ma2211
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Description
Engineering Mathematics Material
2013
SUBJECT NAME
: Transforms and Partial Differential Equations
SUBJECT CODE
: MA2211
MATERIAL NAME
: FormulaMaterial FormulaMaterial
MATERIAL CODE
: JM08AM3005
Unit – I (Fourier Series) 1)
Dirichlet’s Conditions: Any function f ( x ) can be expanded as a Fourier series
a 0
bn sin n x where a0 , an , b n are constants provided the
a n cos n x
2
n 1
n 1
following conditions are true. f ( x ) is periodic, single – single – valued valued and finite.
2)
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.
The Fourier Series in the interval (0,2π): f (x)
a 0
2
Where a0
a n cos nx nx n 1
1
bn sin n x n 1
2
f ( x )dx , an
1
2
f ( x ) cos n x dx , bn 0
0
3)
1
2
f ( x ) sin n x dx 0
The Fourier Series in the interval (-π,π): (-π,π): f (x)
a 0
2
Where a0
a n cos nx nx n 1
2
bn sin n x n 1
f ( x )dx , an 0
2
f ( x ) cos n x dx , bn
2
f ( x ) sin n x dx 0
0
In this interval, we have to verify the function is either odd function or even function. If it is even function then then find only a0 and a n ( b n 0 ). If it is odd function then find only b n ( a0
a n
0 ).
Engineering Mathematics Material
2013
If the function is neither odd nor even then you should find 1
a0 , an and b n by using the following formulas a0
1
an
4)
f ( x ) cos n x dx , bn
1
f ( x )dx ,
f ( x ) sin n x dx .
The half range Fourier Series in the interval (0,π): The half range Cosine Series in the interval (0,π): a 0
f (x)
a n cos n x
2
n 1
2
Where a0
f ( x )dx , an
2
0
f ( x ) cos n x dx 0
The half range Sine Series in the interval (0,π): f (x)
bn sin n x n 1
2
Where bn
f ( x ) sin n x dx 0
5)
The Parseval’s Identity in the interval (0,2π):
1
2
2
2
[ f ( x )] dx 0
6)
4
2
bn ]
n 1
2
a 0
2
[ f ( x )] dx
[an2
4
0
2
bn ]
n 1
The Parseval’s Identity for half range cosine series in the i nterval (0,π): 2
2
a 0
2
[ f ( x )] dx
2
a n
4
0
8)
[an2
The Parseval’s Identity in the interval ((-π,π):
2 7)
a 0
n 1
The Parseval’s Identity for half range sine series in the interval (0,π): 2
2
2
[ f ( x )] dx
bn n 1
0
Change of interval: 9)
The Fourier Series in the interval (0,2ℓ): f (x)
a 0
2
Where a0
a n cos
n x
n 1
1
n 1
2
f ( x )dx , an 0
n x bn sin
1
2
f ( x ) cos 0
n x
dx , bn
1
2
f ( x ) s in 0
n x
dx
Engineering Mathematics Material 10)
2013
The Fourier Series in the interval (-ℓ (-ℓ, ℓ): f (x)
a 0
n x
a n cos
2
n x bn sin
n 1
2
Where a0
n 1
f ( x )dx , an
2
0
f ( x ) cos
n x
2
dx , bn
f ( x ) sin
n x
dx
0
0
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 a n ( b n 0 ). If it is odd odd function then find only b n ( a0
11)
a n
0 ).
The half range Fourier Series in the interval (0, ℓ):
The half range Cosine Series in the interval (0, ℓ): f (x)
a 0
a n cos
2
n x
n 1
2
Where a0
f ( x )dx , an
2
0
f ( x ) cos
n x
dx
0
The half range Sine Series in the interval (0, ℓ): f (x)
b n sin
n x
n 1
2
Where bn
f ( x ) sin
n x
dx
0
12)
The Parseval’s Identity in the interval (0,2ℓ): (0,2ℓ): 1
2
2
2
[ f ( x )] dx 0
13)
4
2
bn ]
n 1
2
2
[ f ( x )] dx 0
a 0
2
[an
4
2
bn ]
n 1
The Parseval’s Identity for half range cosine series in the i nterval (0,ℓ): (0,ℓ): 2
2
2
[ f ( x )] dx 0
15)
2
[ an
The Parseval’s Identity in the interval (-ℓ (-ℓ,ℓ): 2
14)
a 0
a 0
2
an
4
n 1
The Parseval’s Identity for half range sine series in the interval (0,ℓ): (0,ℓ):
2
[ f ( x )]2 dx 0
2
bn 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)
a 0
where 2 a0 n
b1
a n cos nx nx
2
y ,
2
n 1
a1
y sin x ,
n
bn sin n x
n 1
2
y cos x ,
n
b2
2
2
a2
n
y sin 2 x ,
n
y cos 2x , a 3
2
b3
n
When the values of x is given as numbers the
2 n
. .. y cos 3x , ..
, ... y sin 3x ,.
is calculated by
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. Complex form of Fourier Series: 17)
The Complex form of Fourier Series in the interval (0,2π): in x
f (x)
cn e
where cn
n
18)
in x
cn e
where cn
n
2
in x
dx
0
1
f ( x )e
2
in x
dx
The Complex form of Fourier Series in the i nterval (0,2ℓ (0,2ℓ): in x
f (x)
cn e
where cn
n
20)
f ( x )e
The Complex form of Fourier Series in the interval ( -π,π): f (x)
19)
2
1
1
2
2
in x
f ( x )e
dx
0
The Complex form of Fourier Series in the interval ( -ℓ,ℓ): in x
f (x)
cn e
where cn
n
1 2
in
f ( x )e
x
dx
Unit – II (Fourier Transforms) 1)
Fourier Integral theorem The Fourier integral theorem of f ( x ) in the interval f (x)
1
f ( x ) cos ( t 0
)dx d
,
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 )]
3)
F s .G s
The Fourier Transform of a function f ( x ) is given by F [ f ( x )] is denoted by F [ s ] .
1
isx
4)
Fourier Transform F [ s]
5)
Inverse Fourier Transform f ( x )
6)
The Fourier transforms and Inverse Fourier transforms are called Fourier transforms pairs.
7)
Fourier Sine Transform F s [ s]
F [ f ( x )]
f ( x )e dx
2 1
F [ s]e
2
F s [ f ( x )]
2
isx
ds
f ( x ) sin sx dx 0
8)
Fourier Cosine Transform F c [ s]
F c [ f ( x )]
2
f ( x ) cos sx dx 0
9)
If f ( x )
e
ax
then the Fourier Cosine and Sine transforms as follows
a) F c [ f ( x )] b) F s [ f ( x )] 10)
F s [ x f ( x )]
b) F c [ x f ( x )]
a
2
2
2
s s
a
2
2
s
d ds d ds
F c [ f ( x )]
F s [ f ( x )]
Parseval’s Identity 2
a)
F ( s) ds ds
b)
2
f ( x ) dx 2
f ( x ) g( x )dx (Or)
F c ( s)Gc ( s)ds 0
12)
a
Property a)
11)
2
0
)] Condition for Self reciprocal F [ f ( x )]
0
f ( s)
2
F c ( s) ds ds
f ( x ) dx 0
Engineering Mathematics Material
2013
Unit – III (Partial Differential Equation) 1) Lagrange’s Linear equation The equation of the form Pp
Qq
R
dx
dy
dz
P
Q
R
then the subsidiary equation is
2) Homogeneous Linear Partial Differential Equation of higher order with constant coefficients: 2
The equation of the form a
x
2
z
b
2
2
z
c
x y
z
f ( x , y)
2
y
The above equation can be written as aD
2
bD D
2
x
2
f ( x , y ) …………………………….. (1)
z
2
2
where D
cD
2
, D
x
and D
2
y
, D
2
y
The solution of above equation is z = C.F + P.I Complementary Complementary Function (C.F) :
1
m1
To find C.F consider the auxiliary equation by replacing D by m and D by 2 bm c 0 , solving this equation 1.The equation (1) implies that am we get two values of m. m. The following table gives C.F C.F of the above equation. Nature of m Complementary Complementary Function m 2 C.F = f 1 ( y m 1 x ) f 2 ( y m 2 x )
2
m1
m 2
3
m1
m2
4
m1
5
m1
Sl.No.
C.F = f 1 ( y
mx )
m 3
C.F = f 1 ( y
m1 x )
m2
m 3
C.F = f 1 ( y
mx )
m2 ,
m 3 is different different
xf 2 ( y f 2( y xf 2 ( y
C.F = f 1 ( y m x )
xf 2 ( y
2
cD .
Particular Integral (P.I) : To find P.I consider ( D , D ) 0 , then P.I
Type: 1If 1 If
f ( x , y )
Type: 2
If f ( x , y ) P .I
ax by
e
1 ( D , D )
ax by
e
aD
0.
bD D
2
m x ) m2 x ) mx )
mx mx )
f 3( y 2
x f 3(y
f 3( y
m 3 x ) m x )
m 3 x )
Engineering Mathematics Material
2013
Replace D by a and D by b. If If
( D , D )
0 , then it is P.I.
0 , then diff. denominator w.r.t D and multiply x in
( D , D )
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 )
si n ( a x
1
P .I
( D , D )
b y ) ( o r ) c os ( a x
by )
by ) (or ) cos(ax
by )
sin(ax
2 2 2 2 Here replace D by a , D by b and DD by ab . Do not replace for D and D . If the denominator denominator equal to zero, then apply the same producer as in Type: 2.
Type: 4
m
If f ( x , y ) 1
P .I
m
( D , D ) 1
1 1
n
x y
x y m
g ( D , D )
n
n
x y 1
g( D , D )
m
n
x y
Here we can use Binomial formula as follows: 1
i) 1
x
ii) 1
x
iii) 1
x
iv) 1
x
v) (1
x)
vi) (1
x)
1
2 2
3 3
x
x
1
x
x
3
...
4 x 3 ...
1 2x
3x 2
4 x 3 . ..
1 3x
6x
1 3x
6x2
2
10 x
3
...
10 x 3 ...
V , where
by) (or) cos(ax
m
n
by) (or ) x y
a x by
( D , D )
V
e
ax First operate e
a x by
Type: 4.
x
. ..
3x 2
1
e
2
3
1 2x
e
V=sin(ax
P .I
x
ax by
Type: 5 If f ( x , y )
P .I
2
1
(D
by
by replacing D by D a and D by D a . 1 V , Now this will either Type: 3 or a, D b )
Engineering Mathematics Material Type: 6 If f ( x , y )
y sin ax (or) y cos ax
1
P .I
2013
y sin ax
( D , D )
1 D
m1 D
D
1
D
m2D
y
c
m2 x
m 2 x si sin ax dx (Apply Bernouili’s method)
c
m 1 D
y sin ax
3) Solution of Partial Differential Equations: Standard Type: 1
Equation of the form Assume that
z
ax
equation.put p
Standard Type: 2
f ( p, q )
0
c be the solution the above
by
a and q
b in
equation (1), we get
f (a , b )
0 . Now, solve this, we get b
z
(a ) y
ax ax
c which is called
Equation of the form
z
The Complete solution is
Complete solution.
px
qy
z
ax
f ( p, q ) by
Singular integral diff. partially w.r.t and eliminate Standard Type: 3
(Clairaut’s form)
f (a , b) . To find a & b , equate to zero
a and b .
Equation of the form f 1 ( x , p) The solution is z
Standard Type: 4
( a ) .
f 2 ( y , q )
qdy .
pdx
Equation of the form f ( z, p, q ) 0 In this type put u
x
ay a y , then p
dz du
,q
a
Unit – IV (Application of Partial Differential Equation) 1) The One dimensional Wave equation: e quation: 2
t
y 2
2 2
a
y 2
x
The three solutions of the above equation are
px px
i)
y( x , t )
ii)
y( x , t )
A cos px
iii)
y( x , t )
Ax
Ae
Be
B
px
Ce
pat
B sin px
Ct
D
pat
De
C cos pat
D sin pat
dz du
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: 2
u
2
u 2
t
x
k
k
2
c
Thermal Conductivity Density
where
Specific Heat
c
The three solutions of the above equation are px
i)
u( x , t )
Ae
ii)
u( x , t )
A cos px
iii)
u( x , t )
Ax
px
Be
2
Ce
2
p t
B sin px Ce
2
2
p t
B C
But the correct solution is ii), u ( x , t )
A cos px
B sin px Ce
2
2
p t
3) The Two dimensional Heat flow equation: 2
u
2
2
y
x
u 2
0
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
py
De
(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 Z
f (n ) be the sequence defined for all the positive integers n such that n
f (n )
f (n )z n 0
Engineering Mathematics Material
2013
2)
Sl.No
Z f ( n )
1.
Z 1
F [ z ]
z z 1
2.
Z ( 1)
3.
Z a
4.
Z n
z
n
z 1 z
n
z
a z 2
z 1 2
Z n 1
5.
Z
7.
Z sin
z
log
n
Z cos
2
z 1
1
6.
8.
z
z 1
n
z 2
2
1
z
2
n
z 2
2
1
z
3) Statement of Initial value theorem: If Z
f (n )
F [ z] , then L t F [ z] z
L t f ( n )
n
0
4) Statement of Final value theorem: If Z 5)
6)
f (n )
F [ z] , then L t f ( n ) n
n
Z a f (n ) Z n f (n )
Z f (n ) z
d dz
z
L t ( z 1)F ( z)
z
1
z a
Z f (n )
7) Inverse Z-transform 1
Sl.No
Z
1.
Z
1
2.
Z
1
3.
Z
1
F ( z ) z z 1 z z 1
z z
a
f ( n )
1 n
( 1) n
a
Engineering Mathematics Material
4.
5.
Z
6.
Z
7.
Z
z
1
Z
2013
z
a z
1
n
2
z 1 z
1
z
n 1
n a
2
a z
1
z
n
2
a
Z
8.
2
cos
1
z
n
2
Z
9.
10.
z
Z
2
n
a cos
2
a
z
1
Z
11.
z
1
2
sin
1
z
2
z
2 n
2
n
2
z
1
n 1
a
2
z
1
n
a
n 1
a
2
a
sin
n
2
8) Inverse form of Convolution Theorem 1
Z [F ( z).G ( z)]
Z
1
1 [F (z )] Z [G ( z ) ] n
and by the defn. of Convolution of two functions f ( n )
g( n )
f ( r ) g( n r 0
9) a) Z [ y( n )]
F ( z)
b) Z [ y( n
1)]
zF ( z)
c) Z [ y( n
2)]
z F ( z)
d) Z [ y( n
3)]
z F ( z)
zy(0 )
2
z y(0)
3
z y(0)
----
2
zy(1 )
3
z y(1)
2
zy(2 )
----
r )
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