UNIT – I FOURIER SERIES PROBLEM 1 - … – I FOURIER SERIES ... Expand f(x) = cosx in a Fourier series in the interval ... Find the complex form of the Fourier series of f(x) =

UNIT – I

FOURIER SERIES

PROBLEM 1:

The turning moment T lb feet on the crankshaft of steam engine is given for a series of values of the crank angle degrees

0 30 60 90 120 150 180T 0 5224 8097 7850 5499 2626 0

Expand T in a series of sines.

Solution:

Let T = b1 sin + b2 sin2 + b3 sin3 + ………,Since the first and last values of T are repeated neglect last one

T T sin T sin2 T sin3

0 0 0 0 030 5224 2612 4524.11 522460 8097 7012.2 7012.2 090 7850 7850 0 -7850120 5499 4762.27 - 4762.27 0150 2624 1313 - 2274.18 2626

Total 23549.47 4499.86 0

b1 = 2 [mean value of T sin ]

= 2[23549.47

6] = 7849.8

b2 = 2 [mean value of T sin2 ]

= 2[6

86.4499] = 1499.95

b3 = 2 [mean value of T sin2 ] = 0

f ( x ) = 7849.8 sin + 1499.95 sin2

PROBLEM 2:

Analyze harmonically the given below and express y in Fourier series upto the third harmonic.

x 0

3

3

2

3

43

5 2

y 1.0 1.4 1.9 1.7 1.5 1.2 1Solution:

Since the last value of y is a repetition of the first, only the six values will be used. The length of the interval is 2.

Let y = 2

a0+ (a1 cosx + b1 sinx) + (a2 cos2x + b2 sin2x) + (a3 cos3x + b3 sin3x) + … (1)

x y cosx sinx cos2x sin2x cos3x sin3x

0 1.0 1 0 1 0 1 0

3

1.4 0.5 0.866 - 0.5 0.866 -1 0

3

2 1.9 - 0.5 0.866 - 0.5 - 0.866 1 0

1.7 -1 0 1 0 -1 0

3

4 1.5 - 0.5 - 0.866 - 0.5 0.866 1 0

3

5 1.2 0.5 - 0.866 - 0.5 - 0.866 -1 0

a0 = 2 [mean value of y]

= 6

2(8.7) = 2.9

a1 = 2 [mean value of y cosx]

= 6

2[1 (1.0) + 0.5 (1.4) – 0.5 (1.9) – 1 (1.7) – 0.5 (1.5) + 0.5 (1.2)]

= - 0.37b1 = 2 [mean value of y sinx]

= 6

2[0.866 (1.4 + 1.9 -1.5 – 1.2)]

= 0.17a2 = 2 [mean value of y cos2x]

= 6

2[1(1.0 + 1.7) – 0.5 (1.4 + 1.9 + 1.5 + 1.2)]

= - 0.1b2 = 2 [mean value of y sin2x]

= 6

2[0.866(1.4 – 1.9 + 1.5 – 1.2)]

= - 0.06a3 = 2 [mean value of y cos3x]

= 6

2[1.0 – 1.4 + 1.9 – 1.7 + 1.5 – 1.2]

= 0.03b3 = 2 [mean value of y sin3x] = 0

Hence y = 1.45 + (-0.37 cosx + 0.17 sinx) – (0.1 cos2x + 0.06 sin2x) + 0.03 cos3x.

PROBLEM 3:

Find the Fourier series expansion for the function f(x) = x sinx in 0 < x < 2 and deduce

that 4

2.........

7.5

1

5.3

1

3.1

1

SOLUTION:

The Fourier series is

0n n

n 1

af (x) (a cosnx b sin nx)..................(1)

2

Given f(x) = x sinx

a0 = 12

0

dx)x(f

= 12

0

xdxsinx

= 1 2

0 sinx)](-(1)- cosx)(-[x

= 1

(-2) = -2

an = 12

0

nxdxcos)x(f

= 12

0

nxdxcosxsinx

= 2

1

2

0

dx]x)n1sin(x)n1sin(x[

)]BA(Sin)BA(SinSinACosB2[ 2

2 20

1 cos(1 n)x cos(1 n)x sin(1 n)x sin(1 n)xx( ) ( )

2 1 n 1 n ( 1 n) (1 n)

= 1 cos 2(1 n) cos 2(1 n)[ 2 ( )]

2 1 n 1 n

(

1 1[ ]1 n 1 n

21 n 1 n

[ ]1-n

= n 22

a where n 1n 1

a1 = 12

0

xdxcos)x(f

= 12

0

xdxcosxsinx

= 2

12

0

xdx2sinx

= 2

1

2

0

)4

x2sin)(1()

2

x2cos(x

= 2

1[2 (-

2

4cos ) = -

2

1

)2

4cos(2[

2

1

bn = 2

0

1f (x)sin nxdx

= 2

0

1x sin xsin nxdx

= 2

0

1x[cos(1 n)x cos(1 n)x]dx

2

)]BA(Cos)BA(CosSinASinB2[

=

2

2 20

1 sin(1 n)x sin(1 n)x cos(1 n)x cos(1 n)xx( ) ( )

2 1 n 1 n (1 n) (1 n)

= 2 2 2 2

1 cos2(1 n) cos2(1 n) cos0 cos0

2 (1 n) (1 n) (1 n) (1 n)

= 2 2 2 2

1 1 1 1 1

2 (1 n) (1 n) (1 n) (1 n)

= 0 where n 1

b1 = 2

0

1f (x)sin xdx

= 2

0

1x sin x sin xdx

= 2

2

0

1x sin xdx

= 2

0

1 1 cos2xx( )dx

2

= 2

0

1x(1 cos2x)dx

2

22

0

1 sin 2x x cos2xx(x ) (1)( )

2 2 2 4

= 21 1 1[2 ]

2 4 4

=

Substitute in (1), we get

f (x) = 2

n 2

1 21 cos x cosnx sin x .......(2)

2 n 1

+ ……………….(2)

Deduction Part:

Put x [x isa point of continuity]2

[x=

2

is a point of continuity]

f (2

) = f ( ) sin

2 2 2 2

=

2

( sin

2

= 1)

(2) 2

n 2

2 n(2) 1 0 cos

2 2n 1

= -1-0+

2n 2

1 n1 2 cos

2 2n 1

- +1 =

-n 2

1 n1 2 cos

2 (n 1)(n 1) 2

+1 =

-n 2

1 1 ncos

4 2 (n 1)(n 1) 2

=

= n 2

1 n 2cos

(n 1)(n 1) 2 4

4

2........)1(

5.3

1)0(

4.2

1)1(

3.1

1

]4

2[........]

7.5

1

5.3

1

3.1

1)[1(

]4

2[........

7.5

1

5.3

1

3.1

1

Problem:4

Find the Fourier series for f(x) = xsin in - < x < Solution:

Given f(x) = xsin

This is an even function. bn = 0

f(x) = )1.......(..............................Cosnxa2

a

1nn

0

a0 =

0

dx)x(f2

=

0

dxSinx2

=

0

xdxsin2

= 0]xcos[2

= -2

(cos cos0)

= -2

[ 1 1]

[-1-1]

= 4

an =

0

nxdxcos)x(f2

=

0

nxdxcosSinx2

=

0

nxdxcosxsin2

=

0

dx]x)1nsin(x)1n[sin(2

12

=

0]1n

x)1ncos(

1n

x)1ncos([

1

= ]1n

1

1n

1

1n

)1ncos(

1n

)1ncos([

1

= ]1n

1

n1

1

1n

)1(

1n

)1([

1 1n1n

= ]1n

1

1n

1

1n

)1(

1n

)1([

1 nn

= ]1n

1)1(

1n

1)1([

1 nn

= n( 1) 1 1 1

[ ]n 1 n 1

= n

2

( 1) 1 n 1 n 1[ ]

n 1

= n

2

( 1) 1 2[ ]n 1

=

)1n(

]1)1[(22

n

if n 1.

a1 =

0

xdxcos)x(f2

=

0

xdxcosSinx2

=

0

xdxcosxsin2

=

0

xdx2sin2

12

= 0]2

x2cos[

1

= ]2

1

2

1[

1

= 0

Substitute in equation (1), we get

f(x) = nxcos)1n(

]1)1[(20

2

2n2

n

Hence f(x) = nxcos)1n(

]1)1[(22

2n2

n

Problem:5

Expand f(x) = xcos in a Fourier series in the interval (-,)

Solution:

Given f(x) = xcos .

This is an even function, bn = 0

f(x) = )1.......(..............................Cosnxa2

a

1nn

0

a0 =

0

dx)x(f2

=

0

dxxcos2

= /2

0 /2

2[ cos xdx (cosx)dx]

= /2

0 /2

2[ cos xdx cos xdx]

= /20 /2

2[(sin x) (sin x) ]

= 2

[(1 0) (0 1)]

a0 = 4

an =

0

nxdxcos)x(f2

=

0

nxdxcosxcos2

= /2

0 /2

2[ cos x cosnxdx (cos x)cosnxdx]

= /2

0 /2

2 2cos x cosnxdx cos x cosnxdx

= /2

0 /2

1 1[cos(n 1)x cos(n 1)x]dx [cos(n 1)x cos(n 1)x]dx

-

= /20 /2

1 sin(n 1)x sin(n 1)x 1 sin(n 1)x sin(n 1)x[ ] [ ]

n 1 n 1 n 1 n 1

= 1 sin(n 1) / 2 sin(n 1) / 2 1 sin(n 1) / 2 sin(n 1) / 2

[ ] [ ]n 1 n 1 n 1 n 1

= ]1n

2/)1nsin(

1n

2/)1nsin([

2

= ]1n

)2/2/sin(

1n

)2/n2/sin([

2

= ]1n2

ncos

1n2

ncos

[2

= ]1n

1

1n

1[

2

ncos

2

= ]1n

1n1n[

2

ncos

22

= ]1n

2[

2

ncos

22

= ]1n

1[

2

ncos

42

= ]1n

1[

42

if n is even

= 0 if n is odd [n 1]Hence

f(x) = 2

n 2

2 4cosnx

(n 1)

Complex Form of a Fourier series:

Problem 6:

Find the complex form of the Fourier series of the function f(x) = ex when - < x < and f(x+2) = f(x).

Solution:

We know that f(x) =

n

inxneC -----------------------------(1)

Cn =

inxe)x(f

2

1dx

=

inxxee

2

1dx

=

dx(e

2

1 x)in1(

=

in1

e

2

1 x)in1(

=

x)in1(e

)in1(2

1

= ]ee[)in1(2

1 )in1()in1(

= ]eeee[)in1(2

1 inin

nnin )1(0i)1(nsinincose

nnin )1(0i)1(nsinincose

= ])1(e)1(e[)in1(2

1 nn

= ]2

ee[

)in1(2

)1( n

Cn =

sinh)n1(

)in1()1(22

n

(1) f(x) =

n

sinh)n1(

)in1()1(22

n

einx

i.e., ex =

n

inx22

n en1

in1)1(

sinh

Problem7:Find the complex form of the Fourier series of f(x) = Cosax in ( -, ) where ‘a’ is neither zero nor an integer.

Solution:

Here 2c = 2 or c = .

We know that f(x) =

n

inxneC -----------------------------(1)

Cn =

inxe)x(f

2

1dx

= dxCosaxe2

1 inx

=

)]axsinaaxcosin(na

e[

2

122

inx

= )asinaacosin(e)asinaacosin(e[)na(2

1 inin22

= )]ee(asina)ee(acosin[)na(2

1 inininin22

= )]ncos2(asina)nsini2(acosin[)na(2

122

Cn =

asina2)1()na(2

1 n22

Hence (1) becomes

Cosax = inx

n22

n

e)na(

)1(

2

asina

Problem8:

Find the complex form of the Fourier series of f(x) = e-x in -1 ≤ x ≤ 1

Solution:

We know that f(x) =

n

xinneC -----------------------------(1)

Cn = dxe)x(f2

1 1

1

xin

= dxee2

1 1

1

xinx

= dxe2

1 1

1

x)in1(

=

]

)in1(

e[

2

1 x)in1(

= )in1(2

ee )in1(in1

= )in1(2

)nsinin(cose)nsinin(cose 1

= )in1(2

)1(e)1(e n1n

=

= 2

)1)(ee( n1

22n1

)in1(

= 1sinhn1

)in1()1(22

n

Hence (1) becomes e-x = xin22

n

n

e1sinhn1

)in1()1(

UNIT- 2

FOURIER TRANSFORMS

1. Find the Fourier transform of xae , if 0a . Deduce that

0

3222 4)( aax

dx

Sol:

The Fourier transform of the function )(xf is

dxexfxfF isx)(2

1)(

.

)(

2

cos22

1

0cos22

1

sincos2

1

)sin(cos2

1

2

1)(

22

0

0

as

a

sxdxe

sxdxe

sxdxeisxdxe

dxsxisxe

dxeexfF

ax

xa

xaxa

xa

isxxa

By Parseval’s identity

dxxfdssF22

)()(

22

a x

2 2

2 2ax2ax

2 2 20 0

2 ads e dx

(s a )

2a 1 eds 2 e dx 2

(s a ) 2a

10 1

a

2 2 2 30

2 2 2 30

12 ds

(s a ) 2a

dx

(x a ) 4a

2.Show that the Fourier transform of

0;0

;)(

22

ax

axxaxf

is 3

2 sinsa sacossa2

s

. Hence deduce that4

cossin

03

dtt

ttt.

Sol:


dxexfxfF isx)(2

1)(

.

Given 22)( xaxf in axa and 0 otherwise.

3

32

032

22

0

22

2222

22

22

cossin22)(

000sin2cos2

02

sin)2(

cos)2(

sin)(

2

0cos)(22

1

sin)(cos)(2

1

)sin)(cos(2

1

)(2

1)(

s

sasasasF

s

sa

s

saa

s

sx

s

sxx

s

sxxa

sxdxxa

sxdxxaisxdxxa

dxsxisxxa

dxexaxfF

a

a

a

a

a

a

a

a

a

a

isx

By inverse Fourier transform

dsesFxf isx)(2

1)(

03

33

3

3

0coscossin

22

)(

sincossin

coscossin2

)sin(coscossin2

cossin22

2

1)(

sxdss

sasasaxf

sxdss

sasasaisxds

s

sasasa

dssxisxs

sasasa

dses

sasasaxf isx

Put 1&0 ax , we get

4

cossin.

cossin41

03

03

dss

sssei

dss

sss

Replace s by t , we get4

cossin

03

dtt

ttt

3.Find the Fourier transform of

1;0

1;1)(

xif

xifxxf . Hence deduce that

3

sin4

0

dtt

tand dt

t

t2

0

sin

.

Sol:


dxexfxfF isx)(2

1)(

.

Given xxf 1)( in 11 x and 0 otherwise.

2

22

1

02

1

0

1

1

1

1

1

1

1

1

cos12)(

10

cos0

2

cos)1(

sin)1(

2

0cos)1(22

1

sin)1(cos)1(2

1

)sin)(cos1(2

1

)1(2

1)(

s

ssF

ss

s

s

sx

s

sxx

sxdxx

sxdxxisxdxx

dxsxisxx

dxexxfF isx

By inverse Fourier transform

dsesFxf isx)(2

1)(

02

22

2

2

0coscos1

21

)(

sincos1

coscos11

)sin(coscos11

cos12

2

1)(

sxdss

sxf

sxdss

sisxds

s

s

dssxisxs

s

dses

sxf isx

Put 0x , we get

1.

22sin2

.

cos121

02

2

02

dss

sei

dss

s

Let ,2

ts then dtds 2

ts

ts 00

2

sin

22

2

sin2

02

2

02

2

dtt

t

dtt

t

By Parseval’s identity

dxxfdssF22

)()(

2 12

21

2 12

20

2 12

20

2 1 cossds 1 x dx

s

2 1 cossds 2 1 x dx

s

2 1 cossds 2 1 x dx

s

2 1

22

0 0

2 1 coss2 ds 2 1 x 2x dx

s

12 3 2

20 0

2 1 coss x 2xds x

s 3 2

0

2

2

2

0

2

2

32sin2

2

00013

11

cos12

dss

s

dss

s

Let ,2

ts then dtds 2

ts

ts 00

3

sin

3

sin

32

2

sin22

0

4

0

2

2

2

0

2

2

2

dtt

t

dtt

t

dtt

t

4.Find the Fourier transform of22xae . Hence prove that 2

2x

e

is self reciprocal.

Sol:


dxexfxfF isx)(2

1)(

.

dxee

dxe

dxe

dxe

dxeeeF

a

isax

a

s

a

s

a

isax

isxxa

isxxa

isxxaxa

2

2

2

2

22

22

22

2222

24

42

)(

)(

2

1

2

1

2

1

2

1

2

1

Put ya

isax

2, then

a

dydx

yx

yx

2

2

22

2

2

22

2

4

4

4

2

1

2

1

2

1

a

sxa

a

s

ya

s

ea

eF

ea

a

dyee

Put2

1a , then

22

2

2

22

22

2

14

2

1

22

11

sx

s

x

eeF

eeF

2

2x

e

is self reciprocal with respect to Fourier transform.

5. Find Fourier sine and cosine transform of 1nx and hence prove that

C

1 1F

x s

Sol:

We know that

0

1n

nnax

adxxe

Put isa ,

0

1n

nnisx

isdxxe

nn 1n n

0

nn 1 n 1n n

0 0

nn 1 n 1n

0 0 n

cossx isin sx x dxi s

x cossxdx x isin sxdxi s

x cossxdx i x sin sxdx

cos i sin s2 2

n

nn 1 n 1

n0 0

cos i sin2 2

x cossxdx i x sin sxdxs

nn 1 n 1

n0 0

n nn 1 n 1

n n0 0

n ncos isin

2 2x cossxdx i x sin sxdx

s

n ncos i sin

2 2x cossxdx i x sin sxdx

s s

Equating Real and imaginary parts, we get

n

nn

n

nn

s

n

sxdxx

s

n

sxdxx

2sin

sin

2cos

cos

0

1

0

1

n

nn

S

n

nn

C

n

nn

n

nn

s

n

xF

s

n

xF

s

n

sxdxx

s

n

sxdxx

2sin

2

2cos

2

2sin

2sin

2

2cos

2cos

2

1

1

0

1

0

1

Put2

1n in the above results, we get

1

211

2C 1

2

12c o s2

2F x

s

1

2C

cos42

F xs

C

C

11 2 2Fx s

1 1F

x s

Thereforex

1is self reciprocal with respect to Fourier cosine transform.

5. Find Fourier sine transform and Fourier cosine transform of 0, ae xa . Hence

evaluate

0222

2

)(dx

xa

xand

02222 )()( bxax

dx.

Sol:

The Fourier sine transform of the function )(xf is

0

sin)(2

)( sxdxxfxfFS .

22

0

2

sin2

as

s

sxdxeeF axaxS

The Fourier cosine transform of the function )(xf is

0

cos)(2

)( sxdxxfxfFC .

22

0

2

cos2

as

a

sxdxeeF axaxC

By parseval’s identity

0

2

0

2dxxfdsxfFS

0

2

0

2

22

0

2

0

2

22

2

2

dxedsas

s

dxedsas

s

ax

ax

ads

as

s

ads

as

s

a

eeds

as

s

a

eds

as

s

dxedsas

s

aa

ax

ax

4

2

12

2

2

2

2

2

0222

2

0222

2

022

0222

2

0

2

0222

2

0

2

0222

2

Replace s by x , we get

adx

ax

x

40222

2

Let axexf and bxexg

By parseval’s identity

00

dxxgxfdsxgFxfF CS

002222

002222

2..

22

dxedsasas

abei

dxeedsas

b

as

a

xba

bxax

baabasas

ds

baasas

dsab

ba

ee

asas

dsab

ba

e

asas

dsab

baba

xba

2

12

2

2

02222

02222

0

02222

002222

Replace s by x , we get

baabaxax

dx

202222

6. Verify Parseval’s theorem of Fourier transform for the function

0;

0;0)(

xe

xxf

x.

Sol:

By Parseval’s theorem

dssFdsxf22

…………(1)

Given

0;

0;0)(

xe

xxf

x

0

1

0

1

0

0

12

1

2

1

2

1

2

1

is

e

dxe

dxe

dxeesF

xis

xis

isxx

isxx

1

1

2

1

11

1

2

1

1

1

2

1

1

1

1

0

2

1

112

1

2

011

s

issF

isis

is

is

isis

is

e

is

esF

isis

L. H. S. of (1)

22 x

0

2x

0

f x dx e dx

e dx

2x

0

2 0

e

2

e e

2

1

2

R. H. S. of (1)

s

s

ds

dss

s

dss

is

dss

is

dss

isdssF

1

2

22

2

22

2

2

2

2

2

2

tan2

1

12

1

1

1

2

1

1

1

2

1

1

1

2

1

1

1

2

1

2

1

2

1

222

1

tantan2

1

tantan2

1

11

112

dssF

L. H. S of (1) = R. H. S of (1)Hence Parseval’s theorem verified.

7. Solve for )(xf from the integral equation

0

cos)( exdxxf .

Sol:

0

cos)( exdxxf

edxxe2

cos22

0

Inverse Fourier cosine transform is

2

2

2

02

0

1

2

1

12

)1(1

10

2

)sincos.1(1

2

cos22

)(

x

x

x

xxxx

e

dxexf

UNIT-3

PARTIAL DIFFERENTIAL EQUATIONS

Problems on Lagrange’s equations

1. Solve p x q y z

Solution:

p x q y z ------ ( 1 )

This is of the form Pp + Qq = R

Here , , P x Q y R z

Subsidiary equations are

dx dy dz

P Q R

dx dy dz

x y z

Grouping the first two members

dx dy

x y

Integrating

1 1

2 2

1 1

2 2

1 1

2 2

2( )

2

2

dx dx

x y

x dx y dy

x y a

x y a

ax y

ax y c c

u x y

|||ly Grouping the another two members

dy dz

y z

Integrating

2 2

2 2

2( )

2

2

y z b

y z b

by z

by z c c

v y z

The general solution of (1) is (u,v) = 0

, 0 x y y z

2. Obtain the general solution of pzx + qzy = xy.

Solution:

Given pzx + qzy = xy


Here P = zx, Q= zy, R = xy


dx dy dz

P Q R

dx dy dz

zx zy xy

Grouping the first two members

dx dy

zx zy

Integrating

(1)

dx dy

x y

og x og y og a

xog og a

y

xa

y

xu

y

Grouping the another two members

( (1) )

dy dz

zy xy

dy dz

z xdy dz

from x ayz ay

ay dy z dz

Integrating

2 2

2 2

2 2

2 2

2

2

2 2

2

( 2 )

.

ay zc

ay z c

ay z b b c

xy z b

y

xy z b

v xy z

The general solution is (u,v) = 0

2, 0

xxy z

y

3. Solve 2 2 2 2 2 2( ) ( ) ( ) 0 x y z p y z x q z x y

Solution:

Given 2 2 2 2 2 2( ) ( ) ( ) 0 x y z p y z x q z x y


Here 2 2( ) P x y z , 2 2( ) Q y z x , 2 2( ) R z x y


2 2 2 2 2 2(1)

( ) ( ) ( )

dx dy dz

P Q R

dx dy dz

x y z y z x z x y

Choose the set of multipliers x,y,z

0dzzdyydxx

0

dzzdyydxx

)yx(z)xz(y)zy(x

dzzdyydxx

)yx(z

dz

)xz(y

dy

)zy(x

dx222222222222

Integrating we get

2 2 2

2 2 2

2 2 2

2 2 2

( 2 )

x y zc

x y z a et c a

u x y z

|||ly consider another set of multipliers z

1,

y

1,

x

1

Each member of (1)

0z

dz

y

dy

x

dx0

z

dz

y

dy

x

dx

yxxzzy

z

dz

y

dy

x

dx

222222

Integrating we get

log x + log y + log z = log b

log(xyz) = log b

xyz = b

v = xyz


2 2 2, 0 x y z x y z

4. Solve 2 2 2(x y z ) p ( y z x ) q z x y

Solution:

Given 2 2 2(x y z ) p ( y z x ) q z x y


Here 2 P x y z , 2 Q y z x , 2 R z x y


2 2 2(1)

) )

dx dy dz

P Q R

dx dy dz

x y z y z x z x y

Each of (1)

2 2 2 2 2 2

d x d y d y d z d x d z

x y z (x y ) y z x (y z ) x z y (x z )

d ( x y ) d ( y z ) d ( x z )i.e.,

( x y) ( x y z ) ( y z ) ( x y z ) ( x z) ( x y z )

d ( x y ) d ( y z ) d ( x z )i.e.,

x y y z x z

Grouping the members

d ( x y ) d ( y z )

( x y) ( y z )

Integrating we get

1

1

1

log ( x y ) log ( y z) log c

x ylog log c

y z

x yc

y z

x yu

y z

Choose multipliers x , y , z

3 3 3

2 2 2

x dx y dy z dzEach of 1 ( 2 )

x y z 3x y z

Choose another set of multipliers 1,1,1

x dx ydy z dzEach of 1 (3)

x y z y z z x x y

3 3 3 2 2 2

2 2 2 2 2 2

x dx ydy z dz x dx ydy z dz

x y z 3x y z x y z y z z x x y

x dx ydy z dz x dx ydy z dzi.e.,

(x y z ) ( x y z y z z x x y) x y z y z z x x y

x dx ydy z dzdx dy dz

(x y z )

(x y z )d( x y z ) x dx y dy z dz

Integrating weget

( x y z

2 2 2 2

2 2 2 2

2

) x y zb

2 2 2 2

( x y z ) ( x y z )b

2x y y z zx 2b

x y y z zx c

v x y y z zx

x yThe generalsolution is , x y y z zx 0.

y z

5. Solve ( mz – ny ) p + ( nx – lz ) q = ly – mx

Solution:

Given ( mz – ny ) p + ( nx – lz ) q = ly – mx


Here P = mz – ny, Q = nx – lz, R = ly – mx

The Subsidiary equations are

(1)

dx dy dz

P Q R

dx dy dz

m z n y n x l z l y m x

Choose a set of multipliers x, y ,z

x dx ydy z dzEach of 1

x ( m z n y) y ( n x z ) z ( y m x )

x dx ydy z dz

0x dx ydy z dz 0

l l

Integrating we get

x 2 + y 2 + z 2 = a

v = x 2 + y 2 + z 2

Choose a set of multipliers l, m ,n

dx m dy n dzEach of 1

( m z n y) m ( n x z ) n ( y m x )

dx m dy n dz

0dx m dy n dz 0

Integrating we get

x m y n z b

v x m y n z

l

l l l

l

l

l

l


2 2 2, 0 x y z l x m y n z

Equations reducible to standard types:

Type V:

Equations of the form

f ( x m p , y n q ) = 0 or f (x m p, y n q , z ) = 0 ------------(1)

where m and n are constants

This type of equations can be reduced to Type I or Type III by the following

transformations.

Case ( i ):

If m 1 , n 1

Put x 1 – m = X, y 1– n = Y

m

m

n

n

z z X zp (1 m) x P , where P

x X x X

i.e., x p ( 1 m ) P

z z Y zq (1 n ) y Q, where Q

y Y y Y

i.e., y q ( 1 n )Q

m n

m n

The equation f x p, y q 0 reduces to f (1 m) P , (1 n )Q 0 which is a Type I eqn.

The equation f x p, y q , z 0 reduces to f (1 m) P ,(1 n )Q ,z 0 which is aType III eqn.

Case ( ii ):

If m = 1, n = 1 then (1) f (x p ,y q ) = 0 or f (x p ,y q, z ) = 0

Put log x = X, log y = Y.

z z X z 1 1 zp P, where P

x X x X x x Xi.e., x p P

z z Y z 1 1 zq Q, where Q

y Y y Y y y Y

i.e., y q Q

The equation f ( px , qy) = 0 reduces to f ( P, Q) = 0 which Type I eqn.

The equation f ( px , qy , z ) = 0 reduces to f ( P, Q , z ) = 0 which Type III eqn.

Type VI :

Equations of the form f ( z k p, z n q) = 0 or f (z k p , y n q , x , y ) = 0 ------------(1)

Where k is constant,

This can be reduced to Type I or Type IV equations by the following

substitutions.

Case ( i ):

If k –1

Put Z = z k + 1

k

k

k

k

k k

k k

Z Z zP ( k 1) z p

x z xP

z pk 1

Z Z zQ ( k 1) z q

y z y

Qz q

k 1

P QThe equation f z p, z q 0 reduces to f , 0 which is a Type I eqn.

k 1 k 1

PThe equation f z p, z q , x , y 0 reduces to f ,

k 1

Q, x , y 0 which is a Type IV eqn..

k 1

Case ( ii ):

If k = –1

( 1 )

p q p qf , 0 or f , , x , y 0

z z z z

put z log z

z Z z 1P p.

x z x zp

Pz

z Z z 1Q q.

y z y z

qQ

zp q

The eqn f , 0 reduces to Type Ieqn.z z

p qThe eqn f , , x , y 0 reduces to Type IV eqn.

z z

k

k

k

k

k k

k k

Z Z zP ( k 1) z p

x z xP

z pk 1

Z Z zQ ( k 1) z q

y z y

Qz q

k 1

P QThe equation f z p, z q 0 reduces to f , 0 which is a Type I eqn.

k 1 k 1

PThe equation f z p, z q , x , y 0 reduces to f ,

k 1

Q, x , y 0 which is a Type IV eqn..

k 1

1– m

2

1

H ere m – 1 , n 1.

Pu t X x

X = x Y log y

X Y 12 x

x y y

z zp q

x y

z X z Y

X x Y y

1p P 2 x q Q

y

x p 2 P y q Q

2 2 2 1 Reduces to 2 P Q z 2

This is of the form f p, q , z 0 which is TypeIII eqn.

Let z f X a Y be a trial Solution.

z f u

u X a Y

u u1 , a

x y

z zP Q

X Yz u z u

u X u Yz z

P Q au u

2 22

22 2

2 2

2

2

2

( 2 ) r e d u c e s t o

d z d z2 a z

d u d u

d z( 4 a ) z

d u

d z z

d u ( 4 a )

d z z.

d u 4 a

d z d u

z 4 a

Integrating

2

2

2

22

2

dz du

z 4 a

1log z u c

4 a

X a Ylogz c

4 a

x a log ylogz c ( X x , Y log y)

4 a

which is the complete int egral

2 2 2 2 2

2 2 2 2 2

2 2 2 2

k k

2 . Solve z p q x y

Solution :

Given z p q x y

z p z q x y 1

This is of the form f z p, z q , x , y 0 Type VI

1 1

2

Here k 1

Put Z z

Z z

Z zP Q

x y

Z z Z z

z x z y

P 2 z p Q 2 z q

P Qz p z q

2 2

2 2 2 2

2 2 2 2

1 2

Eqn 1 reduces to P Q 4 x y

i.e., P – 4 x 4 y – Q

This is of Standard Type IV f x , p f y , q

2 2 2 2

2 2 2 2

2 2 2 2

L e t P – 4 x 4 y – Q 4 c s a y

P – 4 x 4 c 4 y – Q 4 c

P 4 c 4 x Q 4 y – 4 c

2 2

2 2

2 2

P 4 ( c x ) Q 4( y c)

P 2 ( c x ) Q 2 ( y c)

W e know

Z ZdZ dx dy

x y

Z ZdZ P dx Q dy P , Q

x x

dZ 2 ( c x ) dx 2 ( y c ) dy

2 2

2 1 2 1

2 2 1 2 1 2

dZ 2 (c x ) dx 2 ( y c) dy

x c x y c xZ 2 ( c x ) sinh ( y c) cosh a

2 2 2 2C C

x xz x (c x ) c sinh y ( y c) ccosh a ( Z z )

C C

Examples:

1.Form the p d e by eliminating the arbitrary

function f

from the relation z = f(x2 + y2)

Solution:

Given z = f(x2 + y2)--------------(1)

Differentiating (1) partially w. r.t. x and y,

' 2 2

' 2 2

zp f ( x y ) .2 x, ( 2 )

x

zq f (x y ) .2 y ( 3 )

y

( 2 ) p x

(3) q y

y p x q 0

y p x q 0 is the required partial differential equation.

2.Form the p d e by eliminating the arbitrary functions f and g

from the relation z = f ( x +ct ) + g (x -ct) .

Solution :

Given z = f ( x +ct ) + g (x -ct) -------- ( 1 )

2

2

D i f f e r e n t i a t i n g 1 p a r t i a l ly w i t h r e s p e c t t o x

zf '( x c t ) g '( x c t ) ,

x

A g a i n d i f f e r e n t i a t i n g p a r t i a l ly w i t h r e s p e c t t o x

zf " ( x c t ) g " ( x c t ) ( 2 )

x

22 2

2

22

2

2 2

2 2

D i f f e r e n t i a t i n g 1 p a r t i a l l y w i t h r e s p e c t t o t

zc f '( x i t ) c g '( x i t ) ,

t

A g a i n d i f f e r e n t i a t i n g p a r t i a l l y w i t h r e s p e c t t o t

zc f " ( x i t ) c g " ( x i t )

t

zc ( 3 )

x

z z( 2 ) ( 3 ) 0

t x

UNIT-IV

Applications of Partial Differential Equations

1. A tightly stretched string with fixed end points 0x and lx is initially in a position

given by

l

xyxy

30 sin)0,( . It is released from rest from this position. Find the

displacement at any time ''t .

Sol:

One dimensional wave equation 2

22

2

2

x

ya

t

y

The conditions are

(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(

xt

y

(iv)

l

xyxy

30 sin)0,(

The correct solution which satisfying the given boundary conditions is

)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------------(1)

Apply the boundary condition (i) in (1)

),0( ty 0)sincos( 431 patcpatcc

weget 01 c , then equation (1) becomes

)sincos(sin),( 432 patcpatcpxctxy -------------(2)

Apply the boundary condition (ii) in (2)

0)sincos(sin),( 432 patcpatcplctly

weget 0sin pl npl l

np

Then equation (2) becomes

l

atnc

l

atnc

l

xnctxy

sincossin),( 432 -------------- (3)

Differentiate Partially (3) w.r.to t

),( txt

y

l

atn

l

anc

l

atn

l

anc

l

xnc

cossinsin 432 -------- (4)

Apply the boundary condition (iii) in (4)

0sin)0,( 42

l

anc

l

xncx

t

y


l

atn

l

xnc

l

atnc

l

xnctxy

n

cossin

cossin),( 32

where 32cccn

By the superposition principle, the most general solution is

1

cossin),(n

n l

atn

l

xnctxy

----------------(5)

Apply the boundary condition (iv) in (5)

l

x

l

xyl

xc

l

xc

l

xc

l

xc

l

x

l

xy

l

xy

l

xncxy

nn

3sinsin3

4

..........4

sin3

sin2

sinsin

3sinsin3

4

sin1.sin)0,(

0

4321

0

30

1

0..........,4

,0,4

354

032

01 cc

ycc

yc

The equation (5) becomes

l

at

l

xy

l

at

l

xytxy

3cos

3sin

4cossin

4

3),( 00

2. A tightly stretched string of length l has its ends fastened at 0x and lx . The midpoint of the string is then taken to a height h and then released from rest in that position. Obtain an expression for the displacement of the string at any subsequent time.

Sol:

One dimensional wave equation is 2

22

2

2

x

ya

t

y

The conditions are

(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(

xt

y

(iv) )()0,( xfxy

,0l(0,0),0

2

l

,2

lh

X

Y

Consider the interval ),0( 2l , the end points are ),(),0,0( 2 hl

Using two point formula for the straight line

l

hxy

x

h

y

xx

xx

yy

yy

l

2

0

0

0

0

2

21

1

21

1

Consider the interval ),( 2 ll , the end points are )0,(),,( 2 lhl

Again by two point formula for the straight line

)(2

0 2

2

21

1

21

1

xll

hy

l

x

h

hy

xx

xx

yy

yy

l

l

lxxll

h

xl

hx

xfl

l

2

2

),(2

0,2

)(


)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------- (1)




)sincos(sin),( 432 patcpatcpxctxy ------- (2)



weget 0sin pl npl l

np

, then equation (2) becomes

l

atnc

l

atnc

l

xnctxy

sincossin),( 432 ------- (3)

Differentiate Partially (3) w. r. to t

),( txt

y

l

atn

l

anc

l

atn

l

anc

l

xnc

cossinsin 432 ------- (4)


0sin)0,( 42

l

anc

l

xncx

t

y

We get 04 c then equation (3) becomes

l

atn

l

xnc

l

atnc

l

xnctxy

n

cossin

cossin),( 32

where 32cccn


1

cossin),(n

n l

atn

l

xnctxy

(5)


)0,(xy

1

1.sinn

n l

xnc

)(

),(2

0,2

2

2

xf

lxxll

h

xl

hx

l

l

To find the value of nc , expand )(xf in a half range Fourier sine series

We know that half range Fourier sine series of )(xf is

1

sin)(n

n l

xnbxf

(7)

By comparing (6) and (7), we get nn cb

To find nc it is enough to find nb

l

l

n

l

l

dxl

xnxl

l

hdx

l

xn

l

hx

l

dxl

xnxf

lb

2

2

sin)(2

sin22

sin)(2

0

0

n

l

c

n

n

h

n

n

l

l

h

n

n

ln

n

ln

n

ln

n

l

l

h

l

nl

xn

l

nl

xn

xl

l

nl

xn

l

nl

xn

xl

h

l

l

2sin

82

sin8

2sin

2cos

22sin

2cos

2

4

sin)1(

cos)(

sin1

cos4

22

22

2

2

22

22

22

22

2

2

22

02

222

2

2


1

22cossin

2sin

8),(

n l

atn

l

xnn

n

htxy

3 . The points of trisection of a string are pulled aside through a distance ‘b’ on

opposite sides of the position of equilibrium and the string is released from rest.

Find an expression for the displacement.

Sol:

One dimensional wave equation is 2

22

2

2

x

ya

t

y

The conditions are

(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(

xt

y

(iv) )()0,( xfxy

Equation of OA:

1 1

1 2 1 2

y y x x y 0 x 0 3bxy in 0, 3

y y x x 0 b 0 3

Equation of AB:

1 1

1 2 1 2

y y x x x 3y b 3by 2x in 3,2 3

y y x x b ( b) 3

Equation of BC:

1 1

1 2 1 2

y y x x y 0 x 0 3by x in 2 3 ,

y y x x 0 b 0 3

3bxin 0, 3

3bf (x) 2x in 3,2 3

3bx in 2 3,


)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------- (1)




)sincos(sin),( 432 patcpatcpxctxy ------- (2)



weget 0sin pl npl l

np

, then equation (2) becomes

l

atnc

l

atnc

l

xnctxy

sincossin),( 432 ------- (3)

Differentiate Partially (3) w. r. to t

),( txt

y

l

atn

l

anc

l

atn

l

anc

l

xnc

cossinsin 432 ------- (4)


0sin)0,( 42

l

anc

l

xncx

t

y

We get 04 c then equation (3) becomes

l

atn

l

xnc

l

atnc

l

xnctxy

n

cossin

cossin),( 32

where 32cccn


1

cossin),(n

n l

atn

l

xnctxy

-------------- (5)


)0,(xy

1

1.sinn

n l

xnc

3bxin 0, 3

3bf (x) 2x in 3,2 3

3bx in 2 3,



1

sin)(n

n l

xnbxf

(7)

By comparing (6) and (7), we get nn cb

To find nc it is enough to find nb

n0

2 n xb f (x)sin dx

2

3 3

n20

3 3

2 3bx n x 3b n x 3b n xb sin dx 2x sin dx x sin dx

nn 2 2

18b nb sin 1 ( 1)

3n

n

2

0 if n is odd

36b nb sin if n is even3n

Substitute nb values in equation (5) we get the required solution

2n 2,4

36b n n x n aty x, t sin sin cos

3n

4. A tightly stretched string with fixed end points 0x and lx is initially at rest

in its equilibrium position. If it is set vibrating by giving each point a

velocity )( xlkx . Find the displacement of the string at any time.

Sol:

One dimensional wave equation 2

22

2

2

x

ya

t

y

The conditions are

(i) 0),0( ty (ii) 0),( tly (iii) 0)0,( xy

(iv) )()0,( xlkxxt

y


)sincos)(sincos(),( 4321 patcpatcpxcpxctxy --------(1)

apply the boundary condition (i) in (1)


weget 01 c then equation (1) becomes

)sincos(sin),( 432 patcpatcpxctxy -----------------(2)



weget 0sin pl npl l

np

,

then equation (2) becomes

l

atnc

l

atnc

l

xnctxy

sincossin),( 432 --------------(3)


)0,(xy 0sin 32 cl

xnc

here 03 c

then equation (3) becomes

l

atn

l

xnc

l

atnc

l

xnctxy

n

sinsin

sinsin),( 42

where 42cccn


1

sinsin),(n

n l

atn

l

xnctxy

----------------(4)

Differentiate Partially (4) w.r.to t

),( txt

y

l

an

l

atn

l

xnc

nn

1

cossin ------------------------(5)


)0,(xt

y)()(.sin

1

xfxlkxl

an

l

xnc

nn

-------------(6)



1

sin)(n

n l

xnbxf

--------------------------(7)

By comparing (6) and (7), we get nnnn ban

lc

l

ancb

1..200cos.2002

cos2

sin2

cos2

sin)(2

sin)(2

33

3

33

3

03

33

2

222

0

0

n

ln

n

l

l

k

ln

lxn

ln

lxn

xl

ln

lxn

xlxl

k

dxl

xnxlkx

l

dxl

xnxf

lb

l

l

l

n

oddisnifn

kl

evenisnifn

kl

n

l

l

k

n

n

33

2

33

2

33

3

8

0

)1(14

)1(122

44

3

33

2 88

n

kl

n

kl

an

lb

an

lc nn


5,3,1

44

3

sinsin8

),(n l

atn

l

xn

n

kltxy

4. A rod of length l has its ends A and B kept at c0 and c120 respectively until steady state conditions prevail. If the temperature at B is reduced to c0 and so while that of A is maintained, find the temperature distribution of the rod.

Sol:

One dimensional heat equation 2

22

x

ua

t

u

-----(1)

Steady state equation is 02

2

x

uand the steady state solution is baxxu )( --(2)

The boundary conditions are 120)()(0)0()( luiiui

Apply (i) in (2), 0)0( bu

The equation (2) becomes axxu )( -------- (3)

Apply (ii) in (3)

la

allu

120

120)(

The equation (3) becomesl

xxu

120)(

Now consider the unsteady state condition.

The conditions are

)0,()(0),()(0),0()( xuvtluivtuiiil

x120


tpaecpxcpxctxu22

321 )sincos(),( ----------- (4)


0),0(22

31 tpaecctu

weget 01 c

then equation (4) becomes tpaecpxctxu22

32 sin),( ---------- (5)


0sin),(22

32 tpaecplctlu

weget 0sin pl npl l

np

,


t

l

na

n

tl

na

el

xnc

el

xncctxu

2

222

2

222

sin

sin),( 32

where nccc 32

By the super position principle, the most general solution is

1

2

222

sin),(n

tl

na

n el

xnctxu

------------------ (6)

Apply the boundary condition (v) in (6)

1

)(120

1.sin)0,(n

n xfl

x

l

xncxu

------------------ (7)

To find nc , expand )(xf in half range sine series


1

sin)(n

n l

xnbxf

------------------ (8)

From the equations (7) & (8) we get nn cb

Now

1

2

02

222

0

0

)1(240

)1(240

cos240

sin)1(

cos240

sin1202

sin)(2

n

n

l

l

l

n

n

ln

l

nn

l

l

l

nl

xn

l

nl

xn

xl

dxl

xn

l

x

l

dxl

xnxf

lb

The required solution is

1

1 2

222

sin)1(240

),(n

tl

nan e

l

xn

n

ltxu

5. The ends A and B of a rod l cm long have their temperatures kept at c30 and c80 , until steady state conditions prevail. The temperature of the end B is suddenly reduced to c60 and that of A is increased to c40 . Find the temperature distribution in the rod after time t.

Sol:

One dimensional heat equation2

22

x

ua

t

u

-----(1)

Steady state equation is 02

2

x

uand the steady state solution is baxxu )( --(2)

the boundary conditions are 80)()(30)0()( luiiui

Apply (i) in (2), 30)0( bu

Then the equation (2) becomes 30)( axxu (3)

Apply (ii) in (3)

la

allu

50

8030)(

Then the equation (3) becomes 3050

)( l

xxu

Now consider the unsteady state condition.

In unsteady state the suitable solution which satisfying the given

boundary conditions is

tpaecpxcpxctxu22

321 )sincos(),( (4)

The boundary conditions are

3050

)()0,()(60),()(40),0()( l

xxuxuvtluivtuiii

Since we have all non zero boundary conditions, we write the temperature

distribution function as ),()(),( txuxutxu ts (5)

)(),(),( xutxutxu st

To find )(xus

The solution is baxxus )( (6)

The boundary conditions are 40)(,60)0( luu ss

,40)0( bus

la

al

ballus

20

6040

60)(

the equation (6) becomes 4020

)( l

xxus (7)

To find ),( txut

Given the boundary conditions are

1030

4020

3050

)()0,()0,()(

06060)(),(),()(

04040)0(),0(),()(

l

x

l

x

l

xxuxuxuviii

lutlutluvii

ututxuvi

st

st

st

In unsteady state, the suitable solution which satisfying the given boundary conditions

is

tpaecpxcpxctxu22

321 )sincos(),( (8)

Apply the boundary condition (vi) in (8)

0),0(22

31 tpat ecctu

here 01 c

the equation (8) becomes tpat ecpxctxu

22

32 sin),( (9)

Apply the boundary condition (vii) in (9)

0sin),(22

32 tpat plecctlu

here

l

np

pl

0sin

the equation (9) becomes t

l

na

t ecxl

xnctxu

2

222

32 sin),(

t

l

na

n

tl

na

t

el

xnc

el

xncctxu

2

222

2

222

sin

sin),( 32

where nccc 32

By the super position principle, the most general solution is

1

2

222

sin),(n

tl

na

nt el

xnctxu

(10)

Apply the boundary condition (viii) in (10)

1

)(1030

1.sin)0,(n

nt xfl

x

l

xncxu



1

sin)(n

n l

xnbxf

(11)

From the equations (10) & (11) we get nn cb

1.10

cos202

sin30

cos10

302

sin10302

sin)(2

02

22

0

0

n

ln

n

l

l

l

nl

xn

ll

nl

xn

l

x

l

dxl

xn

l

x

l

dxl

xnxf

lb

l

l

l

n

n

n

cn

)1(2120

the equation (10) becomes

1

2

222

sin)1(2120

),(n

tl

nan

t el

xn

ntxu

Then the required temperature distribution function is

1

2

222

sin)1(2120

4020

),(n

tl

nan e

l

xn

nl

xtxu

6. A rectangular plate is bounded by the lines byaxyx ,,0,0 , ab . Its surfaces are insulated. The temperature along 0x and 0y are kept at C0 and

the others at C100 . Find the steady state temperature at any point of the plate.

Sol:

Two dimensional heat equation is 02

2

2

2

y

u

x

u

Let l be the length of the square plate

Given the boundary conditions are

( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 100

( ) ( , ) 100

i u y ii u x iii u a y

iv u x b

Assume the temperature distribution as

),(),(),( 21 yxuyxuyxu (A)

To find ),(1 yxu

Consider the boundary conditions

1 1 1

1

( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0

( ) ( , ) 100

v u y vi u x vii u x b

viii u a y

The suitable solution which satisfying the given boundary conditions is

)sincos)((),( 43211 pycpycececyxu pxpx (1)

Apply the boundary condition (v) in (1)

0)sincos)((),0( 43211 pycpycccyu

weget 021 cc

12 cc

then the equation (2) becomes

)sincos)((

)sincos)((),(

431

43111

pycpyceec

pycpycececyxupxpx

pxpx

(2)

Apply the boundary condition (vi) in (2)

0)()0,( 311 ceecxu pxpx

weget 03 c


pyceecyxu pxpx sin)(),( 411 (3)

Apply the boundary condition (vii) in (3)

0sin)(),( 411 pbceecbxu pxpx

weget 0sin pb npb b

np


b

yn

b

xnc

b

yn

b

xncc

b

ynceecyxu

n

b

xn

b

xn

sinsinh

sinsinh2

sin)(),(

41

411

where 412 cccn

The most general solution is

b

yn

b

xncyxu

nn

sinsinh),(

11

(4)

Apply the boundary condition (viii) in (4)

)(100sinsinh),(1

1 yfb

yn

b

ancyau

nn

(5)



1

sin)(n

n b

ynbyf

(6)

From the equations (5) & (6) we get

b

anb

cb

ancb n

nnn

sinhsinh

oddisnifn

evenisnifn

nn

b

b

b

nb

yn

b

dyb

yn

b

dyb

ynyf

bb

n

b

b

b

n

400

0

)1(1200

1cos200

cos200

sin1002

sin)(2

0

0

0

nn

cn sinh

400 if n is odd


b

yn

b

xn

b

ann

yxun

sinsinhsinh

400),(

5,3,11

To find ),(2 yxu

Consider the boundary conditions

2 2 2

2

( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0

( ) ( , ) 100

ix u y x u x xi u a y

xii u x b

and the suitable solution in this case is

))(sincos(),( 87652pypy ececpxcpxcyxu (7)

Apply the boundary condition )(ix in (7)

0)(),0( 8652 pypy ececcyu

weget 05 c


)(sin),( 8762pypy ececpxcyxu (8)

Apply the boundary condition )(x in (8)

0)(sin)0,( 8762 ccpxcxu

weget 7887 0 cccc


)(sin

)(sin),(

76

7762

pypy

pypy

eepxcc

ececpxcyxu

(9)

Apply the boundary condition )(xi in (9)

0)(sin),( 762 pypy eepaccyau

weget 0sin pa npa a

np


a

xn

a

ync

a

xn

a

yncc

eea

xnccyxu

n

a

yn

a

yn

sinsinh

sinsinh2

)(sin),(

76

762

where 762 cccn

The most general solution is

a

xn

a

yncyxu

nn

sinsinh),(

12

(11)

Apply the boundary condition )(xii in (11)

)(100sinsinh),(1

2 xfa

xn

a

bncbxu

nn


Half range Fourier sine series of )(xf is

1

sin)(n

n a

xnbxf

(12)

From the equations (11) & (12) we get

a

bnb

ca

bncb n

nnn

sinhsinh

a

a

a

n

a

na

xn

a

dxa

xn

a

dxa

xnxf

ab

0

0

0

cos200

sin1002

sin)(2

oddisnifn

evenisnifn

nn

a

a

n

400

0

)1(1200

1cos200

a

bnn

cn sinh

400 if n is odd


a

xn

a

yn

nnyxu

n

sinsinhsinh

400),(

5,3,12

Then the equation (A) becomes

a

xn

a

yn

nnb

yn

b

xn

nn

yxuyxuyxu

nn

sinsinhsinh

400sinsinh

sinh

400

),(),(),(

5,3,15,3,1

21

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UNIT – I FOURIER SERIES PROBLEM 1 - … – I FOURIER SERIES ... Expand f(x) = cosx in a Fourier series in the interval ... Find the complex form of the Fourier series of f(x) =