( 0,t )  s 2U s 
dt   
 2 
 K su 0  s 2U s 
  
dU s 2
or  s 2 KU s  K su 0
dt 
This is a linear differential equation.

I .F .  e 
s 2 Kdh
 e s Kt
2

2
2 Kt 2 2 2 e s Kt
The solution is U s e s K su 0 e s Kt dt  C1  Ksu0  C1
  s2K

u0 2 2 Kt
 U s ( s ,t )   C1e s ….(1)
s 
Given u(x, 0) = 0  U s ( s, 0 )  0 [Using Fourier sine transform]

u0 2
At t = 0, (1) becomes C1  u0 2
 U s ( s ,0 )  0  C1  
s  s 

Thus by (1), we have U s ( s ,t )  u0 2 1  e s Kt

s 
2
 
Taking inverse Fourier transform we have

u( x , t ) 
0 
2 2

2

1  e s Kt sin sx ds 
or
2u  2
u( x ,t )  0  1  e  s Kt
 0
sin x
s

ds 
or u( x , t ) 
2u0  sin sx 
 s 2 Kt sin sx   2u0     e s Kt sin sx ds 2

 ds   e ds    2 0 
  0 s 0 s s 

Hence u( x , t )  u0 1 
 2 
 s 2 Kt sin sx 
 e ds .
  0 s 

109
 EXERCISE 1.1

Q.1. Find the Fourier transform of f(x) if f ( x )   x, x a

 .
0 , x a

Q.2. Find the Fourier transform of f(x) if e  x .

 ax
Q.3. Find the Fourier sine transform of e .
x

cos sx 
Q.4. Show that  ds  e s , x  0.
0 s 1
2
2
1
Q.5. Find f(x), if its cosine transform is .
1  s2
u  2u
Q.6. Solve  2 2 , given that u( 0, t )  0, u( x ,0 )  e  x , x  0 is
t x
bounded where x>0, t>0.

ANSWERS 1.1
s
Q.1. i 2
 as cos sa  sin sa . Q.2. 2 1
. Q.3. 2
tan 1  .
s2   1  s2  a
  2 s t2
Q.5. e  x . Q.6. 2  se sin sx
ds.
2  0 1 s2

110