Engineering Mathematics Material

2013

SUBJECT NAME

: Transforms and Partial Differential Equations

SUBJECT CODE

: MA2211

MATERIAL NAME

: FormulaMaterial

MATERIAL CODE

: JM08AM3005

Unit I (Fourier Series)

1)

Dirichlets Conditions:
Any function f ( x ) can be expanded as a Fourier

a0
series an cos nx bn sin nx where a0 , an , bn are constants provided the
2 n 1
n 1
following conditions are true.
f ( x ) is periodic, single valued and finite.
f ( x ) has a finite number of discontinuities in any one period.
f ( x ) has at the most a finite number of maxima and minima.

2)

The Fourier Series in the interval (0,2):

a
f ( x ) 0 an cos nx bn sin nx
2 n 1
n 1
Where a0

3)

f ( x )dx , an

f ( x )cos nxdx , bn

f ( x )sin nxdx

The Fourier Series in the interval (-,):

a
f ( x ) 0 an cos nx bn sin nx
2 n 1
n 1
2

0
0
0
In this interval, we have to verify the function is either odd function or
even function. If it is even function then find only a0 and an ( bn 0 ). If it is odd
function then find only bn ( a0 an 0 ).
Where a0

f ( x )dx , an

f ( x )cos nxdx , bn

f ( x )sin nxdx

Engineering Mathematics Material

2013

If the function is neither odd nor even then you should find

1
a0 , an and bn by using the following formulas a0 f ( x )dx ,

an

4)

f ( x )cos nxdx , bn

f ( x )sin nxdx .

The half range Fourier Series in the interval (0,):

The half range Cosine Series in the interval (0,):

a
f ( x ) 0 an cos nx
2 n 1
2

0
0
The half range Sine Series in the interval (0,):
Where a0

f ( x )dx , an

f ( x )cos nxdx

f ( x ) bn sin nx
n 1

7)

8)

f ( x )sin nxdx

a02 2
[an bn2 ]
0
4 n 1
The Parsevals Identity in the interval (-,):

a02 2
2
2
[
f
(
x
)]
dx

[an bn2 ]

0
4 n 1
The Parsevals Identity for half range cosine series in the interval (0,):

a02 2
2
2
[
f
(
x
)]
dx

an
0
4 n 1
The Parsevals Identity for half range sine series in the interval (0,):

2
2
[
f
(
x
)]
dx

bn2

6)

0
The Parsevals Identity in the interval (0,2):
Where bn

5)

2
[ f ( x )] dx

n 1

Change of interval:
9)

The Fourier Series in the interval (0,2):

a
n x
n x
f ( x ) 0 an cos
bn sin
2 n 1
n 1
Where a0

f ( x )dx , an

f ( x )cos

n x

dx , bn

f ( x )sin

n x

dx

Engineering Mathematics Material

10)

2013

The Fourier Series in the interval (-, ):

a
n x
n x
f ( x ) 0 an cos
bn sin
2 n 1
n 1
Where a0

f ( x )dx , a

f ( x )cos

n x

dx , bn

f ( x )sin

n x

dx

In this interval, you have to verify the function is either odd function or even
function. If it is even function then find only a0 and an ( bn 0 ). If it is odd
function then find only bn ( a0 an 0 ).

11)

The half range Fourier Series in the interval (0, ):

The half range Cosine Series in the interval (0, ):

a
n x
f ( x ) 0 an cos
2 n 1
Where a0

f ( x )dx , a

f ( x )cos

n x

dx

The half range Sine Series in the interval (0, ):

n x
f ( x ) bn sin
n 1

Where bn

f ( x )sin

n x

dx

12)

13)

14)

15)

The Parsevals Identity in the interval (0,2):

2
a02 2
1
2
[
f
(
x
)]
dx

[an bn2 ]
0
4 n 1
The Parsevals Identity in the interval (-,):
a02 2
2
2
[
f
(
x
)]
dx

[an bn2 ]
0
4 n 1
The Parsevals Identity for half range cosine series in the interval (0,):
a02 2
2
2
an
0 [ f ( x )] dx 4
n 1
The Parsevals Identity for half range sine series in the interval (0,):

2
2
2
[ f ( x )] dx bn
0

n 1

Engineering Mathematics Material

16)

2013

Harmonic Analysis:
The method of calculation of Fourier constants by means of numerical
calculation is called as Harmonic analysis.
f ( x)

a0
an cos nx bn sin nx
2 n 1
n 1

where
2
2
2
2
a0 y , a1 y cos x , a2 y cos 2 x , a3 y cos 3 x , ...
n
n
n
n

b1

2
2
2
y sin x , b 2 y sin 2 x , b 3 y sin 3 x , ...

n
n
n

2 x
.
T
Where T is period, n is the number of values given. If the first and last y values
are same we can omit one of them.
When the values of x is given as numbers the is calculated by

Complex form of Fourier Series:

17)

18)

19)

20)

The Complex form of Fourier Series in the interval (0,2):

2

1
inx
where cn
f ( x ) cn e
f ( x )e inx dx

n
0
The Complex form of Fourier Series in the interval (-,):

1
inx
where cn
f ( x ) cn e
f ( x )e inx dx

2
n
The Complex form of Fourier Series in the interval (0,2):
2
in x
in x

1
f ( x ) cn e
where cn
f ( x )e
dx
2 0
n
The Complex form of Fourier Series in the interval (-,):
in x
in x

1
f ( x ) cn e
where cn
f
(
x
)
e
dx
2
n

Unit II (Fourier Transforms)

1)

Fourier Integral theorem

The Fourier integral theorem of f ( x ) in the interval ,
f ( x)

f ( x )cos ( t )dxd

is

Engineering Mathematics Material

2)

2013

Convolution Theorem
If F [ s ] and G[ s ] are the Fourier transform of the functions f ( x ) and
g ( x ) respectively, then F [ f ( x )* g( x )] F s .G s

3)

4)
5)
6)
7)
8)
9)

The Fourier Transform of a function f ( x ) is given by F [ f ( x )] is denoted by

F [ s] .
Fourier Transform F [ s ] F [ f ( x )]
Inverse Fourier Transform f ( x )

11)

f ( x )e isx dx

F [ s]e

isx

0
If f ( x ) e ax then the Fourier Cosine and Sine transforms as follows
Fourier Cosine Transform Fc [ s ] Fc [ f ( x )]

f ( x )cos sx dx

a
a s2
2
s
b) Fs [ f ( x )]
2
a s2
Property
d
a) Fs [ xf ( x )] Fc [ f ( x )]
ds
d
b) Fc [ xf ( x )] Fs [ f ( x )]
ds
Parsevals Identity

a)

b)
12)

ds
2
The Fourier transforms and Inverse Fourier transforms are called Fourier
transforms pairs.

2
Fourier Sine Transform Fs [ s ] Fs [ f ( x )]
f ( x )sin sx dx

a) Fc [ f ( x )]

10)

F ( s ) ds

f ( x ) dx

Fc ( s )Gc ( s )ds f ( x ) g( x )dx (Or)

Condition for Self reciprocal F [ f ( x )] f ( s )

Fc ( s ) ds f ( x ) dx
2

Engineering Mathematics Material

2013

Unit III (Partial Differential Equation)

1) Lagranges Linear equation
The equation of the form Pp Qq R
dx dy dz
then the subsidiary equation is

P
Q
R
2) Homogeneous Linear Partial Differential Equation of higher order with constant
coefficients:

2z
2z
2z
The equation of the form a 2 b
c 2 f ( x, y)
x
x y
y
The above equation can be written as
aD2 bDD cD2 z f ( x, y ) .. (1)

where D 2 , D
and D 2 , D
y
y
x
x
The solution of above equation is z = C.F + P.I
Complementary Function (C.F) :
2

Sl.No.
1

To find C.F consider the auxiliary equation by replacing D by m and D by

1.The equation (1) implies that am 2 bm c 0 , solving this equation
we get two values of m. The following table gives C.F of the above
equation.
Nature of m
Complementary Function
C.F = f1 ( y m1 x ) f 2 ( y m2 x )
m1 m2

m1 m2

C.F = f1 ( y mx ) xf 2 ( y mx )

m1 m2 m3

C.F = f1 ( y m1 x ) f 2 ( y m2 x ) f 3 ( y m3 x )

m1 m2 m3

C.F = f1 ( y mx ) xf 2 ( y mx ) x 2 f 3 ( y mx )

m1 m2 , m3 is different

C.F = f1 ( y mx ) xf 2 ( y mx ) f 3 ( y m3 x )

Particular Integral (P.I) :

To find P.I consider ( D, D) aD2 bDD cD 2 .
Type: 1 If f ( x , y ) 0 , then P.I 0 .
Type: 2

If f ( x, y ) e ax by
P .I

1
e ax by
( D , D )

Engineering Mathematics Material

2013

Replace D by a and D by b. If ( D , D ) 0 , then it is P.I.

If ( D , D ) 0 , then diff. denominator w.r.t D and multiply x in
numerator. Again replace D by a and D by b. If again
denominator equal to zero then continue the same procedure.
Type: 3

If f ( x , y ) sin( ax by ) (or ) cos( ax by )

1
sin(ax by ) (or ) cos(ax by )
( D , D )
Here replace D 2 by a 2 , D 2 by b 2 and DD by ab . Do not
replace for D and D . If the denominator equal to zero, then
apply the same producer as in Type: 2.
P .I

Type: 4

If f ( x, y ) x m y n
1
xm yn
( D , D )
1

xm yn
1 g ( D , D )
P .I

1 g ( D , D ) x m y n
1

Here we can use Binomial formula as follows:

1
i) 1 x 1 x x 2 x 3 ...
ii) 1 x 1 x x 2 x 3 ...
1

iii) 1 x 1 2 x 3 x 2 4 x 3 ...
2

iv) 1 x 1 2 x 3 x 2 4 x 3 ...
2

v) (1 x )3 1 3 x 6 x 2 10 x 3 ...
vi) (1 x )3 1 3 x 6 x 2 10 x 3 ...
Type: 5 If f ( x, y ) e ax by V , where

V=sin(ax by ) (or) cos(ax by ) (or) x m y n

1
e ax by V
( D , D )
First operate e ax by by replacing D by D a and D by D a .
1
P . I e ax by
V , Now this will either Type: 3 or
( D a , D b )
Type: 4.
P .I

Engineering Mathematics Material

2013

Type: 6 If f ( x , y ) y sin ax (or) y cos ax

1
y sin ax
( D , D )
1

y sin ax
D m1 D D m2 D

P .I

y c m2 x

1
c m2 x sin ax dx (Apply Bernouilis method)
D m1 D

3) Solution of Partial Differential Equations:

Standard Type: 1

Equation of the form f ( p, q ) 0

Standard Type: 2

Assume that z ax by c be the solution the above

equation.put p a and q b in equation (1), we get
f (a , b ) 0 . Now, solve this, we get b (a ) .
z ax (a ) y c which is called Complete solution.
Equation of the form z px qy f ( p, q ) (Clairauts form)
The Complete solution is z ax by f (a , b) . To find
Singular integral diff. partially w.r.t a & b , equate to zero
and eliminate a and b .

Standard Type: 3

Equation of the form f1 ( x, p) f 2 ( y, q)

The solution is z pdx qdy .

Standard Type: 4

Equation of the form f ( z , p, q ) 0

In this type put u x ay , then p

dz
dz
,q a
du
du

Unit IV (Application of Partial Differential Equation)

1) The One dimensional Wave equation:
2
2 y
2 y

a
t 2
x 2

The three solutions of the above equation are

i)

y( x , t ) Ae px Be px Ce pat De pat

ii)

y( x , t ) A cos px B sin px C cos pat D sin pat

iii)

y( x , t ) Ax B Ct D

Engineering Mathematics Material

2013

But the correct solution is ii),

y( x , t ) A cos px B sin px C cos pat D sin pat .

2) The One dimensional Heat flow equation:

u
2u
2 2
t
x
k

c
2

k Thermal Conductivity
where Density

c Specific Heat

The three solutions of the above equation are

i)

u( x , t ) Ae px Be px Ce

ii)

u( x , t ) A cos px B sin px Ce

iii)

u( x , t ) Ax B C

2 2

p t
2

p2 t

But the correct solution is ii), u( x , t ) A cos px B sin px Ce

p2 t

3) The Two dimensional Heat flow equation:

2u 2u

0
x 2 y 2
The three solutions of the above equation are
i)

u( x , y ) Ae px Be px C cos py D sin py
(Applicable when given value is parallel to y-axies)

ii)

u( x , y ) A cos px B sin px Ce py De py

(Applicable when given value is parallel to x-axies)

iii)

u( x, y ) Ax B Cy D (Not applicable)

Unit V (Z - Transform)
1) Definition of Z-transform:
Let f ( n) be the sequence defined for all the positive integers n such that

Z f ( n ) f ( n ) z n
n 0

Engineering Mathematics Material

2013

2)

Sl.No

Z f ( n )

1.

Z 1

2.

Z ( 1)n

3.

Z a n

4.

Z n

5.

Z n 1

6.
7.
8.

F [z]

z
z 1
z
z 1
z
za
z

z 1
z2

z 1

z
log

z 1
z
2
z 1

Z cos
2

z2
z2 1

If Z f ( n) F [ z ] , then Lt F [ z ] Lt f ( n)
z

n 0

4) Statement of Final value theorem:

If Z f ( n) F [ z ] , then Lt f ( n) Lt ( z 1)F ( z )
n

z 1

Z a n f ( n) Z f ( n) z z

6)

Z nf (n) z

d
Z f ( n)
dz

7) Inverse Z-transform

Sl.No
1.
2.
3.

1
Z
n
n
Z sin
2

3) Statement of Initial value theorem:

5)

Z 1 F ( z )

z
Z 1

z 1
z
Z 1

z 1
z
Z 1

za

f (n)

( 1)n
an

Engineering Mathematics Material

2013

z
Z 1

za

Z 1
2
z 1

Z 1
2
z a

Z 1
2
z a

2
z
Z 1 2

z 1

4.
5.

6.

7.

8.
9.

z2
Z 1 2
2
z a

10.

z
Z 1 2

z 1

11.

z
Z 1 2
2
z a

na n 1

n a

cos

n
2

a n cos

sin

n 1

n
2

n
2
a n1 sin

n
2

8) Inverse form of Convolution Theorem

Z 1[F ( z ).G( z )] Z 1[F ( z )] Z 1[G( z )]

n

and by the defn. of Convolution of two functions f ( n) g ( n) f ( r ) g ( n r )

r 0

9) a) Z [ y ( n )] F ( z )
b) Z [ y ( n 1)] zF ( z ) zy (0)
c) Z[ y(n 2)] z 2 F ( z ) z 2 y(0) zy(1)
d) Z[ y(n 3)] z 3 F ( z ) z 3 y(0) z 2 y(1) zy(2)

----All the Best----