ma 2211 unit 1
DESCRIPTION
transforms and partial differential equationsTRANSCRIPT
1
MISRIMAL NAVAJEE MUNOTH JAIN ENGINEERINGCOLLEGE, CHENNAI - 97
DEPARTMENT OF MATHEMATICS
(MA2211)
TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS
FOR
THIRD SEMESTER ENGINEERING STUDENTSANNA UNIVERSITY SYLLABUS (R-2008-2009)
This text contains some of the most important short-answer (Part A) and long- answerquestions (Part B) and their answers. Each unit contains 30 university questions. Thus, atotal of 150 questions and their solutions are given. A student who studies these modelproblems will be able to get pass mark (hopefully!!).
Prepared by the faculty of Department of Mathematics
AUGUST, 2009
www.engg-maths.com
2
UNIT I FOURIER SERIES
PART – A
Problem 1 Write the formula for finding Euler’s constants of a Fourier series in ( 0, 2 ).Solution:
Euler’s constants of a Fourier series in ( 0, 2 ) is given by
2
0
0
2
0
2
0
1
1cos
1sin
n
n
a f x dx
a f x nxdx
b f x nxdx
Problem 2 Write the formula for Fourier constants for f(x) in the interval , .
Solution:
0
1
1cos
1sin
n
n
a f x dx
a f x nxdx
b f x nxdx
Problem 3 Find the constant a0 of the Fourier series for the function f(x) = k , 0 2x .Solution:
2
0
0
2
0
2
0
1
1
1
2
a f x dx
kdx
kx
k
Problem 4 If f(x) = e x in x , find a n .Solution:
1cosx
na e nxdx
2
1cos sin
1
xenx n nx
n
3
2 2
11 1
1 1
n ne e
n n
2
1
1
n
na e en
.
Problem 5 Write the formula’s for Fourier constants for f(x) in (c, c+2l).Solution:
2
0
2
2
1
1cos
1sin
c
c
c
n
c
c
n
c
a f x dx
a f x nx dx
b f x nx dx
Problem 6 Write the formulas for Fourier constants for f(x) in (-l, l).Solution:
0
1
1cos
1sin
n
n
a f x dx
a f x nx dx
b f x nx dx
Problem 7 What is the sum of Fourier series at a point x = x 0, where the function f(x) has afinite discontinuity ?Solution:
Sum of the Fourier series at 0x x is 0 0
2
f x f x
Problem 8 If
22
2 2 2
cos cos2 cos34 .......... (1)
3 1 2 3
x x xx to
in x , find2 2 2
1 1 1.............
1 2 3to .
Solution:
Put2
x
a point of continuity
2 2
2 2
1 1(1) 4 ........
4 3 1 2
4
2 2
2 2
1 14 .......
4 3 1 2
2
2 2
1 1 1. .......
12 4 1 2
2
2 2 2
1 1 1.....
1 2 3 48
Problem 9 Check whether the function is odd or even, where f(x) is defined by
21 0
21 0
xx
f xx
x
Solution:
2 2
0, 1 1 , 0x x
For x f x f x where x
f x is an even function.
Problem 10 When an even function f(x) is expanded in a Fourier series in the intervalx , show that bn = 0.
Solution:
1
sinn
n
n
b f x nxdxn
Given f x is even and
sin nx is odd function
even X odd = odd. Therefore sinf x nx is odd function.
1
sinnb f x nx dx
= 0.
Problem 11 Find the Fourier constant bn for x sin x in x , when expressed as a
Fourier series.Solution:
sin
sin
sin
f x x x
f x x x
x x f x
Here f x is an even function
0nb
Problem 12 If f(x) is a function defined in 2 2x , what is the value of bn ?Solution:
5
2
2
1sin
2 2n
n xb f x dx
Problem 13 Explain half range cosine series in 0, .
Solution:
Half range cosine series in 0, is given by
0
1
cos2
n
af x a n
0
0
2a f x dx
0
2cosna f x nxdx
Problem 14 Find the sine series of f(x) = k in 0, .
Solution:
1
sinnf x b nx
0
2sinnb k nxdx
0
2 cosk nx
n
1 12n
n
kb
n
1
1 12sin
nk
f x nxn
.
Problem 15 Write Parseval’s formula in the interval , 2c c n .
Solution:
2 2
2 2 20
1
1 1
2 4 2
c
n n
c
af x dx a b
PART – B
Problem 16 If 2
2
xf x
in 0 2x . Hence show that
(a)2
2 2 2
1 1 1.............
1 2 3 6
.
(b)2
2 2 2 2
1 1 1 1...
1 2 3 4 12
.
6
Solution:We know that
0
1
cos sin2
n n
af x a nx b nx
22
0
0
1
2
xa dx
2
2
0
1
4x dx
31
4 3
x
3 3 21
4 3 3 6
22
0
1cosn
xa nxdx
2
2
0
1cos
4x nxdx
2
2
2 3
0
1 sin cos sin2 2
4
nx nx nxx x
n n n
2 2
1 cos 2 20 2 0 0
4
n
n n
2 2
1 4 1
4 n n
2
2
0
1sin
4nb x nxdx
2
2
2 3
0
1 cos sin cos2 2
4
nx nx nxx x
n n n
2 2
3 3
1 2 20
4 n n n n
2
21
1cos .
12f x n x
n
2
2 2
1 1cos cos .... 1
12 1 2x x
0Put x
2
2 2
1 10 ... 2
12 1 2f
0x is a pt of discontinuity.
7
2 2 21
02 4 4 4
f
2 2
2 2
1 12 ...
4 12 1 2
2 2
2 2
1 1......
4 12 1 2
2
2 2
1 1.....
1 2 6
Put x in (1)
2
21
13
12
n
f xn
Here is a pt of continuity.
0.f
2
2 2 2
1 1 13 0 .....
12 1 2 3
2
2 2 2
1 1 1.....
12 1 2 3
2
2 2 2
1 1 1.....
12 1 2 3
Problem 17 Find the Fourier series of f (x) = xsin x in 0 2x .
Solution:
0
1
cos sin2
n n
af x a nx b n x
2
0
0
1sina x x dx
2
0
1 1cos 1 sin 2 1 2x x x
2
0
1sin sinnb x x nxdx
2
0
1cos 1 cos 1
2x n n dx
2
cos 1sin 111
2 1 1
n xn xx
n n
2
2
0
sin 1 cos 11
1 1
n x n xx
n n
2 2 22
1 1 1 1 1
2 11 1 1nn n n
0 1n
8
2
1
0
1sin sinb x x x dx
2
2
0
1sinx x dx
2
0
1 cos 21
2
xx dx
22
0
1 sin 2 cos 21
2 2 2 4
x x xx
21 4
0 02 2
2
0
1sin cosna x x nx dx
2
0
1sin 1 sin 1
2x n x n x dx
2
2 2
0
cos 1 sin 1 cos 1 sin 111 1
2 1 11 1
n x n x n x n xx x
n nn n
1 2 2
2 1 1n n
2
1 1 2, 1
1 1 1n
n n n
2 2
1
0 0
1 1sin cos sin 2
2a x x x dx x x dx
2
0
1 cos 2 sin 21
2 2 4
x xx
1 2
2 2
1
1
2a
22
2 1 1cos 2 cos sin
2 2 1f x x nx x
n
22
1 11 cos 2 cos sin .
2 1f x x nx x
n
Problem 18 Find the Fourier series for f(x) = x2 in x and deduce that
(c)2
2 2 2
1 1 1.............
1 2 3 6
.
(d)2
2 2 2
1 1 1.............
1 2 3 12
.
9
(e)2
2 2 2
1 1 1.............
1 3 5 8
.
Solution:
Given: 2f x x
2f x x f x
f x is an even function. Hence 0nb
1
cos2
nn
af x a nx
3 3 22
0 0
2 2 2 2.
3 3 3n
xa x dx
2 2
2
0 0
2 2 sin cos sincos 2 2 .n x
nx nx nxa x nx dx x x
n n n
2
2 120
n
n
2 2
4 1 4 1n n
n n
2
21
14 1 cos 1
3
nf x nx
n
Put x in (1)
2
21
14 1 1
3
n nf
n
x is a point of continuity.
2
2
2 2
1 12 4 ....
3 1 2
2
2
2 2
1 14 ....
3 1 2
2
2 2
2 1 14 ...
3 1 2
2
2 2
1 1.... 3
6 1 2
(ii) put x = 0, a point of continuity
2
21
10 4 1
3
n
n n
2
21
14 1
3
n
n n
2
21
11
3
n
n n
10
2
2 2 2
1 1 1.......
12 1 2 3
2
2 2 2
1 1 1
12 1 2 3
2
2 2 2
1 1 1........ 4
1 2 3 12
Add (3) and (4)2 2
2 2
1 1....
6 12 1 3
2
2 2
1 12 ....
4 1 3
2
2 2
1 1....
8 1 3
Problem 19 If f (x) = x + x2 in x . Hence show that
i.2
21
1
6n
ii.2
2 2 2
1 1 1.............
1 2 3 12
.
Solution:
Given: 2f x x x
2f x x x
f x is neither even nor odd
0
1
cos sin2
n n n
af x a x b x
2 3
20
1 1
2 3
x xa x x dx
2 3 2 3 3 21 1 2 2
2 3 2 3 3 3
21cosna x x nxdx
2
2 3
1 sin cos sin1 2 2
nx nx nxx x x
n n n
2 2
1 111 2 1 2
n n
n n
2 2
1 4 14
n n
n n
11
21sinnb x x nx dx
2
2 3
1 cos sin cos1 2 2
nx nx nxx x x
n n n
2 2
3 3
1 11 2 21 1
n nn n
nn n n n
2 2
2 3
11 2 2n
n n n
2
21 1
2
21 1
2
21
2 1
4 1 2 1cos sin
3
1 14 cos 2 sin 1
3
0 1
10 4 2
3
n
n
n
n
n
bn
f x nx nxn n
f x nx nxn n
Put x in
fn
Here 0 is a pt of continuity
2
2 2
2
2 2 3
2
2 2 3
0 0
1 12 0 4 ....
3 1 2
1 1 14 ...
3 1 2 3
1 1 1......
12 1 2 3
f
Hence (ii)Put x in (1)
2
21
14 1 3
3
nn
fn
x , is a pt of discontinuity
2 21
2 2
f ff x
2 212
2
2
2
21
13 4
3 n
2
2 2 3
2 1 1 1...
3 4 1 2 3
12
2
2 2
1 1...
6 1 2
Hence (i)
Problem 20 a. Expand 2f x x x as a Fourier series in 0, 2 .
Solution:We know that
0
1
cos sin2
n n
af x a nx b nx
22 3 3 2
23 2
0 00
1 1 1 82 2 4 8 4
2 3 3 3
x xa x x dx
2
0
4
3a s
2
2
0
12 cosna x x nxdx
2
2
2 2
0
1 sin cos sin2 2 2 2
nx nx nxx x x
n n n
2 2 2
1 2 2 4
n n n
2
2
0
12 sinnb x x nx dx
2
2
2 3
0
1 cos sin cos2 2 2 2
nx nx nxx x x
n n n
3 3
1 2 20
n n
2
21
2 4cos .
3f x nx
n
b. Find the Fourier series for f (x) = ex defined in , .
Solution:
1
cos sin2n
n n
af x a nx b nx
0
1 xa e dx
1 xe
1e e
1cosx
na e nx dx
13
2
1cos sin
1
xenx n nx
n
2 2
1cos cos
1 1
e en n
n n
2
1
1
n
e en
1sinx
nb e nx dx
2
1sin cos
1
enx n x
n
2 2
11 cos
1 1
ne en n nx
n n
2 2
111
1 1
nnn e e
nn n
2
1
1
n
nb n e en
2 21
1 11cos sin .
2 1 1
n nn
f x e e e e nx e e nxn n
Problem 21 Obtain the Fourier series expansion of f (x) where
21 , 0
21 0
xx
f xx
x
and hence deduce that2
2 2 2
1 1 1.............
1 3 5 8
.
Solution:
2 2
1 1x x
f x f x
0 0 .x x The given function is an even function.Hence 0nb
0
1
cosn
af x a nx
0
0
2 21
xa dx
2
0
2 20
2
xx
14
0
2 21 cosn
xa nx dx
2
0
2 2 sin 2 cos1
x nx nx
n n
2 2
2 12 2n
n n
2 2
41 1
n
n
0na if n is even
2 2
8na
n is n is odd.
2 21,3,5
8cosf x nx
n
2 2 2
8 cos cos3...
1 3
x x
2 2 2
8 cos cos30 ...
1 3
x xf
0Putx
2 2 2
8 1 10 ... 1
1 3f
0 is a pt of continuity 0 1f
2 2 2
8 1 11 1 .....
1 3
2
2 2 2
1 1 1....
8 1 3 5
Problem 22 Obtain the Fourier series to represent the function f(x) = | x | is x and
deduce that2
2 2 2
1 1 1.............
1 3 5 8
.
Solution:
Given f x x
f x f x
The given function is an even function.Hence 0nb
0
1
cos2
n
af x a nx
0
2na x dx
15
2
0
2
2
x
0
2cosna x nxdx
2 2 2 20
2 sin cos 2 cos 1 21 1
nnx nx nx
n n n n n
0na if n is even
2
4na
n if n is odd
21,3
4cos
2f x nx
n
2
4 cos3cos ...
2 3
xx
0Put x
2 2
4 1 10 1 ...
2 3 5f
Here 0 is a pt of continuity
0 0f
2 2
4 1 10 1 ...
2 3 5
2 2
4 1 11 ...
2 3 5
2
2 2
1 11 ... .
8 3 5
Problem 23 Find the Fourier series expansion of period 2 for the function f(x) = 2
x in
the range 0, 2 . Deduce the sum of the series
21
1
n.
Solution: The Fourier series of f x in 0,2 is given
0
1
cos sin2
n n
a n x n xf x a b
l l
232 22 2
0
0 00
1 1 2
3 3
ll l l x
a f x dx l x dx ll l l
2
2
0
1cos
l
n
n xa l x dx
l l
16
2
2
2 2 3 3
2 30
sin cos sin1
2 1 2
l
n x n x n x
l l ll x l xn n nll l l
2 2 2 2
2 2
1 2 cos 2 2l n l
n nl
l l
2
2 2
4n
la
n
2
2
0
1sin
l
n
n xb l x dx
l l
2
2
2 2 3 3
2 3
0
sin cos1
cos 2 1 2
l
n x n x n x
l l ll x l xn n nll l l
22
3 3 3 3
3 3
1 cos 2 2cos 2 2n n ll
n nn nll ll l
2 210
l l
n nll l
2 2
2 21
4cos 1
3
l l n xf x
n l
0x in (1)
2 2
2 21
40 2
3
lf
n
Here 0 is a Pt of discontinuity
0 2
02
f f lf
2 2 21
2l l l
2 2
2
2 21
42
3
l ll
n
2 2
2
2 21
4 1
3
l ll
n
2 2
2 2 2
2 4 1 1...
3 1 2
l l
17
2
2 2 2
1 1 1... .
6 1 2 3
Problem 24 a. Find the Fourier expansion of f x if
0 , 2 1
1 , 1 0( )
1 , 0 1
0 , 1 2
x
x xf x
x x
x
.
Solution:
0
1
cos sin2
n n
a n x n xf x a b
l l
2
0
2
1
2a f x dx
1 0 1 2
2 1 0 1
10 1 1 0
2dx x dx x dx dx
0 12 2
1 0
1
2 2 2
x xx x
1 1 10 1 1 0
2 2 2
0 1
1 0
11 cos 1 cos
2 2 2n
n x n xa x dx x dx
0 1
2 2 2 2
1 0
sin cos sin cos1 2 2 2 21 1 1 12
2 24 4
n x n x n x n x
x xn nn n
0
0 1
1 0
11 sin 1 sin
2 2 2n
n x n xb x dx x dx
0 1
2 2 2 2
1 0
cos sin cos sin1 2 2 2 2
1 1 1 12
2 24 4
n x n x n x n x
x xn nn n
2 2 2 2
1 4 4 10 sin sin 0
2 2 2
n n l
n n n n
1
1sin .
2
n xf x
n
18
b. Find the Fourier series for f (x) where 0, 1 0
1, 0 1
xf x
x
.
Solution:
0
1
cos sin2
n n
a x n xf x a b
l l
0
1a f x dx
l
0 1
1 0
0 dx dx
1
01x
0 1
1 0
10 cos
1na dx n x dx
1
0
sinn x
n
0sin sin 0
n n
n n
1
0
11sin
1nb n x dx
1
0
1 1cos cos 1n
n n
n n n n
1,3
1 2sin .
2f x n x
n
Problem 25 Find the half – range cosine series for f (x) = (x – 1)2 in (0, 1). Hence show that2
2 2 2
1 1 1...
1 2 3 6
.
Solution:Here 1l
0
1
cos2
n
af x a n x
1212
0
00
12 1 2
3
xa x dx
0
2
3a
1
2
0
2 1 cosna x n xdx
1
2
2 2 3 3
0
sin cos sin2 1 2 1 2
n x n x n xx x
n n n
19
2 2 2 2
2 42
n n
2 21
1 4cos
3f x n x
n
2 21
1 4 1cos 1
3f x n x
n
0 1Put x in
2 21
1 4 10 2
3f
n
Here 0 is o pt of discontinuity
0 0
0 12
f ff
2 21
1 4 11
3 n
2 2 2
2 4 1 1...
3 1 2
2
2 2
1 1.......... .
1 2 6
Problem 26 a. Express
1, 02
1,2
ax
f xa
x a
as a cosine series
Solution: 0
1
cos2
n
a n xf x a
a
2
0
0
2
2 21 0 0
2 2
a
a
a
a aa dx dx a
a a
2
0
2
2cos cos
a
a
n
a
nx nxa dx dx
a a a
2
02
sin sin2
aa
a
n x n x
a an naa a
2 4sin sin sin
2 2 2
a n a n n
a n n n
1
4sin cos .
2
n nf x
n a
20
b. Express f x as a Fourier sine series where
1 1, 0,
4 2
3 1, ,1
4 2
x
f x
x
.
Solution:
We know that 1
sinn
n xf x b
l
1
12
10
2
2 1 3sin sin
1 4 4nb x n x dx x n x dx
11
2
2 2 2 21
02
1 cos sin 3 cos sin2 1 (1)
4 4n
n x n x n x n xb x x
n n n n
2 2
2 2 2 2
cos sin1 1 1 12 22 04 2 4
cos sin3 cos sin 1 3 2 22 14 2 4
n n
n n n
nn
n n
n n n n
2 2
sin12
24
n
n n
2 2
sin1 cos 22 04
nn
n n
2 2
4sin 1122 2
nn
n n n
2 2
4sin12 isn
n
b if n oddn n
= 0 if even
2 21/ 3
4sin12 sin .
n
f x n xn n
Problem 27 a. Find the Fourier cosine series for x x in 0 x .
21
Solution:
0
1
cos2
n
af x a nx
0
0
2a x x dx
2 2
0
2
2 3
x x
3 3 22
2 3 3
0
2cosna x x nx dx
2 3
0
2 sin cos sin2 2
nx nx nxx x x
n n n
2 2
12n
n n
2
21 1
n
n
2
4na
n If n is even
0na If n is odd.
2
22,4
4cos .
6f x nx
n
b. Prove that complex form of the Fourier series of the function , 1 1xf x e x is
2 2
11 sin 1.
1
n in xinf x h e
n
.
Solution:Here 2 2, 1l l
in xnf x C e
1
1
1
2x in x
nC e e dx
1
1
1
1
2
in xe dx
11
1
1
1
2
in x
in
e
1 11
2 1
in in xe ein
22
1 1
2 2
1cos sin cos sin
2 1
ine n i n e n i n
n
1 1
2 2
1cos
2 1
inn e e
n
1
2 2
11
2 1
nine e
n
2 2
11 sinh 1
2 1
nin
n
2 2
11 sinh 1 .
2 1
n in xinf x e
n
Problem 28 Find the cosine series for f (x) = x in (0, ) and then using Parseval’s theorem,
show that4
4 4
1 1....
1 3 96
.
Solution:
0
1
cos2
n
af x a nx
0
0
2a f x dx
0
2x dx
2
0
2
2
x
0
2cosna x nx dx
2
0
2 sin cos1
x nx nx
n n
2 20
2 cos 10
n
n n
2
2 cos 1n
n
2
21 1
n
n
2
4na
n
if n is odd
0na if n is even
2
21,3
4cos
2f x nx
n
23
By Parseval’s theorem
2 2
20
0
1 1
4 2n
af x dx a
222
21,30
1 1 4
4 2x dx
n
2 2
2 41,30
1 1 16
3 4 2
x
n
3 2
2 4 4
1 8 1 1....
3 4 1 3
2 2
2 4 4
8 1 1....
3 4 1 3
2 2
4 4
1 1.....
12 8 1 3
4
4 4
1 1.....
96 1 3
Problem 29 a. Find the complex form of Fourier series of f x if
sinf x ax in x .
Solution: in xnf x C e dx
1sin
2inx
nC ax e dx
2 2
1sin cos
2
inxein ax a ax
a n
2 2
1sin cos sin cos
2
in ine in a a a e in a a aa n
2 2
1sin 2cos cos 2isin
2in a n a a n
a n
1
2 2 2 2
2 1 sin 1 sin
2
n nin a in a
a n a n
1
2 2
1sin.
n
inxinaf x e
a n
b. Find the first two harmonic of the Fourier series of f (x) given by
x 0 1 2 3 4 5f (x) 9 18 24 28 26 20
24
Solution:Here the length of the in level is 2 6, 3l l
01 1 2 2
2 2cos sin cos sin
2 3 3 3 3
a x x x xf x a b a b
x
3
x 2
3
x ycos
3
xy
sin
3
xy
2cos
3
xy
2sin
3
xy
0 0 0 9 9 0 9 01
3
2
3
18 9 15.7 -9 15.6
2 2
3
4
3
24 -12 20.9 -12 -20.8
3 2 28 -28 0 28 04 4
3
8
3
26 -13 -22.6 -13 22.6
5 5
3
10
3
20 10 17.6 -10 -17.4
125 -25 -3.4 -7 0
0
2 1252 41.66
6 6
ya
1
1
2cos 8.33
6 3
2sin 1.15
6 3
xa y
xb y
2
2 2cos 2.33
6 3
xa y
2
2 2sin 0
6 3
xb y
41.66 2
8.33cos 2.33cos 1.15sin .2 3 3 3
x x xf x
Problem 30 a. Find the first two harmonic of the Fourier series of f (x). Given by
x 0
3
2
3
4
3
5
3
2
f (x) 1 1.4 1.9 1.7 1.5 1.2 1.0
Solution: The last value of y is a repetition of the first; only the first six values will be used
The values of cos , cos 2 , sin , sin 2y x y x y x y x as tabulated
25
x f x cos x sin x cos 2x sin 2x
0 1.0 1 0 1 0
3
1.4 0.5 0.866 -0.5 0.866
2
3
1.9 -0.5 0.866 -0.5 0.866
1.7 -1 0 1 0
4
3
1.5 -0.5 -0.866 -0.5 -0.866
5
3
1.2 0.5 -0.866 -0.5 -0.866
0 2 2.96
ya
1
cos2 0.37
6
y xa
2
cos 22 0.1
6
y xa
1
sin2 0.17
6
y xb
2
sin 22 0.06
6
y xb
b. Find the first harmonic of Fourier series of f x given by
x 0
6
T
3
T
2
T 2
3
T 5
6
T T
f (x) 1.98 1.30 1.05 1.30 -0.88 -0.35 1.98
Solution:
First and last valve are same Hence we omit the last valve
x 2 x
T
y cos sin cosy siny
0 0 1.98 1.0 0 1.98 0
6
T
3
1.30 0.5 0.866 0.65 1.1258
3
T 2
3
1.05 -0.5 0.866 -0.525 0.9093
2
T 1.30 -1 0 -1.3 0
26
2
3
T 4
3
-0.88 -0.5 -0.866 0.44 0.762
5
6
T 5
3
-0.25 0.5 -0.866 -0.125 0.2165
4.6 1.12 3.013
0
2 4.61.5
6 3a y
1
2 cos 21.12 0.37
6 6
ya
1
23.013 1.005
6b
0.75 0.37 cos 1.005sinf x