Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 1
SUBJECT NAME : Transforms and Partial Differential Equations
SUBJECT CODE : MA 2211
MATERIAL NAME : Problem Material
MATERIAL CODE : JM08AM3006
Name of the Student: Branch: Unit – I (Fourier Series)
1) Develop a Fourier series for the function ( ) ( )f x x xππππ= −= −= −= − in the interval (0,2 )ππππ .
2) Find the Fourier series for the function 2( ) ( )f x xππππ= −= −= −= − in the interval (0,2 )ππππ and
deduce sum of series as 2 2 2
1 1 11 ...
2 3 4+ + + ++ + + ++ + + ++ + + + and also find 4 4 4
1 1 11 ...
2 3 4+ + + ++ + + ++ + + ++ + + +
3) Find the Fourier series of periodic 2π for in (0, )
( )2 in ( ,2 )
xf x
x
πππππ π ππ π ππ π ππ π π
==== −−−−
and deduce that
2
2 2 2
1 1 1...
1 3 5 8ππππ+ + + =+ + + =+ + + =+ + + = .
4) Obtain the Fourier expansion of periodicity 2π for ( )f x x==== when xπ ππ ππ ππ π− < <− < <− < <− < < .
Deduce that
2
2 2
1 11 ...
3 5 8ππππ+ + + =+ + + =+ + + =+ + + = and also find the value of 4 4
1 11 ...
3 5+ + ++ + ++ + ++ + +
5) Find the Fourier series of 2( )f x x==== in the interval ( , )π ππ ππ ππ π−−−− . Hence deduce that
2 2 2
1 1 1...
1 2 3+ + ++ + ++ + ++ + + also find the value of 4 4 4
1 1 1...
1 2 3+ + ++ + ++ + ++ + +
6) Find the Fourier series for the function2( )f x x x= += += += + in the interval ( , )π ππ ππ ππ π−−−− and also
deduce the value of
2
2
16n
ππππ====∑∑∑∑ .
7) Show that when 0 x ππππ< << << << < , sin 2 sin4 sin6
...2 1 2 3
x x xx
ππππππππ − = + + +− = + + +− = + + +− = + + +
8) Develop a sine series of the function 0 / 2
( ) / 2
x xf x
x x
πππππ π ππ π ππ π ππ π π
< << << << <==== − < <− < <− < <− < <
9) Find the Fourier series of ( ) cosf x x x==== in the interval ( , )π ππ ππ ππ π−−−− .
10) Find the Fourier series of ( ) cosf x x==== in the interval [-π, π].
11) Obtain the Fourier series of ( )f x period 2 � and defined as follows
0( )
0 2
x xf x
x
− < <− < <− < <− < <==== < << << << <
� �� �� �� �
� �� �� �� � and hence deduce the following
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 2
i) 1 1 1
1 ...3 5 7 4
ππππ− + − + ∞ =− + − + ∞ =− + − + ∞ =− + − + ∞ =
ii) 2
2 2 2
1 1 1...
1 3 5 8ππππ+ + + ∞ =+ + + ∞ =+ + + ∞ =+ + + ∞ =
12) Find the Fourier series of 2( )f x x==== in the interval ( , )−−−−� �� �� �� � . Hence deduce the sum of
series 2 2 2
1 1 1...
1 2 3+ + + ∞+ + + ∞+ + + ∞+ + + ∞ .
13) Find the Fourier series of ( ) (2 )f x x x= −= −= −= − in 0 2x< << << << < . Deduce the sum of the
series 2 2 2 2
1 1 1 1...
1 2 3 4+ + + + ∞+ + + + ∞+ + + + ∞+ + + + ∞ .
14) Find the complex form of the Fourier series of the function ( ) xf x e==== when
xπ ππ ππ ππ π− < <− < <− < <− < < and ( 2 ) ( )f x f xππππ+ =+ =+ =+ = .
15) Find the complex form of the Fourier series of the periodic function ( ) sinf x x==== when
0 x ππππ< << << << < .
16) Find the Fourier Series up to three harmonic for y = f(x)in (0,2π) for the following data
x 0 π/3 2π/3 π 4π/3 5π/3 2π
y 1.0 1.4 1.9 1.7 1.5 1.2 1.0
17) Find the Fourier Series up to two harmonic for y = f(x)in (0,2π) for the following data
x 0° 60° 120° 180° 240° 300° 360°
y 40.0 31.0 -13.7 20.0 3.7 -21.0 40.0
18) Find the Fourier Series up to two harmonic for y = f(x)in (0,2π) for the following data
x 0 1 2 3 4 5
y 9 18 24 28 26 20
19) The values of x and the corresponding values of f(x) over a period T are given below
Show that ( ) 0.75 0.37cos 1.004sinf x θ θθ θθ θθ θ= + += + += + += + + where2 xTππππθθθθ ==== .
x 0 T/6 T/3 T/2 2T/3 5T/6 T
f(x) 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 3
Unit – II (Fourier Transforms) Part – I
1) Show that the Fourier transform of
2 2, ( )
0 ,
a x x af x
x a
− <− <− <− <==== >>>>
is
3
2 sin cos2
as as assππππ
−−−− .
Hence deduce that 30
sin cos4
t t tdt
tππππ∞∞∞∞ −−−− ====∫∫∫∫ . Using
Parseval’s identity show that
2
30
sin cos15
t t tdt
tππππ∞∞∞∞ −−−− ====
∫∫∫∫ .
2) Find the Fourier transform of
21 , 1( )
0 , 1
x xf x
x
− <− <− <− <==== >>>>
. Hence prove that
30
sin cos 3cos
2 16s s s s
dss
ππππ∞∞∞∞ −−−− ====∫∫∫∫ .
3) Find the Fourier transform of ( )f x if1 ,
( )0 , 0
x af x
x a
<<<<==== > >> >> >> >
. Hence deduce that
i)
0
sin2
tdt
tππππ∞∞∞∞
====∫∫∫∫ ii)
2
0
sin2
tdt
tππππ∞∞∞∞
==== ∫∫∫∫
4) Find the Fourier transform of ( )f x if1 , 1
( )0 , 1
x xf x
x
− <− <− <− <==== >>>>
. Hence deduce that
2
0
sin2
tdt
tππππ∞∞∞∞
==== ∫∫∫∫
and
4
0
sin3
tdt
tππππ∞∞∞∞
==== ∫∫∫∫ .
Note: The same problem they may ask ,
( )0 ,
a x x af x
x a
− <− <− <− <==== >>>>
with same
deduction.
Part – II
5) Find the Fourier cosine and sine transform of the function ( ) , 0axf x e a−−−−= >= >= >= > .
6) Evaluate
(((( ))))22 20
, 0dx
if aa x
∞∞∞∞
>>>>++++
∫∫∫∫ using Parseval’s identity.
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 4
7) Evaluate
(((( ))))2
22 20
, 0x
dx aa x
∞∞∞∞
>>>>++++
∫∫∫∫ using Parseval’s identity.
8) Evaluate (((( )))) (((( ))))2 2 2 20
, , 0dx
if a bx a x b
∞∞∞∞
>>>>+ ++ ++ ++ +∫∫∫∫ using Parseval’s identity.
9) Evaluate (((( )))) (((( ))))2
2 2 2 20
, , 0x
dx if a bx a x b
∞∞∞∞
>>>>+ ++ ++ ++ +∫∫∫∫ using Parseval’s identity.
10) Evaluate (((( )))) (((( ))))2 20 1 4
dx
x x
∞∞∞∞
+ ++ ++ ++ +∫∫∫∫ using transforms method.
Part – III (Special Problems)
11) Find the Fourier cosine and sine transform of ( ) axf x xe−−−−==== .
12) Find the Fourier cosine transform of ( )axe
f xx
−−−−
==== .
13) Find the Fourier sine transform of ( )axe
f xx
−−−−
==== .
14) Find the Fourier cosine transform of2
( ) xf x e−−−−==== .
15) Find the Fourier cosine transform of 2 2
, 0a xe a−−−− >>>> and hence deduce that sine
transform of 2 2a xxe−−−−
.
16) Show that the Fourier sine transform of
2
2x
xe−−−−
is self reciprocal.
17) Find the Fourier sine and cosine transform of xe−−−−
also find the Fourier sine transform of
21xx++++
and Fourier cosine transform of 2
11 x++++
.
18) State and Prove Convolution theorem in Fourier transform.
Unit – III (Partial Differential Equations)
• Higher Order Homogeneous PDE
1) Solve (((( ))))2 26 5 sinhxD DD D z e y xy′ ′′ ′′ ′′ ′− + = +− + = +− + = +− + = + .
Ans.:
3 4
1 2
1( ) ( 5 )
8 24 6 4x y x yx x y x
z f y x f y x e e+ −+ −+ −+ −= + + + − − + += + + + − − + += + + + − − + += + + + − − + + .
2) Solve (((( ))))3 2 24 4 6sin(3 6 )D D D DD z x y′ ′′ ′′ ′′ ′− + = +− + = +− + = +− + = + .
Ans.: 1 2 3
2( ) ( 2 ) ( 2 ) cos(3 6 )
81z f y f y x xf y x x y= + + + + + += + + + + + += + + + + + += + + + + + + .
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 5
3) Solve (((( ))))2 sin sin 2D DD z x y′′′′− =− =− =− = .
Ans.: 1 2
1 1( ) ( ) cos( 2 ) cos( 2 )
6 2z f y f y x x y x y= + + − − − += + + − − − += + + − − − += + + − − − + .
4) Solve
3 3 33
3 2 37 6 sin( 2 ) x yz z zz x y e
x x y y++++ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂
− − = + +− − = + +− − = + +− − = + + ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂ .
Ans.: 3
1 2 3
1( ) ( 2 ) ( 3 ) cos( 2 )
75 20x yx
z f y x f y x f y x x y e ++++= − + − + + − + += − + − + + − + += − + − + + − + += − + − + + − + + .
5) Solve (((( ))))2 26 cosD DD D z y x′ ′′ ′′ ′′ ′+ − =+ − =+ − =+ − = .
Ans.: 1 2( 2 ) ( 3 ) cos sinz f y x f y x y x x= + + − − += + + − − += + + − − += + + − − + .
6) Solve (((( ))))2 22 2 2 sin( 2 )D DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′+ + − − = ++ + − − = ++ + − − = ++ + − − = + .
7) Solve (((( ))))2 2 23 3 7x yD D D D z e xy++++′ ′′ ′′ ′′ ′− − + = + +− − + = + +− − + = + +− − + = + + .
8) Solve (((( ))))2 22 3 3 2 cosh(2 )D DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′− + − + + = +− + − + + = +− + − + + = +− + − + + = + .
9) Solve (((( ))))2 22 2 2 cos 2 cosD DD D D D z x y′ ′ ′′ ′ ′′ ′ ′′ ′ ′− − + + =− − + + =− − + + =− − + + =
• Standard Types
1) Solve z px qy pq= + += + += + += + + . Ans.: 0z xy+ =+ =+ =+ =
2) Solve2 2z px qy p q= + + −= + + −= + + −= + + − . Ans.:
2 24z y x= −= −= −= −
3) Solve2 21z px qy p q= + + + += + + + += + + + += + + + + . Ans.:
2 2 2 1x y z+ + =+ + =+ + =+ + = .
4) Solve2 21z px qy c p q= + + + += + + + += + + + += + + + + . Ans.:
2 2 2 2x y z c+ + =+ + =+ + =+ + = .
5) Solvez x y
pqpq q p
= + += + += + += + + . Ans.:
Hint: Multiply pq on both side, we get (((( ))))3/2z px qy pq= + += + += + += + +
6) Solve (((( ))))2 2 21p y x qx+ =+ =+ =+ = .
Hint: Rearrange the above equation, we get (((( ))))2 2
2
1 (says)
p x qk
x y
++++= == == == = .
This is of the form 1 2( , ) ( , )f x p f y q==== .
7) Solve (((( ))))2 29 4p z q+ =+ =+ =+ = . Ans.: (((( )))) (((( ))))3 22z a x ay b+ = + ++ = + ++ = + ++ = + + .
Hint: This is of the form ( , , ) 0f z p q ==== .
8) Solve2 2 21z p q= + += + += + += + + . Ans.: (((( ))))1
2
1cosh
1z x ay b
a−−−− = + += + += + += + +
++++.
Hint: This is of the form ( , , ) 0f z p q ==== .
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 6
9) Solve (((( ))))1p q qz+ =+ =+ =+ = . Ans.: (((( ))))log 1x ay c az + = −+ = −+ = −+ = − .
Hint: This is of the form ( , , ) 0f z p q ==== .
10) Solve (((( )))) (((( )))) (((( ))))x y z p y z x q z x y− + − = −− + − = −− + − = −− + − = − . Ans.: (((( )))), 0x y z xyzφφφφ + + =+ + =+ + =+ + =
11) Solve (((( )))) (((( )))) (((( ))))2 2 2 2 2 2x y z p y z x q z x y− + − = −− + − = −− + − = −− + − = − . Ans.: (((( ))))2 2 2, 0x y z xyzφφφφ + + =+ + =+ + =+ + =
12) Solve (((( )))) (((( )))) (((( ))))m n n mz y p x z q y x− + − = −− + − = −− + − = −− + − = −� �� �� �� � .
Ans.: (((( ))))2 2 2, m n 0x y z x y zφφφφ + + + + =+ + + + =+ + + + =+ + + + =����
13) Solve (((( )))) (((( )))) (((( ))))3 4 4 2 2 3z y p x z q y x− + − = −− + − = −− + − = −− + − = − .
Ans.: (((( ))))2 2 2,2 3 4 0x y z x y zφφφφ + + + + =+ + + + =+ + + + =+ + + + =
14) Solve (((( )))) (((( )))) (((( ))))2 2 2x yz p y zx q z xy− + − = −− + − = −− + − = −− + − = − .
Ans.: , 0x y
xy yz zxy z
φφφφ −−−− + + =+ + =+ + =+ + = −−−−
Unit – IV (Application of PDE) • One dimensional Wave equation.
1) Problems on Zero initial velocity (Pre work: 1).
For the following ( )f x
a) ( ) sinx
f x aππππ====����
(another name Sinusoid of length a)
b) 30( ) sin
xf x y
ππππ====����
c) ( ) ( )f x x xλλλλ= −= −= −= −����
d) The midpoint of the string is displaced to a small height ‘h’ or ‘b’ is given.
e) Any problems with string of length 2ℓ.
2) Problems on Non – Zero initial velocity (Pre work: 2).
For the following ( )f x
f) ( ) sinx
f x aππππ====����
(another name Sinusoid of length a)
g) 30( ) sin
xf x v
ππππ====����
h) ( ) ( )f x x xλλλλ= −= −= −= −����
i) ,0 / 2
( ), / 2
cx xv
c x x
≤ ≤≤ ≤≤ ≤≤ ≤==== − ≤ ≤− ≤ ≤− ≤ ≤− ≤ ≤
����
� � �� � �� � �� � �
j) Any problems with string of length 2ℓ.
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 7
• One dimensional Heat flow equation.
3) Problems on zero boundary conditions. (Pre work: 3)
4) Problems on One end zero and another end non – zero boundary condition and
reduced to 0°C. (Pre work: 4)
5) Problems on both ends non – zero boundary condition and reduced to 0°C.
6) Problems on both ends non – zero boundary condition and reduced to non – zero
temperature.
Unit – V (Z – Transform) • Problems on Z – transform
1) Find the Z – transform of the function cos , sin , cos , sin ,n nn n a n a nθ θ θ θθ θ θ θθ θ θ θθ θ θ θ
cos / 2, sin / 2, cos / 2, sin / 2.n nn n a n a nπ π π ππ π π ππ π π ππ π π π
2) State and Prove final value theorem in Z – transform.
3) State and Prove Initial value theorem in Z – transform.
4) Find the Z – transform of the following functions (i) {{{{ }}}}na (ii) {{{{ }}}}nna (iii) 1n
(iv)
1!n
(v) !
nan
(vi) 1
1n ++++
(vii) {{{{ }}}}n (viii) 1n
(ix) (((( )))) (((( ))))1 2n n+ ++ ++ ++ + (x) (((( ))))1
1n n ++++
5) Find the Z – transform of the function (((( )))) (((( ))))2 3
( )1 2n
f nn n
++++====+ ++ ++ ++ +
.
• Problems on Inverse Z – transform using Partial fraction
method
1) Find the inverse Z – transform of (((( )))) (((( )))) (((( ))))2 3
2 1 5z z
z z z++++
+ − ++ − ++ − ++ − + using partial fraction
method.
2) Find (((( ))))
(((( )))) (((( ))))
2
12
2
1 1
z z zZ
z z−−−− − +− +− +− + + −+ −+ −+ −
by using partial fraction method.
3) Find the inverse Z – transform of(((( )))) (((( ))))
2
22 4
z
z z+ ++ ++ ++ + using partial fraction method.
4) Find 1
2 7 10z
Zz z
−−−− + ++ ++ ++ +
by using partial fraction method.
5) Find the inverse Z – transform of
2
3 2
33 4
z zz z
++++− +− +− +− +
using partial fraction method.
Engineering Mathematics Material 2010
Prepared by C.Ganesan, M.Sc., M.Phil., (Ph: 9841168917) Page 8
6) Find (((( ))))
(((( ))))1
3
1
1
z zZ
z−−−− ++++
−−−− by using reduces method.
• Problems on Convolution Theorem
1) State Convolution theorem and use it to evaluate (((( ))))2
1
( )z
Zz a z b
−−−− − −− −− −− −
.
2) Find
21
2( )z
Zz a
−−−− ++++
using convolution theorem.
3) Find
21 8
(2 1)(4 1)z
Zz z
−−−− − +− +− +− +
using convolution theorem.
4) Using Convolution theorem, find (((( )))) (((( ))))2
1
1 3z
Zz z
−−−− − −− −− −− −
.
5) Find the inverse Z – transform of
2z
z a −−−−
, using Convolution theorem.
6) Use Convolution theorem to find the inverse Z – transform (((( )))) (((( ))))1 1
12
3 4z z− −− −− −− −− −− −− −− −.
• Solving Difference Equation
1) Using Z – transform, solve 2 16 9 2kk k ky y y+ ++ ++ ++ ++ + =+ + =+ + =+ + = given 0 1 0y y= == == == = .
2) Solve 2 13 2 0n n nu u u+ ++ ++ ++ ++ + =+ + =+ + =+ + = given 0 11, 2u u= == == == = , using Z – transform.
3) Solve ( 3) 3 ( 1) 2 ( ) 0y n y n y n+ − + + =+ − + + =+ − + + =+ − + + = given that (0) 4, (1) 0y y= == == == = and
(2) 8.y ====
4) Solve the system 2 15 6 ,n n n ny y y u+ ++ ++ ++ +− + =− + =− + =− + = with 0 10, 1y y= == == == = and 1nu ==== for
0,1,2, ...n ==== by Z – transform method.
Hint: Substitute 1nu ==== in given equation, we get 2 15 6 1n n ny y y+ ++ ++ ++ +− + =− + =− + =− + = and do as
usual method.
5) Using Z – transform solve ( ) 3 ( 1) 4 ( 2) 0,y n y n y n+ − − − =+ − − − =+ − − − =+ − − − = 2n ≥≥≥≥ given that
(0) 3y ==== and (1) 2.y = −= −= −= −
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