Fourier Series | Engineering Mathematics - Civil Engineering (CE) PDF Download

 Page 1


Fourier Series
Periodic Function A function f(x) is said to be periodic 
if f(x + a) = f(x) for all x. The least value of a is called the 
period of f(x).
Example: sin x, cos x are periodic functions with period 
2p .
 1.  f (x) and g (x) are periodic functions with period k 
then af (x) + bg (x) is also a periodic function with 
period k. 
 2.  If f (x) is a periodic function with period k, then the 
period of f (ax) is 
k
a
.
 3.  If the periods of functions f (x), g(x) and h(x) are a, b, 
c, respectively, then the period of f (x) + g (x) + h (x) is 
the lcm of a, b and c.
NOTES
Euler’s Formula for the 
 F ourier  Coe?  cient s
Let f (x) is a periodic function whose period is 2p and is 
integrable over a period. Then f (x) can be represented by 
trigonometric series.
fx
a
anxb nx
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
where a
0 
 a
n
, b
n
 are called Fourier coeffi  cients and these are 
obtained by
af xdx
0
1
=
-
?
p
p
p
() ,
af xnxdxn
n
==
-
?
1
p
p
p
()cosfor 1, 2, 3,… 
bf xnxdxn
n
==
-
?
1
p
p
p
()sinfor 1, 2, 3,… 
SOLVED EXAMPLES
Example 1
Obtain the Fourier series expansion of f (x) = e
x
 in (0, 2p ).
Solution
af xdx
0
0
2
1
=
?
p
p
()
     =
?
1
0
2
p
p
edx
x
=
?
?
?
=-
11
1
0
2
2
pp
p
p
ee
x
() (1)
  
Partial Diff erential 
Equations
Chapter 03.indd   58 5/31/2017   12:42:27 PM
Page 2


Fourier Series
Periodic Function A function f(x) is said to be periodic 
if f(x + a) = f(x) for all x. The least value of a is called the 
period of f(x).
Example: sin x, cos x are periodic functions with period 
2p .
 1.  f (x) and g (x) are periodic functions with period k 
then af (x) + bg (x) is also a periodic function with 
period k. 
 2.  If f (x) is a periodic function with period k, then the 
period of f (ax) is 
k
a
.
 3.  If the periods of functions f (x), g(x) and h(x) are a, b, 
c, respectively, then the period of f (x) + g (x) + h (x) is 
the lcm of a, b and c.
NOTES
Euler’s Formula for the 
 F ourier  Coe?  cient s
Let f (x) is a periodic function whose period is 2p and is 
integrable over a period. Then f (x) can be represented by 
trigonometric series.
fx
a
anxb nx
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
where a
0 
 a
n
, b
n
 are called Fourier coeffi  cients and these are 
obtained by
af xdx
0
1
=
-
?
p
p
p
() ,
af xnxdxn
n
==
-
?
1
p
p
p
()cosfor 1, 2, 3,… 
bf xnxdxn
n
==
-
?
1
p
p
p
()sinfor 1, 2, 3,… 
SOLVED EXAMPLES
Example 1
Obtain the Fourier series expansion of f (x) = e
x
 in (0, 2p ).
Solution
af xdx
0
0
2
1
=
?
p
p
()
     =
?
1
0
2
p
p
edx
x
=
?
?
?
=-
11
1
0
2
2
pp
p
p
ee
x
() (1)
  
Partial Diff erential 
Equations
Chapter 03.indd   58 5/31/2017   12:42:27 PM
    
af xnxdx
n
=
?
1
0
2
p
p
()cos
     =
?
1
0
2
p
p
enxdx
x
cos
we know that ·
?
ebxdx
ax
cos 
=
+
+
e
ab
abxb bx
ax
22
[cos sin]   
?=
+
+
?
?
?
?
?
?
a
e
n
nx nnx
n
x
1
1
2
0
2
p
p
(cos sin) 
=
+
-
+
?
?
?
?
?
?
1
1
2
1
1
2
22
p
p
p
e
n
n
n
(cos )
  
bf xnxdxe nxdx
n
x
==
? ?
11
0
2
0
2
pp
p p
()sinsin  
 =
-
+
1
1
0
2
2
p
p
enxn nx
n
x
(sin cos]
 
? ebxdx
e
ab
abxb bx
ax
ax
sin( sincos )) =
+
?
?
?
?
?
?
-
? 22
=
+
-
11
1
2
2
2
p
p
p
n
ne nn (cos ) 
=
+
-
n
n
en
p
p
p
()
(cos )
1
12
2
2
 
?= ++
=
8
?
fx
a
anxb nx
n
n
n
() (cos sin)
0
1
2
 =- +
=
8
?
1
2
1
2
1
p
p
() e
n
 
11
1
21
1
12
2
2
2
2
p
p
p
p
pp
+
-+
+
-
?
?
?
?
?
?
n
en
n
n
en (cos )
()
(cos ) 
Even and Odd Functions
Even function: A function f (x) is said to be even if f (–x)  = 
f (x) for all x.
Example: x
2
, cos x 
Odd function: A function f (x) is said to be odd if f (–x) 
= – f (x) for all x 
Example: x
3
, sin x 
 1. The sum of two odd functions is odd.
 2.  The product of an odd function and an even function 
is odd.
 3. Product of two odd functions is even.
NOTES
Fourier Series for Odd Function and Even Function
Case 1: Let f (x) is an even function in (–p , p ). Then the 
Fourier series of the even function contains only cosine 
terms and is known as Fourier cosine series and it is 
  fx
a
anx
n
n
() cos, =+
=
8
?
0
1
2
  where
  af xdxa fx nxdx
n 0
00
22
==
??
pp
pp
() ,( )cos
Case 2: If f (x) is an odd function, then the Fourier series of 
an odd function contains only sine terms, and is known as 
Fourier sine series. 
fx bnx
n
n
() sin, =
=
8
?
1
where                         bf xnxdx
n
=
?
2
0
p
p
()sin 
Example 2
Expand the function fx
x
()=-
p
22
24 8
 in Fourier series in 
the interval (–p , p ).
Solution
   fx
x
()=-
p
22
24 8
fx
xx
fx ()
()
() -= -
-
=- =
pp
22 22
24 8248
 
\ f (x) is an even function.
fx
a
anx
n
n
() cos =+
=
8
?
0
1
2
    af xdx
x
dx
0
0
2
0
2
22
2248
== -
??
pp
p
pp
()
         =
-
?
?
?
?
?
?
?
?
?
=
1
24 24
0
23
0
p
p
p
xx
 
    af xnxdxs
n
=
?
2
0
p
p
()cos
         =-
?
?
?
?
?
?
?
2
24 8
22
0
p
p
p
x
nxdx cos  
=
-
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
-
-- ?
?
?
?
?
?
2
24 8
1
8
2
22
0
p
p
p
xnx
n
xnx
n
dx
sin
()sin
0 0
p
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Chapter 03.indd   59 5/31/2017   12:42:30 PM
Page 3


Fourier Series
Periodic Function A function f(x) is said to be periodic 
if f(x + a) = f(x) for all x. The least value of a is called the 
period of f(x).
Example: sin x, cos x are periodic functions with period 
2p .
 1.  f (x) and g (x) are periodic functions with period k 
then af (x) + bg (x) is also a periodic function with 
period k. 
 2.  If f (x) is a periodic function with period k, then the 
period of f (ax) is 
k
a
.
 3.  If the periods of functions f (x), g(x) and h(x) are a, b, 
c, respectively, then the period of f (x) + g (x) + h (x) is 
the lcm of a, b and c.
NOTES
Euler’s Formula for the 
 F ourier  Coe?  cient s
Let f (x) is a periodic function whose period is 2p and is 
integrable over a period. Then f (x) can be represented by 
trigonometric series.
fx
a
anxb nx
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
where a
0 
 a
n
, b
n
 are called Fourier coeffi  cients and these are 
obtained by
af xdx
0
1
=
-
?
p
p
p
() ,
af xnxdxn
n
==
-
?
1
p
p
p
()cosfor 1, 2, 3,… 
bf xnxdxn
n
==
-
?
1
p
p
p
()sinfor 1, 2, 3,… 
SOLVED EXAMPLES
Example 1
Obtain the Fourier series expansion of f (x) = e
x
 in (0, 2p ).
Solution
af xdx
0
0
2
1
=
?
p
p
()
     =
?
1
0
2
p
p
edx
x
=
?
?
?
=-
11
1
0
2
2
pp
p
p
ee
x
() (1)
  
Partial Diff erential 
Equations
Chapter 03.indd   58 5/31/2017   12:42:27 PM
    
af xnxdx
n
=
?
1
0
2
p
p
()cos
     =
?
1
0
2
p
p
enxdx
x
cos
we know that ·
?
ebxdx
ax
cos 
=
+
+
e
ab
abxb bx
ax
22
[cos sin]   
?=
+
+
?
?
?
?
?
?
a
e
n
nx nnx
n
x
1
1
2
0
2
p
p
(cos sin) 
=
+
-
+
?
?
?
?
?
?
1
1
2
1
1
2
22
p
p
p
e
n
n
n
(cos )
  
bf xnxdxe nxdx
n
x
==
? ?
11
0
2
0
2
pp
p p
()sinsin  
 =
-
+
1
1
0
2
2
p
p
enxn nx
n
x
(sin cos]
 
? ebxdx
e
ab
abxb bx
ax
ax
sin( sincos )) =
+
?
?
?
?
?
?
-
? 22
=
+
-
11
1
2
2
2
p
p
p
n
ne nn (cos ) 
=
+
-
n
n
en
p
p
p
()
(cos )
1
12
2
2
 
?= ++
=
8
?
fx
a
anxb nx
n
n
n
() (cos sin)
0
1
2
 =- +
=
8
?
1
2
1
2
1
p
p
() e
n
 
11
1
21
1
12
2
2
2
2
p
p
p
p
pp
+
-+
+
-
?
?
?
?
?
?
n
en
n
n
en (cos )
()
(cos ) 
Even and Odd Functions
Even function: A function f (x) is said to be even if f (–x)  = 
f (x) for all x.
Example: x
2
, cos x 
Odd function: A function f (x) is said to be odd if f (–x) 
= – f (x) for all x 
Example: x
3
, sin x 
 1. The sum of two odd functions is odd.
 2.  The product of an odd function and an even function 
is odd.
 3. Product of two odd functions is even.
NOTES
Fourier Series for Odd Function and Even Function
Case 1: Let f (x) is an even function in (–p , p ). Then the 
Fourier series of the even function contains only cosine 
terms and is known as Fourier cosine series and it is 
  fx
a
anx
n
n
() cos, =+
=
8
?
0
1
2
  where
  af xdxa fx nxdx
n 0
00
22
==
??
pp
pp
() ,( )cos
Case 2: If f (x) is an odd function, then the Fourier series of 
an odd function contains only sine terms, and is known as 
Fourier sine series. 
fx bnx
n
n
() sin, =
=
8
?
1
where                         bf xnxdx
n
=
?
2
0
p
p
()sin 
Example 2
Expand the function fx
x
()=-
p
22
24 8
 in Fourier series in 
the interval (–p , p ).
Solution
   fx
x
()=-
p
22
24 8
fx
xx
fx ()
()
() -= -
-
=- =
pp
22 22
24 8248
 
\ f (x) is an even function.
fx
a
anx
n
n
() cos =+
=
8
?
0
1
2
    af xdx
x
dx
0
0
2
0
2
22
2248
== -
??
pp
p
pp
()
         =
-
?
?
?
?
?
?
?
?
?
=
1
24 24
0
23
0
p
p
p
xx
 
    af xnxdxs
n
=
?
2
0
p
p
()cos
         =-
?
?
?
?
?
?
?
2
24 8
22
0
p
p
p
x
nxdx cos  
=
-
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
-
-- ?
?
?
?
?
?
2
24 8
1
8
2
22
0
p
p
p
xnx
n
xnx
n
dx
sin
()sin
0 0
p
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Chapter 03.indd   59 5/31/2017   12:42:30 PM
    
          =-
?
22
8
0
p
p
xnx
n
dx
sin
 
          
=
- ?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
22
8
0
0
p
p p
n
xnx
n
nx
n
dx
cos cos
          =
-4
8
2
p
pp
n
n (cos )
          
=
-
=
=
-
+
1
2
12 3
1
2
2
1
2
n
nn
n
n
(con ), ,, ,
()
p …
?=
-
+
=
8
?
fx
n
nx
n
n
()
()
cos
1
2
1
2
1
 
            =- +-
?
?
?
?
?
?
1
2
2
2
3
3
22
cos
coscos
. x
xx
 
Function of Any Period (P = 2L)
If the function f (x) is of period P = 2L has a Fourier series, 
then f (x) can be expressed as,
fx
a
a
n
L
xb
n
L
x
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
pp
 
where the Fourier coefficients are as follows:
a
L
fx dx
L
L
0
1
=
-
?
() 
a
L
fx
n
L
xdx
n
L
L
=
-
?
1
()cos
p
 b
L
fx
n
L
xdx
n
L
L
=
-
?
1
()sin
p
Fourier Series of Even and Odd Functions Let f (x) be an 
even function in (–L, L), then the Fourier series is 
fx
a
a
nx
L
n
n
() cos =+
=
8
?
0
1
2
p
 
Where                   a
L
fx dx
L
fx dx
L
L
L
0
0
12
==
? ?
-
() () 
                              a
L
fx
nx
L
dx
n
L
L
=
-
?
2
()cos
p
Let f (x) be an odd function in (–L, L) then Fourier series is 
fx b
nx
L
n
n
() sin =
=
8
?
p
1
 
where b
L
fx
n
L
dx
n
L
=
?
2
0
()sin.
p
 
Half Range Expansion
In the pervious examples we define the function f (x) with 
the period 2L. 
Suppose f(x) is not periodic function and defined in half 
the interval say (0, L) of lengths L. such expansions are 
known as half range expansions or half range Fourier series. 
In particular a half range expansion containing only cosine 
series of f(x) in the interval (0, L) in a similar way half range 
Fourier sine series contains only sine terms. To find the 
Fourier series of f(x) which is neither periodic nor even nor 
odd we obtain Fourier cosine series and Fourier sine series 
of f(x) as follows. We define a function g(x) such that g (x) = 
f (x) in the interval from (0, L) and g (x) is an even function 
in (–L, L) and is periodic with period 2L and g(x) is obtained 
by previous methods which are discussed earlier. Similarly 
we can obtain a fourier sine series as follows. Assume f (x) = 
h(x) in (0, L) and h (x) is an odd function in the interval (–L, 
L) with period 2L and evaluate h (x) by pervious methods 
which are discussed earlier.
Example 3
If f (x) = 1 – x in 0 < x < 1 find Fourier cosine series and 
Fourier sine series. 
Solution
Given f (x) = 1 – x in 0 < x < 1 since f (x) is neither periodic 
nor even nor odd function. 
Let us assume g(x) = f (x) = 1 – x in 0 < x < 1
= 1 + x in – 1< x < 0
\ g (x) is even function in (–1, 1) 
?= +
=
8
?
gx
a
a
nx
L
n
n
() cos
0
1
2
p
 
       a
L
fx dx fx dx L
L
0
00
1
2
21 == =
??
() () (here)
            =- =-
?
?
?
?
?
?
=× =
?
21 2
2
1
2
21
2
0
1
0
1
() xdxx
x
 
       a
L
fx
nx
L
dx
n
L
=
?
2
0
()cos
p
 
            
=-
=- --
?
?
?
?
?
?
?
?
21
21 1
0
1
0
1
0
1
()cos
()
sin
()
sin
xn xdx
x
nx
n
nxx
n
p
p
p
p
p
? ?
?
?
?
Chapter 03.indd   60 5/31/2017   12:42:32 PM
Page 4


Fourier Series
Periodic Function A function f(x) is said to be periodic 
if f(x + a) = f(x) for all x. The least value of a is called the 
period of f(x).
Example: sin x, cos x are periodic functions with period 
2p .
 1.  f (x) and g (x) are periodic functions with period k 
then af (x) + bg (x) is also a periodic function with 
period k. 
 2.  If f (x) is a periodic function with period k, then the 
period of f (ax) is 
k
a
.
 3.  If the periods of functions f (x), g(x) and h(x) are a, b, 
c, respectively, then the period of f (x) + g (x) + h (x) is 
the lcm of a, b and c.
NOTES
Euler’s Formula for the 
 F ourier  Coe?  cient s
Let f (x) is a periodic function whose period is 2p and is 
integrable over a period. Then f (x) can be represented by 
trigonometric series.
fx
a
anxb nx
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
where a
0 
 a
n
, b
n
 are called Fourier coeffi  cients and these are 
obtained by
af xdx
0
1
=
-
?
p
p
p
() ,
af xnxdxn
n
==
-
?
1
p
p
p
()cosfor 1, 2, 3,… 
bf xnxdxn
n
==
-
?
1
p
p
p
()sinfor 1, 2, 3,… 
SOLVED EXAMPLES
Example 1
Obtain the Fourier series expansion of f (x) = e
x
 in (0, 2p ).
Solution
af xdx
0
0
2
1
=
?
p
p
()
     =
?
1
0
2
p
p
edx
x
=
?
?
?
=-
11
1
0
2
2
pp
p
p
ee
x
() (1)
  
Partial Diff erential 
Equations
Chapter 03.indd   58 5/31/2017   12:42:27 PM
    
af xnxdx
n
=
?
1
0
2
p
p
()cos
     =
?
1
0
2
p
p
enxdx
x
cos
we know that ·
?
ebxdx
ax
cos 
=
+
+
e
ab
abxb bx
ax
22
[cos sin]   
?=
+
+
?
?
?
?
?
?
a
e
n
nx nnx
n
x
1
1
2
0
2
p
p
(cos sin) 
=
+
-
+
?
?
?
?
?
?
1
1
2
1
1
2
22
p
p
p
e
n
n
n
(cos )
  
bf xnxdxe nxdx
n
x
==
? ?
11
0
2
0
2
pp
p p
()sinsin  
 =
-
+
1
1
0
2
2
p
p
enxn nx
n
x
(sin cos]
 
? ebxdx
e
ab
abxb bx
ax
ax
sin( sincos )) =
+
?
?
?
?
?
?
-
? 22
=
+
-
11
1
2
2
2
p
p
p
n
ne nn (cos ) 
=
+
-
n
n
en
p
p
p
()
(cos )
1
12
2
2
 
?= ++
=
8
?
fx
a
anxb nx
n
n
n
() (cos sin)
0
1
2
 =- +
=
8
?
1
2
1
2
1
p
p
() e
n
 
11
1
21
1
12
2
2
2
2
p
p
p
p
pp
+
-+
+
-
?
?
?
?
?
?
n
en
n
n
en (cos )
()
(cos ) 
Even and Odd Functions
Even function: A function f (x) is said to be even if f (–x)  = 
f (x) for all x.
Example: x
2
, cos x 
Odd function: A function f (x) is said to be odd if f (–x) 
= – f (x) for all x 
Example: x
3
, sin x 
 1. The sum of two odd functions is odd.
 2.  The product of an odd function and an even function 
is odd.
 3. Product of two odd functions is even.
NOTES
Fourier Series for Odd Function and Even Function
Case 1: Let f (x) is an even function in (–p , p ). Then the 
Fourier series of the even function contains only cosine 
terms and is known as Fourier cosine series and it is 
  fx
a
anx
n
n
() cos, =+
=
8
?
0
1
2
  where
  af xdxa fx nxdx
n 0
00
22
==
??
pp
pp
() ,( )cos
Case 2: If f (x) is an odd function, then the Fourier series of 
an odd function contains only sine terms, and is known as 
Fourier sine series. 
fx bnx
n
n
() sin, =
=
8
?
1
where                         bf xnxdx
n
=
?
2
0
p
p
()sin 
Example 2
Expand the function fx
x
()=-
p
22
24 8
 in Fourier series in 
the interval (–p , p ).
Solution
   fx
x
()=-
p
22
24 8
fx
xx
fx ()
()
() -= -
-
=- =
pp
22 22
24 8248
 
\ f (x) is an even function.
fx
a
anx
n
n
() cos =+
=
8
?
0
1
2
    af xdx
x
dx
0
0
2
0
2
22
2248
== -
??
pp
p
pp
()
         =
-
?
?
?
?
?
?
?
?
?
=
1
24 24
0
23
0
p
p
p
xx
 
    af xnxdxs
n
=
?
2
0
p
p
()cos
         =-
?
?
?
?
?
?
?
2
24 8
22
0
p
p
p
x
nxdx cos  
=
-
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
-
-- ?
?
?
?
?
?
2
24 8
1
8
2
22
0
p
p
p
xnx
n
xnx
n
dx
sin
()sin
0 0
p
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Chapter 03.indd   59 5/31/2017   12:42:30 PM
    
          =-
?
22
8
0
p
p
xnx
n
dx
sin
 
          
=
- ?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
22
8
0
0
p
p p
n
xnx
n
nx
n
dx
cos cos
          =
-4
8
2
p
pp
n
n (cos )
          
=
-
=
=
-
+
1
2
12 3
1
2
2
1
2
n
nn
n
n
(con ), ,, ,
()
p …
?=
-
+
=
8
?
fx
n
nx
n
n
()
()
cos
1
2
1
2
1
 
            =- +-
?
?
?
?
?
?
1
2
2
2
3
3
22
cos
coscos
. x
xx
 
Function of Any Period (P = 2L)
If the function f (x) is of period P = 2L has a Fourier series, 
then f (x) can be expressed as,
fx
a
a
n
L
xb
n
L
x
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
pp
 
where the Fourier coefficients are as follows:
a
L
fx dx
L
L
0
1
=
-
?
() 
a
L
fx
n
L
xdx
n
L
L
=
-
?
1
()cos
p
 b
L
fx
n
L
xdx
n
L
L
=
-
?
1
()sin
p
Fourier Series of Even and Odd Functions Let f (x) be an 
even function in (–L, L), then the Fourier series is 
fx
a
a
nx
L
n
n
() cos =+
=
8
?
0
1
2
p
 
Where                   a
L
fx dx
L
fx dx
L
L
L
0
0
12
==
? ?
-
() () 
                              a
L
fx
nx
L
dx
n
L
L
=
-
?
2
()cos
p
Let f (x) be an odd function in (–L, L) then Fourier series is 
fx b
nx
L
n
n
() sin =
=
8
?
p
1
 
where b
L
fx
n
L
dx
n
L
=
?
2
0
()sin.
p
 
Half Range Expansion
In the pervious examples we define the function f (x) with 
the period 2L. 
Suppose f(x) is not periodic function and defined in half 
the interval say (0, L) of lengths L. such expansions are 
known as half range expansions or half range Fourier series. 
In particular a half range expansion containing only cosine 
series of f(x) in the interval (0, L) in a similar way half range 
Fourier sine series contains only sine terms. To find the 
Fourier series of f(x) which is neither periodic nor even nor 
odd we obtain Fourier cosine series and Fourier sine series 
of f(x) as follows. We define a function g(x) such that g (x) = 
f (x) in the interval from (0, L) and g (x) is an even function 
in (–L, L) and is periodic with period 2L and g(x) is obtained 
by previous methods which are discussed earlier. Similarly 
we can obtain a fourier sine series as follows. Assume f (x) = 
h(x) in (0, L) and h (x) is an odd function in the interval (–L, 
L) with period 2L and evaluate h (x) by pervious methods 
which are discussed earlier.
Example 3
If f (x) = 1 – x in 0 < x < 1 find Fourier cosine series and 
Fourier sine series. 
Solution
Given f (x) = 1 – x in 0 < x < 1 since f (x) is neither periodic 
nor even nor odd function. 
Let us assume g(x) = f (x) = 1 – x in 0 < x < 1
= 1 + x in – 1< x < 0
\ g (x) is even function in (–1, 1) 
?= +
=
8
?
gx
a
a
nx
L
n
n
() cos
0
1
2
p
 
       a
L
fx dx fx dx L
L
0
00
1
2
21 == =
??
() () (here)
            =- =-
?
?
?
?
?
?
=× =
?
21 2
2
1
2
21
2
0
1
0
1
() xdxx
x
 
       a
L
fx
nx
L
dx
n
L
=
?
2
0
()cos
p
 
            
=-
=- --
?
?
?
?
?
?
?
?
21
21 1
0
1
0
1
0
1
()cos
()
sin
()
sin
xn xdx
x
nx
n
nxx
n
p
p
p
p
p
? ?
?
?
?
Chapter 03.indd   60 5/31/2017   12:42:32 PM
    
            
=- --
?
?
?
?
?
?
?
?
?
21 1
0
1
0
1
()
sin
()
sin
x
nx
n
nx
n
dx
p
p
p
p
=-
?
?
?
=-
?
?
?
?
?
?
22
1
22
0
1
22 22
coscos nx
nn
n
n
p
pp
p
p
 
=- -
2
11
22
n
n
p
(( )).
\ Fourier cosine series is
gx
n
nx
n
()
()
cos =+
--
=
8
?
1
2
21 1
2
2
2
1
p
p  
Fourier sine series in (0, 1)
\ h(x) = f (x) = 1 – x; 0 < x < 1 
            = – (1 + x); –1 < x < 0 
h(x) is an odd function 
?=
=
8
?
hx b
nx
L
n
n
() sin
p
1
       b
L
fx
nx
L
dx
n
L
=
?
2
0
()sin
p
           =-
?
21
0
1
()sin xn xdx p
           =-
-
?
?
?
-=
?
21 22
0
1
0
1
()
coscos
x
nx
n
nx
n
dx n
p
p
p
p
p /
?=
=
8
?
hx
n
nx
n
() /sin() 2
1
1
pp 
Partial Differential Equations (PDE)
An equation involving two or more independent variables 
and a dependent variable and its partial derivatives is called 
a partial differential equation.
?
?
?
?
?
?
?
?
?
?
?
= fx yz
z
z
z
y
,, ,, . … 0
Standard Notation
       
?
?
==
?
?
==
z
x
pz
z
y
qz
xy
,
    
?
?
==
?
?
==
2
2
2
2
z
x
rz
z
y
tz
xx yy
,
?
??
==
2
z
xy
zs
xy
Formation of Partial Differential Equations
Partial differential equation can be formed by two ways.
 1. By eliminating arbitrary constants.
 2. By eliminating arbitrary functions.
Formation of PDE by Eliminating Arbitrary Constants
Consider a function f (x, y, z, a, b) = 0
where a, b, are arbitrary constants.
Differentiating this partially wrt, x and y eliminate a, b 
from these equations we get an equation f(x, y, z, p, q) = 0, 
which is partial differential equation of first order. 
Example 4
z = ax 
2
 – by
2
, a, b are arbitrary constants.
Solution
Given
          z = ax 
2
 – by
2
 (1)  
Differentiating z partially wrt x,
?
?
=? =? =
z
x
ax paxa
p
x
22
2
Differentiate z partially wrt y,
     
?
?
=
z
y
by 2 , i.e., q = –2by
?=
-
b
q
y 2
Substituting the values of a and b in Eq. (1), we get 
z
p
x
x
q
y
y =+
22
22
, 
2z = px + qy which is a partial differential equation of order 1. 
Formation of PDE by Eliminating Arbitrary Function  
Consider z = f (u) (1) 
f is an arbitrary function in u and u is function in x, y, z.
Now differentiate Eq. (1) wrt x , y partially by chain rule 
we get
         
?
?
=
?
?
·
?
?
+
?
?
·
?
?
·
?
?
z
x
f
u
u
x
f
u
u
z
z
x
 (2)
         
?
?
=
?
?
·
?
?
+
?
?
·
?
?
·
?
?
z
y
f
u
u
y
f
u
u
z
z
y
 (3)
by eliminating the arbitrary functions from Eqs. (1), (2), (3) 
we get a PDE of first order.
F ormation of PDE when T wo Arbitrary Functions are Involved
When two arbitrary functions are involved, we differentiate the 
given equation two times and eliminate the two arbitrary func-
tions from the equation obtained. 
Example 5
Form the partial differential equation of
z
fx
gy
=
()
()
Chapter 03.indd   61 5/31/2017   12:42:34 PM
Page 5


Fourier Series
Periodic Function A function f(x) is said to be periodic 
if f(x + a) = f(x) for all x. The least value of a is called the 
period of f(x).
Example: sin x, cos x are periodic functions with period 
2p .
 1.  f (x) and g (x) are periodic functions with period k 
then af (x) + bg (x) is also a periodic function with 
period k. 
 2.  If f (x) is a periodic function with period k, then the 
period of f (ax) is 
k
a
.
 3.  If the periods of functions f (x), g(x) and h(x) are a, b, 
c, respectively, then the period of f (x) + g (x) + h (x) is 
the lcm of a, b and c.
NOTES
Euler’s Formula for the 
 F ourier  Coe?  cient s
Let f (x) is a periodic function whose period is 2p and is 
integrable over a period. Then f (x) can be represented by 
trigonometric series.
fx
a
anxb nx
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
where a
0 
 a
n
, b
n
 are called Fourier coeffi  cients and these are 
obtained by
af xdx
0
1
=
-
?
p
p
p
() ,
af xnxdxn
n
==
-
?
1
p
p
p
()cosfor 1, 2, 3,… 
bf xnxdxn
n
==
-
?
1
p
p
p
()sinfor 1, 2, 3,… 
SOLVED EXAMPLES
Example 1
Obtain the Fourier series expansion of f (x) = e
x
 in (0, 2p ).
Solution
af xdx
0
0
2
1
=
?
p
p
()
     =
?
1
0
2
p
p
edx
x
=
?
?
?
=-
11
1
0
2
2
pp
p
p
ee
x
() (1)
  
Partial Diff erential 
Equations
Chapter 03.indd   58 5/31/2017   12:42:27 PM
    
af xnxdx
n
=
?
1
0
2
p
p
()cos
     =
?
1
0
2
p
p
enxdx
x
cos
we know that ·
?
ebxdx
ax
cos 
=
+
+
e
ab
abxb bx
ax
22
[cos sin]   
?=
+
+
?
?
?
?
?
?
a
e
n
nx nnx
n
x
1
1
2
0
2
p
p
(cos sin) 
=
+
-
+
?
?
?
?
?
?
1
1
2
1
1
2
22
p
p
p
e
n
n
n
(cos )
  
bf xnxdxe nxdx
n
x
==
? ?
11
0
2
0
2
pp
p p
()sinsin  
 =
-
+
1
1
0
2
2
p
p
enxn nx
n
x
(sin cos]
 
? ebxdx
e
ab
abxb bx
ax
ax
sin( sincos )) =
+
?
?
?
?
?
?
-
? 22
=
+
-
11
1
2
2
2
p
p
p
n
ne nn (cos ) 
=
+
-
n
n
en
p
p
p
()
(cos )
1
12
2
2
 
?= ++
=
8
?
fx
a
anxb nx
n
n
n
() (cos sin)
0
1
2
 =- +
=
8
?
1
2
1
2
1
p
p
() e
n
 
11
1
21
1
12
2
2
2
2
p
p
p
p
pp
+
-+
+
-
?
?
?
?
?
?
n
en
n
n
en (cos )
()
(cos ) 
Even and Odd Functions
Even function: A function f (x) is said to be even if f (–x)  = 
f (x) for all x.
Example: x
2
, cos x 
Odd function: A function f (x) is said to be odd if f (–x) 
= – f (x) for all x 
Example: x
3
, sin x 
 1. The sum of two odd functions is odd.
 2.  The product of an odd function and an even function 
is odd.
 3. Product of two odd functions is even.
NOTES
Fourier Series for Odd Function and Even Function
Case 1: Let f (x) is an even function in (–p , p ). Then the 
Fourier series of the even function contains only cosine 
terms and is known as Fourier cosine series and it is 
  fx
a
anx
n
n
() cos, =+
=
8
?
0
1
2
  where
  af xdxa fx nxdx
n 0
00
22
==
??
pp
pp
() ,( )cos
Case 2: If f (x) is an odd function, then the Fourier series of 
an odd function contains only sine terms, and is known as 
Fourier sine series. 
fx bnx
n
n
() sin, =
=
8
?
1
where                         bf xnxdx
n
=
?
2
0
p
p
()sin 
Example 2
Expand the function fx
x
()=-
p
22
24 8
 in Fourier series in 
the interval (–p , p ).
Solution
   fx
x
()=-
p
22
24 8
fx
xx
fx ()
()
() -= -
-
=- =
pp
22 22
24 8248
 
\ f (x) is an even function.
fx
a
anx
n
n
() cos =+
=
8
?
0
1
2
    af xdx
x
dx
0
0
2
0
2
22
2248
== -
??
pp
p
pp
()
         =
-
?
?
?
?
?
?
?
?
?
=
1
24 24
0
23
0
p
p
p
xx
 
    af xnxdxs
n
=
?
2
0
p
p
()cos
         =-
?
?
?
?
?
?
?
2
24 8
22
0
p
p
p
x
nxdx cos  
=
-
?
?
?
?
?
?
-
?
?
?
?
?
?
?
?
?
-
-- ?
?
?
?
?
?
2
24 8
1
8
2
22
0
p
p
p
xnx
n
xnx
n
dx
sin
()sin
0 0
p
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Chapter 03.indd   59 5/31/2017   12:42:30 PM
    
          =-
?
22
8
0
p
p
xnx
n
dx
sin
 
          
=
- ?
?
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
22
8
0
0
p
p p
n
xnx
n
nx
n
dx
cos cos
          =
-4
8
2
p
pp
n
n (cos )
          
=
-
=
=
-
+
1
2
12 3
1
2
2
1
2
n
nn
n
n
(con ), ,, ,
()
p …
?=
-
+
=
8
?
fx
n
nx
n
n
()
()
cos
1
2
1
2
1
 
            =- +-
?
?
?
?
?
?
1
2
2
2
3
3
22
cos
coscos
. x
xx
 
Function of Any Period (P = 2L)
If the function f (x) is of period P = 2L has a Fourier series, 
then f (x) can be expressed as,
fx
a
a
n
L
xb
n
L
x
nn
n
() (cos sin) =+ +
=
8
?
0
1
2
pp
 
where the Fourier coefficients are as follows:
a
L
fx dx
L
L
0
1
=
-
?
() 
a
L
fx
n
L
xdx
n
L
L
=
-
?
1
()cos
p
 b
L
fx
n
L
xdx
n
L
L
=
-
?
1
()sin
p
Fourier Series of Even and Odd Functions Let f (x) be an 
even function in (–L, L), then the Fourier series is 
fx
a
a
nx
L
n
n
() cos =+
=
8
?
0
1
2
p
 
Where                   a
L
fx dx
L
fx dx
L
L
L
0
0
12
==
? ?
-
() () 
                              a
L
fx
nx
L
dx
n
L
L
=
-
?
2
()cos
p
Let f (x) be an odd function in (–L, L) then Fourier series is 
fx b
nx
L
n
n
() sin =
=
8
?
p
1
 
where b
L
fx
n
L
dx
n
L
=
?
2
0
()sin.
p
 
Half Range Expansion
In the pervious examples we define the function f (x) with 
the period 2L. 
Suppose f(x) is not periodic function and defined in half 
the interval say (0, L) of lengths L. such expansions are 
known as half range expansions or half range Fourier series. 
In particular a half range expansion containing only cosine 
series of f(x) in the interval (0, L) in a similar way half range 
Fourier sine series contains only sine terms. To find the 
Fourier series of f(x) which is neither periodic nor even nor 
odd we obtain Fourier cosine series and Fourier sine series 
of f(x) as follows. We define a function g(x) such that g (x) = 
f (x) in the interval from (0, L) and g (x) is an even function 
in (–L, L) and is periodic with period 2L and g(x) is obtained 
by previous methods which are discussed earlier. Similarly 
we can obtain a fourier sine series as follows. Assume f (x) = 
h(x) in (0, L) and h (x) is an odd function in the interval (–L, 
L) with period 2L and evaluate h (x) by pervious methods 
which are discussed earlier.
Example 3
If f (x) = 1 – x in 0 < x < 1 find Fourier cosine series and 
Fourier sine series. 
Solution
Given f (x) = 1 – x in 0 < x < 1 since f (x) is neither periodic 
nor even nor odd function. 
Let us assume g(x) = f (x) = 1 – x in 0 < x < 1
= 1 + x in – 1< x < 0
\ g (x) is even function in (–1, 1) 
?= +
=
8
?
gx
a
a
nx
L
n
n
() cos
0
1
2
p
 
       a
L
fx dx fx dx L
L
0
00
1
2
21 == =
??
() () (here)
            =- =-
?
?
?
?
?
?
=× =
?
21 2
2
1
2
21
2
0
1
0
1
() xdxx
x
 
       a
L
fx
nx
L
dx
n
L
=
?
2
0
()cos
p
 
            
=-
=- --
?
?
?
?
?
?
?
?
21
21 1
0
1
0
1
0
1
()cos
()
sin
()
sin
xn xdx
x
nx
n
nxx
n
p
p
p
p
p
? ?
?
?
?
Chapter 03.indd   60 5/31/2017   12:42:32 PM
    
            
=- --
?
?
?
?
?
?
?
?
?
21 1
0
1
0
1
()
sin
()
sin
x
nx
n
nx
n
dx
p
p
p
p
=-
?
?
?
=-
?
?
?
?
?
?
22
1
22
0
1
22 22
coscos nx
nn
n
n
p
pp
p
p
 
=- -
2
11
22
n
n
p
(( )).
\ Fourier cosine series is
gx
n
nx
n
()
()
cos =+
--
=
8
?
1
2
21 1
2
2
2
1
p
p  
Fourier sine series in (0, 1)
\ h(x) = f (x) = 1 – x; 0 < x < 1 
            = – (1 + x); –1 < x < 0 
h(x) is an odd function 
?=
=
8
?
hx b
nx
L
n
n
() sin
p
1
       b
L
fx
nx
L
dx
n
L
=
?
2
0
()sin
p
           =-
?
21
0
1
()sin xn xdx p
           =-
-
?
?
?
-=
?
21 22
0
1
0
1
()
coscos
x
nx
n
nx
n
dx n
p
p
p
p
p /
?=
=
8
?
hx
n
nx
n
() /sin() 2
1
1
pp 
Partial Differential Equations (PDE)
An equation involving two or more independent variables 
and a dependent variable and its partial derivatives is called 
a partial differential equation.
?
?
?
?
?
?
?
?
?
?
?
= fx yz
z
z
z
y
,, ,, . … 0
Standard Notation
       
?
?
==
?
?
==
z
x
pz
z
y
qz
xy
,
    
?
?
==
?
?
==
2
2
2
2
z
x
rz
z
y
tz
xx yy
,
?
??
==
2
z
xy
zs
xy
Formation of Partial Differential Equations
Partial differential equation can be formed by two ways.
 1. By eliminating arbitrary constants.
 2. By eliminating arbitrary functions.
Formation of PDE by Eliminating Arbitrary Constants
Consider a function f (x, y, z, a, b) = 0
where a, b, are arbitrary constants.
Differentiating this partially wrt, x and y eliminate a, b 
from these equations we get an equation f(x, y, z, p, q) = 0, 
which is partial differential equation of first order. 
Example 4
z = ax 
2
 – by
2
, a, b are arbitrary constants.
Solution
Given
          z = ax 
2
 – by
2
 (1)  
Differentiating z partially wrt x,
?
?
=? =? =
z
x
ax paxa
p
x
22
2
Differentiate z partially wrt y,
     
?
?
=
z
y
by 2 , i.e., q = –2by
?=
-
b
q
y 2
Substituting the values of a and b in Eq. (1), we get 
z
p
x
x
q
y
y =+
22
22
, 
2z = px + qy which is a partial differential equation of order 1. 
Formation of PDE by Eliminating Arbitrary Function  
Consider z = f (u) (1) 
f is an arbitrary function in u and u is function in x, y, z.
Now differentiate Eq. (1) wrt x , y partially by chain rule 
we get
         
?
?
=
?
?
·
?
?
+
?
?
·
?
?
·
?
?
z
x
f
u
u
x
f
u
u
z
z
x
 (2)
         
?
?
=
?
?
·
?
?
+
?
?
·
?
?
·
?
?
z
y
f
u
u
y
f
u
u
z
z
y
 (3)
by eliminating the arbitrary functions from Eqs. (1), (2), (3) 
we get a PDE of first order.
F ormation of PDE when T wo Arbitrary Functions are Involved
When two arbitrary functions are involved, we differentiate the 
given equation two times and eliminate the two arbitrary func-
tions from the equation obtained. 
Example 5
Form the partial differential equation of
z
fx
gy
=
()
()
Chapter 03.indd   61 5/31/2017   12:42:34 PM
    
Solution
Given         z
fx
gy
=
()
()
            pz
fx
gy
x
==
'()
()
 (1)
    qz
fx
gy
gy
y
==
-
· '
()
[()]
()
2
 (2)
             s
z
xy
fx
gy
gy =
?
??
=
- '
· '
2
2
()
[()]
() (3)
 
Eq.Eq. () ()
()
()
() ()
[()]
12
2
×= =
'
·
-· ' ?
?
?
?
?
?
=- · pq
fx
gy
fx gy
gy
sz
            pq + sz = 0 
F orming PDE by the Elimination of Arbitrary Function of 
Specific Functions
Consider f (u, v) = 0 
Where u, v are the functions in x, y, z.
Differentiate the above equation wrt x and y by chain rule 
and eliminate the
?
?
?
?
F
u
F
v
, and convert them in the form Pp 
+ Qq = R, which is a first order linear PDE where P, Q, R 
are functions of x, y, z.
Linear Equation of First Order
Linear equation of first order is Pp + Qq = R. This is also 
called Lagrange’ s equation, where P, Q, R are the functions 
in x, y, and z.
Procedure For solving Lagrange’s Equations
Take the auxiliary equation as 
dx
P
dy
Q
dz
R
= = .
Solve any two equations and take the solutions as u and v. 
The complete solution is f (u, v) = 0 or u = f (v).
Example 6
Solve (z – y)p + (x – z)q = y – x. 
Solution
Auxiliary equation is
dx
zy
dy
xz
dz
yx -
=
-
=
-
.
Using the multipliers as x, y, z we get 
xdxydy zdz
xz yy xz zy x
++
-+ -+ - () () ()
= xdx + ydy + zdz = 0
\ x
2
 + y
2
 + z
2
 = 0 
and also
dx dy dz
zy xz yz
++
-+ -+ -
=0
dx + dy + dz = 0, x + y + z = 0. 
\ The required solution is x
2
 + y
2
 + z
2
 = f (x + y + z).
Non-linear Equations of First Order
There are four types of non linear equations of first order. 
Type 1:
f (p, q) = 0.
If the given equation contains only p and q then the solution 
is taken as z = ax + by + c. Where a, b and c are arbitrary, 
such that f(a, b) = 0.
Example 7
Solve 2p + 3q = 5 
Solution
Given 2p + 3q = 5
z = ax + by + c. 
Where      2a + 3b = 5,
b
a
=
- 52
3
\ The solution is zax
a
yc =+
- ?
?
?
?
?
?
+
52
3
.
Type 2:
f (z, p, q) = 0
When the equation is not containing x and y then to solve the 
equation assume u = x + ay and substitute p
dz
du
qa
dz
du
= = ,. 
Solve the resulting equation and replace u by x + ay.
Type 3:
f (x, p) = g (y, q). 
The equation is not containing z.
Assume f (x, p) = a and g (y, q) = a.
Solve the equations for p and q  and then write the solution. 
Example 8
Solve p
2
 – q
2
 = x
2
 – y
2
. 
Solution
                    p
2
 – q
2
 = x
2
 – y
2
 
 p
2
 – x
2
 = –y
2
 + q
2
 
Let                         p
2
 – x
2
 = a
2
 = –y
2
 + q
2
 
                   p
2
 = a
2
 + x
2
    q
2
 = y
2
 + a
2
 
Chapter 03.indd   62 5/31/2017   12:42:36 PM