Engineering Mathematics 2014

SUBJECT NAME : Transforms and Partial Diff. Eqns.

SUBJECT CODE : MA6351
MATERIAL NAME : Formula Material
MATERIAL CODE : JM08AM3005
REGULATION : R2013
UPDATED ON : April-May 2014

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Name of the Student: Branch:

Unit – I (Partial Differential Equations)

1) Lagrange’s 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  b  c  f ( x, y)
x 2 xy y 2
The above equation can be written as
 aD2  bDD  cD2  z  f ( x, y) …………………………….. (1)
2  2 
where D 2  , D  and D  2
 , D 
x 2
x y 2
y
The solution of above equation is z = C.F + P.I

 Complementary Function (C.F) :

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.

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Sl.No. Nature of m Complementary Function

1 m1  m2 C.F = f1 ( y  m1 x )  f 2 ( y  m2 x )
2 m1  m2 C.F = f1 ( y  mx )  xf 2 ( y  mx )
3 m1  m2  m3 C.F = f1 ( y  m1 x )  f 2 ( y  m2 x )  f 3 ( y  m3 x )
4 m1  m2  m3 C.F = f1 ( y  mx )  xf 2 ( y  mx )  x 2 f 3 ( y  mx )
5 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

1
P .I  e ax  by
 ( D, D)

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
P .I  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.

Type: 4 If f ( x, y )  x m y n

1
P .I  xm yn

 ( D, D )
1
 xm yn
1  g( D, D)
 1  g( D, D)  x m y n
1

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Here we can use Binomial formula as follows:

i)  1  x   1  x  x 2  x 3  ...
1

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
P .I  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.

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

1
P .I  y sin ax
 ( D, D)
1
 y sin ax y  c  m2 x
 D  m1 D  D  m2 D 
1
 c  m2 x  sin ax dx (Apply Bernouili’s method)
 D  m1 D  


3) Non Homogeneous Linear Partial Differential Equation of higher order with

constant coefficients:

The equation of the form  D  m1 D  1  D  m2 D   2  z  f ( x, y )

The solution of above equation is z  C .F  P . I

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 Complementary Function (C.F) :

If m1  m2 then

C .F  e1 x f1  y  m1 x   e2 x f 2  y  m2 x 

If m1  m2 then

C .F  e x f1  y  mx   xe x f 2  y  mx 

 Particular Integral (P.I):

P.I is same as Homogeneous Linear Partial Diff. Equn.

4) Solution of Partial Differential Equations:

Standard Type: 1 Equation of the form f ( p, q)  0

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.
Standard Type: 2 Equation of the form z  px  qy  f ( p, q) (Clairaut’s 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

dz dz
In this type put u  x  ay , then p  ,q  a
du du

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Engineering Mathematics 2014

Unit – II (Fourier Series)

1) Dirichlet’s Conditions:

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

 
a
series 0   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 (0,2π):

 
a
f ( x )  0   an cos nx   bn sin nx
2 n 1 n 1
2 2 2
1 1 1
Where a0 
 
0
f ( x )dx , an 
 
0
f ( x )cos nxdx , bn 
 
0
f ( x )sin nxdx

3) The Fourier Series in the interval (-π,π):

 
a
f ( x )  0   an cos nx   bn sin nx
2 n 1 n 1
  
2 2 2
 0  0  0
Where a0  f ( x )dx , an  f ( x )cos nxdx , bn  f ( x )sin nxdx

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 ).
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 ,

 
1 1
an 
  f ( x )cos nxdx , b

n 
 

f ( x )sin nxdx .

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

 The half range Cosine Series in the interval (0,π):

a
f ( x )  0   an cos nx
2 n 1
 
2 2
Where a0 
 
0
f ( x )dx , an 
  f ( x )cos nxdx
0

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


f ( x )   bn sin nx
n 1

2
 0
Where bn  f ( x )sin nxdx

5) The Parseval’s Identity in the interval (0,2π):

2
1 a02  2
 [ f ( x )] dx 
2
  [an  bn2 ]
 0 4 n 1
6) The Parseval’s Identity in the interval (-π,π):

2 a02  2
 [ f ( x )] dx 
2
  [an  bn2 ]
 0 4 n 1
7) The Parseval’s Identity for half range cosine series in the interval (0,π):

2 a02  2
 a
 0
[ f ( x )] dx 
2

4 n 1 n
8) The Parseval’s Identity for half range sine series in the interval (0,π):
 
2
  [ f ( x )]2
dx   bn2n 1
0

Change of interval:

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

a 
n x  n x
f ( x )  0   an cos   bn sin
2 n 1 n 1

n x n x
2 2 2
1 1 1
Where a0  
0
f ( x )dx , an  
0
f ( x )cos dx , bn  
0
f ( x )sin dx

10) The Fourier Series in the interval (-ℓ, ℓ):

a 
n x  n x
f ( x )  0   an cos   bn sin
2 n 1 n 1

2 2 n x 2 n x
Where a0   f ( x )dx , a
0
n   f ( x )cos
0
dx , bn   f ( x )sin
0
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, ℓ):

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

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a0  n x
f ( x)    an cos
2 n 1
2 2 n x
Where a0   f ( x )dx , a
0
n   f ( x )cos
0
dx

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


n x
f ( x )   bn sin
n 1

2 n x
Where bn   f ( x )sin
0
dx

12) The Parseval’s Identity in the interval (0,2ℓ):

1
2
a02  2
0 [ f ( x )] dx  4   [an  bn2 ]
2

n 1

13) The Parseval’s Identity in the interval (-ℓ,ℓ):

2 a02  2
0 [ f ( x )] dx  4   [an  bn2 ]
2

n 1

14) The Parseval’s Identity for half range cosine series in the interval (0,ℓ):
2 a02  2
0 [ f ( x )]2
dx   a
4 n 1 n
15) The Parseval’s Identity for half range sine series in the interval (0,ℓ):

2
 [ f ( x )]2
dx   bn2
n 1
0
16) Harmonic Analysis:

The method of calculation of Fourier constants by means of numerical

calculation is called as Harmonic analysis.

a0  
f ( x)    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

2 2 2
b1 
n
 y sin x , b 2   y sin 2 x , b 3   y sin 3 x ,...
n n

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2 x
When the values of x is given as numbers the  is calculated by   .
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.

Complex form of Fourier Series:

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

 2
1
f ( x )   cn e inx
where cn   f ( x )e  inx dx
n  2 0
18) The Complex form of Fourier Series in the interval (-π,π):
 
1
f ( x )   cn e inx
where cn   f ( x )e  inx dx
n  2 
19) The Complex form of Fourier Series in the interval (0,2ℓ):
 in x 2  in x
1
f ( x )   cn e
2 0
where cn  f ( x )e dx
n 

20) The Complex form of Fourier Series in the interval (-ℓ,ℓ):

 in x  in x
1
f ( x )   cn e
2 
where cn  f ( x )e dx
n 

Unit – III (Applications of Partial Differential Equations)

1) Classification of Partial Differential Equations of the second order

The equation is of the form
 2u  2u  2u  u u 
A( x , y )  B ( x , y )  C ( x , y )  f  x , y, u, ,   0
x 2
xy y 2
 x y 
i) If B 2  4 AC  0 then the above equation is elliptic.
ii) If B 2  4 AC  0 then the above equation is Parabolic.
iii) If B 2  4 AC  0 then the above equation is Hyperbolic.

2) The One dimensional Wave equation:

2 y 2  y
2
 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 

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But the correct solution is ii),

y( x, t )   A cos px  B sin px  C cos pat  D sin pat  .
3) The One dimensional Heat flow equation:
u  2u
 2 2
t x
k  Thermal Conductivity
k
 
2
where   Density
c
c  Specific Heat
The three solutions of the above equation are

i) u( x , t )  Ae px  Be  px Ce 
2 2
p t

ii) u( x, t )   A cos px  B sin px  Ce 

2 2
p t

iii) u( x, t )   Ax  B  C

But the correct solution is ii), u( x, t )   A cos px  B sin px  Ce 

2 2
p t

4) 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 – IV (Fourier Transforms)

1) Fourier Integral theorem

The Fourier integral theorem of f ( x ) in the interval   ,  is

 
1
f ( x) 
 
0 
f ( x )cos  ( t   )dxd 

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

F [ s] .

1
4) Fourier Transform F [ s]  F [ f ( x )] 
2


f ( x )e isx dx

1
 F [ s]e
 isx
5) Inverse Fourier Transform f ( x )  ds
2 
6) The Fourier transforms and Inverse Fourier transforms are called Fourier
transforms pairs.

2
7) Fourier Sine Transform Fs [ s]  Fs [ f ( x )]   f ( x )sin sx dx
 0

2
 0
8) Fourier Cosine Transform Fc [ s]  Fc [ f ( x )]  f ( x )cos sx dx

9) If f ( x )  e  ax then the Fourier Cosine and Sine transforms as follows

2
a
a) Fc [ f ( x )] 
 a  s2 2

2 s
b) Fs [ f ( x )] 
 a  s2
2

10) Properties
d
a) Fs [ xf ( x )]  F [ f ( x )]
ds c
d
b) Fc [ xf ( x )]  Fs [ f ( x )]
ds

11) Parseval’s Identity

 

 F ( s ) ds  
2 2
a) f ( x ) dx
 
   

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

2 2
b)
0 0 0 0

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

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Unit – V (Z – Transforms and Difference Equations)

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

2)

Sl.No Z  f ( n ) F [z]

Z 1
z
1.
z 1
z
2. Z ( 1)n 
z 1
z
3. Z  a n 
za
z
4. Z  n
 z  1
2

z2
5. Z  n  1
 z  1
2

1  z 
6. Z  log  
n  z 1
 n  z
7. Z sin
 2  z 1
2

 n  z2
8. Z cos
 2  z2  1
a  z 2  az 
9. Z  n a 
2 n

 z  a
3

3) Statement of Initial value theorem:

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

5) Z  a f (n)   Z  f (n)  z  z
n

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d
6) Z  nf ( n)   z Z  f ( n )
dz

7) Inverse Z-transform s

Sl.No Z 1  F ( z ) f ( n)
 z 
1. Z 1   1
 z 1
 z 
2. Z 1   ( 1)n
 z 1
 z 
3. Z 1   an
 za
 z 
Z 1   a 
n
4. 
 za
 z 
5. Z 1   n
  z  1 2 
 
 z 
6. Z 1   na n1
  z  a 2 
 
 z 
n  a 
n 1
7. Z 1  
  z  a 2 
 
 z  2
n
8. Z 1  2  cos
 z 1 2
 z2  n
9. Z 1  2 2  a n cos
 z a  2

 z 
10. Z 1  2  n
 z 1 sin
2
 z  n
11. Z 1  2 2  a n1 sin
 z a  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

Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 12

Engineering Mathematics 2014

9) Solving Difference Equations

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----

Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 13