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TRANSFORMS & PDE (MA6351)

Question Bank with Answers

UNIT I PARTIAL DIFFERENTIAL EQUATIONS PART-A

1. Obtain partial differential equation by eliminating arbitrary constant ‘a’ and ‘b’ from (x–a) 2 +(y–b) 2= z A.U May/June2013

Solution: Given (x–a)2 +(y–b)2=z -----(1) Diff Partially w.r.t x and y 2(x–a) = p => x-a = p/2 2(y–b)=q => y–b= q/2 Substituting in (1), we get 2 2 4p q z+ =

2. Form the partial differential equation by eliminati ng the arbitrary constants ‘a’ & ‘b’ from z= (x2+a)(y2+b) A.U April/May 2011

Solution: Given z= (x2+a)(y2+b) -----(1) Diff partially w.r.t x and y

ax

y2

q,by

x2

p

),ax(y2qyz

),by(x2pxz

22

22

+=+=

+==∂∂

+==∂∂

Substituting in (1) we get pq= 4xyz 3. Find the PDE of all planes having equal intercepts on the x and y axis. Solution:

The equation of plane is 1 .............(1)x y z

a a b+ + =

Diff. w.r.t. x and y partially respectively

10 .............(2)

p

a b+ =

10 .............(3)

q

a b+ =

Solving (2) & (3) we get, p=q

4. Obtain the partial differential equation by eliminating arbitrary constants ‘a’ and ‘ b’ from

( ) ( )2 2 2 1x a y b z− + − + = .

Solution:

( ) ( )2 2 2 1x a y b z− + − + = - (1)

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

( )2 2 0x a zp− + = x a zp⇒ − = − - (2)

( )2 2 0y b zq− + = y b zq⇒ − = − - (3)

Substituting (2) & (3) in (1) we get

2 2 2 2 2 1z p z q z+ + = i.e., ( )2 2 2 1 1z p q+ + =

www.Vidyarthiplus.com



5. Eliminate the arbitrary function f from y

z fx

=

and form the PDE. (Nov/Dec 2012)

Solution:

2' .............(1)z y y

p fx x x

∂ − = = ∂

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

z yq f

y x x

∂ = = ∂

From (1) & (2) 0p y

px qyq x

−= ⇒ + =

6. Form the partial differential equation by eliminati ng the arbitrary constants ‘a’ & ‘b’ from z= ax+by

Solution: Given z= ax+by -----(1) Diff partially w.r.t x and y

,bqyz

,apxz

==∂∂

==∂∂

Substituting in (1) we get z = px+qy. 7. Form the partial differential equation by eliminati ng f from 2 1

2 logz x f xy

= + +

.

Solution:

Let 2 12 logz x f x

y

= + +

- (1)

Differentiate (1) partially w.r. t x and y

' 1 12 2 log

zp x f x

x y x

∂ = = + + ∂ - (2)

'2

1 12 log

zq f x

y y y

∂ −= = + ∂

- (3)

Eliminating 'f from (2) & (3)

( )22 1p xy

q x

− −∴ = ⇒ 2 22px x qy− = − ⇒ 2 22px qy x+ =

8. From the partial differential equation by eliminati ng g from ( )2 2 2, 0g x y z x y z+ + + + = .

Solution:

We know that if g(u, v) = 0 then u = f (v)

( )2 2 2x y z f x y z∴ + + = + + - ---- (1)

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

We get




( )( )'2 2 1x zp f x y z p+ = + + + ----- (2)

( )( )'2 2 1y zq f x y z q+ = + + + …..(3)

Divide (2) & (3) 1

1

x zp p

y zq q

+ +=

+ + ⇒ x qx zp zpq y py zq zpq+ + + = + + +

( ) ( )z y p x z q y x− + − = −

9. Find the complete integral of p+q = pq Solution:

Given p+q = pq -----(1) This is of the type f(p,q)=0 Let z= ax+by+c be the solution for (1) Partially Diff w.r.t ‘x ‘ and ‘y’ we get p= a and q = b

=> a+b = ab => b= a/(a–1)

Therefore z= ax+ (a/a–1)y+c is the complete solution.

10. Find the singular integral of pq2+qy+px =z

Solution:

This of Clairaut’s type and the complete solution is ab2+by+ax =z

Diff w.r.t ‘a ‘ and ‘b’ we get

0b2

a2x

bz

,0a2

b2x

az

=+=∂∂

=+=∂∂

� b

ay&

a

bx −=−= ⇒ 1

b

a

a

bxy ==

xy= 1 is singular integral. 11. Find the complete integral of p+q=0 Solution:

Given p +q = 0 -----(1) This is of the type f(p,q)=0 Let z= ax+by+c be the solution for (1) Partially Diff w.r.t ‘x ‘ and ‘y’ we get p=a and q=b

=> a+b = 0 => b= –a Therefore z= ax– ay +c be the complete solution.

12. Solve ( )1p q qz+ = .

Solution:

( )1p q qz+ = - (1)

This equation is of the form ( ), , 0f z p q =

( )z g x ay= + be the solution Let ( )x ay u z g u+ = =

dz adzp q

du du= =

(1) reduces to 1

1 1 11

dz dz dz dz dz dz dza a z a az a az z du

du du du du du du a za

+ = ⇒ + = ⇒ = − ⇒ = − ⇒ = −




Integrating 1

log z u ba

− = +

i.e., 1

log z x ay ba

− = + +

is the complete solution.

13. Solve yp= 2yx+log q Solution:

The equation can be reduced to the type F1(x,p)= F2(y,q) � y(p–2x) = log q � p–2x = (log q)/ y =k

� p= 2x+k ; q = kye

� z = ∫ ∫+ qdypdx

� z= ∫ ∫++ dyedx)kx2( ky

z = (x2+kx)+ k

eky

+c

14. Solve '( 1)( 1) 0D D D z− − + = (Nov/Dec 2012) Solution:

'( 1)( 1) 0D D D z− − + = Here c1=1, c2=-1 , m1=0, m1=1

General solution 1 2( ) ( )x xz e f y e f y x−= + +

15. Solve 1p q+ = .

Solution:

This is of the form ( ), 0f p q = .

The solution is given by z ax by c= + + where

( )2111 ababba −=⇒−=⇒=+

∴ The complete solution is ( )2

1z ax a y c= + − +

16. Solve ( )( ) 21 2 x yD D D D z e −′ ′− − − − = .

Solution:

This is of the form ( ) ( ) ( )1 1 2 2 .... 0n nD m D C D m D C D m D C Z′ ′ ′− − − − − − =

Hence 1 1 2 11 1 1 1m c m c= = = = 2 2c =

Hence the C.F. is ( ) ( )21 2

x xz e y x e y xφ φ= + + +

P.I. ( )( )

2

1 2

x ye

D D D D

−

=′ ′− − − − ( )( )

221

2 1 1 2 1 2 2

x yx ye

e−

−= =+ − + −

Hence, the complete solution is ( ) ( )2 21 2

1

2x x x yZ e y x e y x eφ φ −= + + + +

17. Solve ( )3 2 33 2 0D DD D Z′ ′− + = .

Solution:




Substituting D = m, & D1 = 1

The Auxiliary equation is m3 - 3m + 2 = 0

m = 1, 1, -2

Complimentary function is ( ) ( ) ( )1 2 3 2y x x y x y xφ φ φ+ + + + −

i.e., ( ) ( ) ( )1 2 3 2z y x x y x y xφ φ φ= + + + + −

18. Find the particular integral of yx22 ez)'D5'DD2D( −=++ Solution:

P.I = yx22

e'D5'DD2D

1 −

++

Replace D by 1 and 'D by 1 ,we get

= yx2

yxyx e2

xe

'D2D2

xe

121

1 −−− =+

=+−

19. Find the general solution of 0

y

z8

yx

z4

yx

z2

x

z3

3

2

3

2

3

3

3

=∂

∂+

∂∂

∂−

∂∂

∂−

∂

∂

Solution:

Auxiliary Equations is m3– 2m

2 –4m+8=0 � m= 2,2,–2

General solution z= ( ) ( ) ( )x2yfx2yxfx2yf 321 −++++ 20. Find the Particular integral of yx222 ez)1'D2D2'DD2'DD( −=+++++ Solution:

P.I = yx222

e)1'DD(

1 −

++

Replace D by 2 and D’ by –1 ,we get

P.I = yx2yx22

e4

1e

)112(

1 −− =+−

PART - B 1. a) Find the general solution of )y-z(x=)qz-y(x+)pz-x(y 222222

[A.U Apr/ May 2008 ]

Solution:

The equation is of the form Pp+Qq=R

where P= )z-x(y 22 ,Q= )z-y(x 22 ,R= )y-z(x 22

The auxiliary equation is )()()( 222222 yxz

dz

zxy

dy

yzx

dx

−=

−=

−

Using the multipliers x, y, z , we get each ratio is equal to

0)()()( 222222222=

−+−+−

++

yxzzxyyzx

zdzydyxdx

i.e. zdzydyxdx ++ =0

Integrating we get 222 zyx ++ 1c=




using the another set of multipliers zyx

1,

1,

1

we get each of the above ratio is equal to 0)()()(

111

222222=

−+−+−

++

yxzxyz

dzz

ydyy

dxx

i.e. dzz

ydyy

dxx

111++ =0

Integrating we get 2c log =(xyz) log

The solution of the linear p.d.e is ϕ( u, v)=0

i.e. ϕ( 222 zyx ++ , xyz) =0

b) Solve yx3222 eyxz)'D6'DDD( ++=−+ Solution:

The auxiliary equation is 062 =−+mm 3,2 −=m

C.F= )3()2( 21 xyxy −++ φφ

P.I= )('6'

1 3222

yxeyxDDDD

++−+

= yxeDDDD

yx

D

D

D

DD

+

−++

−+

322

2

2

22 '6'

1

'6

'1

1

yxeyxD

D

D

D

D+

−

−++

−+= 32

1

2

2

2 )1(6)1(39

1'6

'1

1

652

1 354

yxexyx

+

+

−=

The general Solution is )3()2( 21 xyxyy −++= φφ +652

1 354

yxexyx

+

+

−=

2. a) Solve ( ) ( )2 2 2x yz p y zx q z xy− + − = −

This is a Lagrange’s linear equation of the form Pp Qq R+ =

The subsidiary equations are 2 2 2

dx dy dz

x yz y zx z xy= =

− − − - (1)

( ) ( ) 2 2 2 22 2

dx dy dy dz dx dz

y zx z xy x yz z xyx yz y zx

− − −= =

− − + − − +− − −

i.e., ( )

( ) ( )( )

( )( )

( )2 2 2 22 2 1

d x y d y z d x z

y z x y z x z y zx y z x y

− − −= =

− + − − + −− + −




i.e., ( )

( )( )( )

( )( )( )

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

x y x y z y z x y z x z x y z

− − −= =

− + + − + + − + + - (2)

i.e., ( )( )

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

x y y z x z

− − −= =

− − − - (3)

Taking the first two ratios, and integrating ( ) ( )log log logx y y z a− = − +

x ya

y z

−∴ =

− - (4)

Similarly taking the last two ratios of (3) we get,

y zb

x z

−∴ =

− - (5)

But x y

y z

−−

and y z

x z

−−

are not independent solutions for 1x y

y z

−+

− gives

x z

y z

−−

which is the

reciprocal of the second solution.

Therefore solution given by (4) and (5) are not independent. Hence we have to search for

another independent solution.

Using multipliers x, y, z in equation (1) each ratio is 3 3 3 3

xdx ydy zdz

x y z xyz

+ +=

+ + −

Using multipliers 1, 1, 1 each ratio is 2 2 2

dx dy dz

x y z xy yz zx

+ +=

+ + − − −

3 3 3 2 2 23

xdx ydy zdz dx dy dz

x y z xyz x y z xy yz zx

+ + + +=

+ + − + + − − −

( )( )( )

( )2 2 2

2 2 22 2 2

12

d x y z d x y z

x y z xy yz zxx y z x y z xy yz zx

+ + + +=

+ + − − −+ + + + − − −

Hence ( ) ( ) ( )2 2 21

2d x y z x y z d x y z+ + = + + + +

Integrating ( ) ( )22 2 21

2 2

x y zx y z k

+ ++ + = +

( ) ( )22 2 2

2 2 2 2 2 2

2

2 2 2 2

x y z x y z k

x y z x y z xy yz zx k

∴ + + = + + +

+ + = + + + + + +




i.e., xy yz zx b+ + =

∴ the general solution is , 0x y

xy yz zxy z

φ −

+ + = −

b) Solve 22222 )( yxqpz +=+ [A.U Nov/ Dec 2007]

Solution:

The given equation can be written as )1.......()()( 2222 yxzqzp +=+

This is of the type of f (x, zmp)=f(y, zmq)

2

Q =zq &

2

P=zp 2zq =Q ,2zp=P and Z= z Put 2 ⇒

Equation (1) can be written as )y+4(x =Q+P 2222

� 2222 4y-Q4x-P =

The equation is of the form f(x,p) =f (y,q)

� 24a4y-Q4x-P 2222 == ;

22 44 a4y-Q anda4x-P 2222 ==

� 222 axP += and

222 ayQ −=

We know that dZ= Pdx + Qdy

+

+= dxaxdZ 222 dyay

− 222

Integrating we get Z = 22 axx + + +

−

a

xa 12

sinh2

−− −

a

yayy 122 cosh

222 axxz += + +

−

a

xa 12

sinh2

−− −

a

yayy 122 cosh

3. a) Find the singular solution of the equation 22qp+qypx=z + Solution:

The equation is of the form z= px+qy+f(p,q)

For complete integral replace ‘p’ by ‘a’ and ‘q’ by b ,we get 22ba+byax=z +

� Complete integral = 22ba+byax=z + -----(1)

Diff w.r.t ‘a’ and ‘b’ we get

)3......(02

)2......(02

2

2

=+=∂

∂

=+=∂∂

bayb

z

abxa

z

Solving we get )4......(2 2abx −= & )5......(2 2bay −=




The equation (4) can be written as )6......(2 2b

xa

−=

Substituting (6) in (5) we get )7......(2

3/12

−=

y

xb and )8......(

2

3/12

−=

x

ya

Substituting (7) & (8) in (1) we get 0y27x16z 223 =+ which is singular integral.

b) Solve yx22 e)yx2sin(5z)'D2'DD5D2( +−++=+− Solution:

The auxiliary equation is 0252 2 =+− mm

2

1,2=m

C.F=

+++ xyxy2

1)2( 21 φφ

P.I= ( )yxeyxDDDD

+−+++−

)2sin(5'2'52

122

yxeDDDD

yxDDDD

+−

+−++

+−=

2222 '2'52

1)2sin(5

'2'52

1

yxeyx +−

+−−++

−+−=

)1(2)1(5)1(2

1)2sin(

2108

15

yxeyxDD

x +−++−

=9

1)2sin(

'545

yxeyxDD

DD

DD

DDx +−++++

−+

=9

1)2sin(

'54

'54

'54

)'54(5

yxeyxDD

DDx +−++−

+=

9

1)2sin(

'2516

)'54(5

22

[ ] yxeyxxyxx +−++−+=9

1)2cos(5)2cos(8

39

5

The General Solution is

+

+++= xyxyy2

1)2( 21 φφ [ ] yxeyxxyxx +−++−+

9

1)2cos(5)2cos(8

39

5

4. a) Solve 2 2( 2 ) ( )z yz y p xy zx q xy zx− − + + = − .

Solution:

This is a Lagrange’s linear equation of the form Pp Qq R+ =

The subsidiary equations are 2 22

dx dy dz

z yz y xy zx xy zx= =

− − + −

Taking x, y, z as multipliers, we have each fraction 0

xdx ydy zdz+ +=

0xdx ydy zdz∴ + + =




Integrating

2 2 2

2 2 2

x y zc+ + =

i.e., 2 2 2

1x y z C+ + = - (1)

Again, taking the last two members, we have ( ) ( )

dy dz

x y z x y z=

+ −

i.e., dy dz

y z y z=

+ −

( ) ( )

( )( )

0

0

y z dy y z dz

ydy zdy ydz zdz

ydy zdy ydz zdz

ydy d yz zdz z

− = +

− = +

− + − =

− − =

Integrating we get 2 2

22y yz z C− − = - (2)

From (1) & (2) the general solution is

( )2 2 2 2 2, 2 0x y z y yz zφ + + − − =

b) Solve 7xyz)'D3D3'DD( 22 +=+−− ( Nov/ Dec 2009 )

Solution:

The given equation is 0z)'D3D3'DD( 22 =+−−

This can be factorized as 0= z3)- (-1)D' - D'-0)(D -(D

It is of the form 0)c'DmD)(c'DmD( 2211 =−−−− where m1 =1 c1 = 0 ,m2 = -1 c2 = 3.

C.F= )()( 23

1 xyfexyf x −++

P.I = 3)- D' )(DD' -(D

7xy

+

+7xy

3

'DD13

1

)'DD(

1+

+−−

−=

)7xy(...3

'DD

3

'DD1

)'DD(3

1)7xy(

3

'DD1

)'DD(3

1 21

+

+

+

++

+−−

=+

+−

−−

=−

)7('9

2

3

'1

)'(3

1+

+

++

−−

= xyDDDD

DD

+++++++

−−

= )7xy('DD9

2)7xy('D

3

1)7xy(D

3

1)7xy(

)'DD(3

1




++++

−−

=9

2)yx(

3

1)7xy(

)'DD(3

1=

++++

−

−=

9

2)yx(

3

1)7xy(

D

'D1D3

1

++++

−

−=

−

9

2)yx(

3

1)7xy(

D

'D1

D3

1 1

++++

+

++

−=

9

2)(

3

1)7(....

''1

3

12

yxxyD

D

D

D

D

+++++++++

−=

9

2)yx(

3

1)7xy(

D

'D

9

2)yx(

3

1)7xy(

D3

1

++++++−

=3

11

9

2)(

3

1)7(

3

1x

Dyxxy

D

++++++−

= ∫ dxxyxxyD 3

1

9

2)(

3

1)7(

3

1

++++++

−=

329

2)(

3

1)7(

3

1 2 xxyxxy

D =

∫

++++++

−dx

xxyxxy

32(

9

2)(

3

1)7(

3

1 2

=

+++

+++

−

669

2

23

1)7

23

1 2322 xxxxy

xx

yx

++++

−=

6

x

3

xy

3

x

9

x65

2

yx

3

1 322

++++−−++=

6339

65

23

1)()(

322

23

1xxyxxyx

xyfexyfz x

5. a) Find the singular solution of the equation 22 qpqp+qypx=z +++ Solution:

The equation is of the form z = px +qy + f(p,q)

For complete integral replace ‘p’ by ‘a’ and ‘q’ by b ,we get 22 baba+byax=z +++

� Complete integral = 22 baba+byax=z +++ -----(1)

Diff w.r.t ‘a’ and ‘b’ we get )3......(02)2......(02 =++=∂∂

=++=∂∂

bayb

zbax

a

z

Solving we get )4......(2

)( bxa

+−= substituting in (3) we get )5......(

3

)2( yxb

−=

Substituting (5) in (4) we get )6......(3

)2( xya

−=

Substituting (5) & (6) in (1) we get 0= y-x-xy-3z 22 which is singular integral.




b) Solve 4ez)'D2D2'D'DD2D( yx322 +=−−++ + Solution:

4ez)'D2D2'D'DD2D( yx322 +=−−++ +

( ) 4)'(2)'( 32 +=+−+ + yxezDDDD ( ) 4)2')('( 3 +=−++ + yxezDDDD C.F= )()( 2

21 xyexy x −+− φφ

P.I= ( )yxyx eeDDDDDD

00322

4'22''2

1 ++ +−−++

yxyx eDDDDDD

eDDDDDD

0022

322

4'22''2

1

'22''2

1 ++

−−+++

−−++=

yxyx eDD

xe 003 4

2'2226169

1 ++

−++

−−++=

yxyx ex

e 003 428

1 ++

−+=

xe yx 28

1 3 −= +

UNIT II FOURIER SERIES PART-A

1. State Dirichlet’s conditions for the existence of Fourier series of f(x) in the interval . (Nov./Dec.2011,Reg 2008)

A function f(x) can be expanded as a Fourier series in the interval if the following conditions are

satisfied. (i) f(x) is periodic, single valued and finite in

(ii) f(x) has finite number of finite discontinuities and no infinite discontinuity in

(iii) f(x) has a finite number of maxima and minima in

2. Does possess a Fourier expansion?

does not possess a Fourier expansion because the function has the infinite

discontinuity at the point .

3. If

2 n 22

2 2 2 2n 1

π ( 1) 1 1 1 πx 4 cosnx ,deduce that ... [ , ]

3 n 1 2 3 6in π π

∞

=

−= + + + + = −∑

2 n 2 2n

22 2

n 1 n 1

2

2 2 2

P ut x π then

π ( 1 ) π ( 1 )π 4 cosn 4

3 n 3 n

1 1 1 π...

1 2 3 6

π∞ ∞

= =

=

− −= + = +

∴ + + + =

∑ ∑

4. The cosine series for f(x) = x sin x in 0 < x < π is given as

x sin x = 22

1 ( 1)1 cos 2 cos

2 1

n

n

x nxn

∞

=

−− −

−∑ Deduce that .




Given x sin x = 2

2

1 ( 1)1 cos 2 cos

2 1

n

n

x nxn

∞

=

−− −

−∑

Put x = , a point of continuity => 22

1 ( 1)sin 1 cos 2 cos

2 2 2 2 1 2

n

n

n

n

π π π π∞

=

− = − − − ∑

i.e. 22

( 1)1 2 cos

2 1 2

n n

n

π π∞ − = − − ∑

i.e.

5. Find the value of the Fourier series for f(x) = at x=1. (Nov./Dec.2011,Reg 2010).

x=1 is a point of discontinuity, value of Fourier series of f(x) is .

2)1()1( ++− ff

f(1-) =

0 0lim (1 ) lim (1 ) 1h h

f h h→ →

− = − = .

f(1+) = .∴f(x) at x = 1 is =1.5

6. Find the constant term of the Fourier series for the function f(x) = x2 , – ππππ < x < ππππ

Since f(x) is an even function in

– ππππ < x < ππππ

2 20

0 0

2 2 2( )

3a f x dx x dx

π

ππ

= = =∫ ∫�

� ∴ Constant term =

32

20 π=

a

7. To which value, the half range sine series corresponding to f(x) = x2 expressed in the interval (0, 5) converges at x =5 ?

At the end points half range sine series converges to zero. ∴ The half range sine series corresponding to f(x) = x2 converges to zero at x = 5.

8. What you meant by Harmonic analysis ? The process of finding the harmonics in the Fourier series expansion of a function numerically is known as

harmonic analysis. 9. Find the Fourier constant bn for x sin x in xπ π− < < , when expressed as a Fourier series. Solution:

( ) ( ) ( ) ( )sin sin sinf x x x f x x x x x f x= ⇒ − = − − ⇒ =

Here ( )f x is an even function 0nb∴ =

10. Find the root mean square value of the function f(x) = x in .

RMS value = RMS value = [ ]

2 21( )

b

ay f x dx

b a= =

− ∫

11. Obtain the constant term of the Fourier cosine series of y = f(x) in (0,6) using the following table

x: 0 1 2 3 4 5 y: 4 8 15 7 6 2

ao = 2 x Mean value of y =2(Σy)/6= 14 ∴ Constant term = 72

a0 =

12. Find the value of an in the cosine series expansion of f(x) = K in the interval (0,10)

10

0 0

2 1( ) cos cos 0

5 10n

n x n xa f x dx K dx

π π= = =∫ ∫

�

� �




13. If f(x) = x2+x is expressed as a Fourier series in the interval (-2,2) to which value this series converges at x = 2?

Since x = 2 is an end point then f(x) converges to 4

2

26

2

)2()2(=

+=

+− ff

14. State Parseval’s theorem in the interval (c,c+2l).

If the Fourier series corresponding to f(x) converges uniformly to f(x) in (c,c+2l)

then [ ] +=∫+

2

adx)x(f

1 20

2c

c

2�

�( )∑

∞

=

+1n

2n

2n ba

15. Write down the complex form of Fourier Series in (c,c+2l)

∑∞

∞−=

π

=n

xni

nec)x(f � where dxe)x(f2

1c

2c

c

xni

n ∫+ π−

=�

�

�

16. If f(x) = e x in xπ π− < < , find a n .

Solution:

1cosx

na e nxdxπ

ππ −

= ∫ ( )2

1cos sin

1

xenx n nx

n

π

ππ

−

= +

+ ( ) ( )2 2

11 1

1 1n ne e

n n

π π

π

− = − − −

+ +

( )( ) { }2

1

1

n

na e en

π π

π−−

= −+

.

17. Determine bn in the Fourier series expansion of

2)(

xxf

−=π

in 0< x < 2ππππ.

∫∫−

==ππ

ππ 2

0

2

0

sin2

1sin)(

1dxnx

xdx

xnxfbn

�

��

nnn

nx

n

nxx

12

2

1sin)1(

cos)(

2

12

02

==

−−−

−−=π

ππ

π

π

18. Find the constant term in the expression of x2cos as a Fourier series in the interval ),( ππ− . Solution:

12

2sin

2

1

2

2cos11cos

1 20 =

+=+

==−

−− ∫∫π

π

π

π

π

π πππx

xdxx

xdxa ∴∴∴∴Constant term = 2

1

20 =

a

19. Give the expression for the Fourier series coefficient bn for the function f(x) defined in 2 2x− ≤ ≤ .

Solution:

( )2

2

1sin

2 2n

n xb f x dx

π

−

= ∫

20. Determine the value of 0& aan in the Fourier series expansion of 3)( xxf = in ππ <<− x

Solution: 3)( xxf = ⇒⇒⇒⇒ )()()( 33 xfxxxf ==−=− ⇒⇒⇒⇒ )(xf is an odd function ∴ 00 == aan

PART – B 1.

a) If ( )

2

2x

f xπ −

=

in 0 2x π< < . Hence show that




(i) 2

2 2 2

1 1 1.............

1 2 3 6

π+ + + = . (ii)

2

2 2 2 2

1 1 1 1...

1 2 3 4 12

π− + − + = .

Solution:

We know that

( ) ( )0

1

cos sin2 n n

af x a nx b nx

∞

= + +∑

22

0

0

12

xa dx

π ππ

− = ∫

( )2

2

0

1

4x dx

π

ππ

= −∫

( )( )

31

4 3

xππ

−=

−

3 3 21

4 3 3 6

π π ππ −

= + = −

22

0

1cosn

xa nxdx

π ππ α

− = ∫

( )2

2

0

1cos

4x nxdx

π

ππ

= −∫

( ) ( )2

2

2 3

0

1 sin cos sin2 2

4

nx nx nxx x

n n n

π

π ππ − − = − − − − +

2 2

1 cos2 20 2 0 0

4

n

n n

π ππ

π = + + − +

2 2

1 4 1

4 n n

ππ = =

( )2

2

0

1sin

4nb x nxdxπ

ππ

= −∫

( ) ( ) ( )2

2

2 3

0

1 cos sin cos2 2

4

nx nx nxx x

n n n

π

π ππ − − = − − − − +




2 2

3 3

1 2 20

4 n n n n

π ππ

= − + + − =

( )2

21

1cos .

12f x n x

n

π ∞

= +∑

( )2

2 2

1 1cos cos .... 1

12 1 2x x

π = + + + −

0Put x=

( ) ( )2

2 2

1 10 ... 2

12 1 2f

π = + + + −

0x = is a pt of discontinuity.

( )2 2 21

02 4 4 4

fπ π π

∴ = + =

( )2 2

2 2

1 12 ...

4 12 1 2

π π ∴ => = + + +

2 2

2 2

1 1......

4 12 1 2

π π− = + + ∞

2

2 2

1 1.....

1 2 6

π∴ + + ∞ =

Put x π= in (1)

( ) ( ) ( )2

21

13

12

n

f xn

π ∞ −= + −∑

Here π is a pt of continuity.

( ) 0.f π∴ =

( )2

2 2 2

1 1 13 0 .....

12 1 2 3

π=> = − + − + ∞

2

2 2 2

1 1 1.....

12 1 2 3

π− = − + − + ∞




2

2 2 2

1 1 1.....

12 1 2 3

π− = − + + ∞

b) Find the Fourier series for in π .

Solution:

Given

Since is an even function




Substitute in equation (1) we get ∑∞

= π−

+−−+

π=

2n2

n

nxcos)1n(

]1)1[(22)x(f

2. a) Obtain the Fourier series to represent the function f(x) = | x | is xπ π− < < and deduce that 2

2 2 2

1 1 1.............

1 3 5 8

π+ + + = .

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

π

π= ∫

2

0

22x

π

ππ

= =

0

2cosna x nxdx

π

π= ∫

( ){ }2 2 2 20

2 sin cos 2 cos 1 21 1

nnx nx nx

n n n n n

π ππ π π = + = − = − −

0na∴ = if n is even

2

4na

n π= − if n is odd

( ) 21,3

4cos

2f x nx

n

ππ

∞ −∴ = +∑

2

4 cos3cos ...

2 3

xx

ππ− = − + +

0Put x=




( ) 2 2

4 1 10 1 ...

2 3 5f

ππ = − + + +

Here 0 is a pt of continuity

( )0 0f∴ =

2 2

4 1 10 1 ...

2 3 5

ππ = − + + +

2 2

4 1 11 ...

2 3 5

ππ = + + +

2

2 2

1 11 ... .

8 3 5

π ∴ = + + +

b) Find the first two harmonic of the Fourier series of f (x) given by

x 0 1 2 3 4 5 f (x) 9 18 24 28 26 20

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

π π π π = + + + +

( )0

2 1252 41.66

6 6

ya = = =∑

x

3

xπ

2

3

xπ

y cos

3

xy

π sin

3

xy

π

2cos

3

xy

π

2sin

3

xy

π

0 0 0 9 9 0 9 0

1

3

π

2

3

π

18 9 15.7 -9 15.6

2 2

3

π

4

3

π

24 -12 20.9 -12 -20.8

3 π 2π 28 -28 0 28 0

4 4

3

π

8

3

π

26 -13 -22.6 -13 22.6

5 5

3

π

10

3

π

20 10 17.6 -10 -17.4

125 -25 -3.4 -7 0




1

1

2cos 8.33

6 3

2sin 1.15

6 3

xa y

xb y

π

π

= = −

= = −

∑

∑

2

2 2cos 2.33

6 3

xa y

π = = − ∑

2

2 2sin 0

6 3

xb y

π= =∑

( ) 41.66 28.33cos 2.33cos 1.15sin .

2 3 3 3

x x xf x

π π π∴ = − − −

3. a) Find the Fourier series for the function

Solution:

Given

Therefore is an odd function. Therefore

Substitute in (1) we get

b) Find the first two harmonic of the Fourier series of f (x). Given by

x 0

3π

23π

π 4

3π

53π

2π

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 , cos2 , sin , sin 2y x y x y x y x as tabulated

x ( )f x cosx sinx cos2x sin 2x

0 1.0 1 0 1 0

3

π

1.4 0.5 0.866 -0.5 0.866

2

3

π

1.9 -0.5 0.866 -0.5 0.866

π 1.7 -1 0 1 0

4

3

π

1.5 -0.5 -0.866 -0.5 -0.866

5

3

π

1.2 0.5 -0.866 -0.5 -0.866

0 2 2.96

ya = =∑

, 1

cos2 0.37

6

y xa = = −∑

, 2

cos22 0.1

6

y xa = = −∑

1

sin2 0.17

6

y xb = =∑

, 2

sin 22 0.06

6

y xb = = −∑

4. a) Obtain the Fourier series for of period and defined as follows

Hence deduce that

Solution:

Given

The Fourier series is




Deduction (i)

Put

Deduction (ii)

Put

b) Expand as a cosine series in and deduce the value of




Solution: Given The cosine series is

2 2

0

4

if n is even

lif n is odd

n π

= −

Substituting in equation (1) we get

Deduction (i)

By Parseval’s identity




Deduction (ii)

9616

15S

9616

11S

44 ππ=

⇒=

−

5. a) Find the complex form of the Fourier series of in

Solution:

The complex form of the Fourier series in is given by

Where

Hence (1) becomes




b) Find the half range Fourier sine series for ( ) ( )f x x xπ= − in the interval (0, )π and deduce that

∞−+− ......5

1

3

1

1

1333

Solution:

The sine series is ∑∞

=

=1

sin)(n

n nxbxf

dxnxxfbn ∫=π

π 0

sin)(2

dxnxxx∫ −=π

ππ 0

2 sin)(2

π

πππ

032

2 cos)2(

sin)2(

cos)(

2

−+

−−−

−−=n

nx

n

nxx

n

nxxx

[ ]

=−−=

oddisnifn

evenisnif

nn

33 8

0)1(1

4

ππ

∑∞

=

=,....3,2,1

3sin

8)(

n

nxn

xfπ

Deduction:

Put 2

π=x is point of continuity

−+−=

...5

1

3

1

1

18

2 333ππ

f

−+−= ...5

1

3

1

1

18

4 333

2

ππ

32...

5

1

3

1

1

1 2

333

π=−+−

UNIT - III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS PART – A 1. State any two assumptions made in the derivation of the one dimensional wave equation.

i. The string is perfectly elastic and does not offer any resistance to bending.

ii. Deflection y and slope dy/dx at every point of the string are small, so that their higher powers may be neglected.




2. Write down the partial differential equation governing one dimensional wave equation?

2

22

2

2

x

yC

t

y

∂∂

=∂∂

where y(x,t) is the displacement of the string .

3. In the wave equation

2

22

2

2

x

yC

t

y

∂∂

=∂∂

, what does C2 stands for?

C2 = T/m, where T is the tension and m is the mass of the string. 4. Write all possible solutions of the transverse vibration of the string in one dimension.

(i) y(x ,t) = (Aepx + Be – px) (Cepat + De – pat) (ii) y(x,t) = (A cos px + Bsin px) (C cos pat + sin pat). (iii) y(x,t) = (Ax +B) (Ct + D)

5. Write down the appropriate solution of the vibration of string equation. How is it chosen?

The appropriate solution of the vibration of string equation is y(x,t) = (A cos px+ B sin px) ( C cos pat+ D sin pat) . When we deal with the vibration of an elastic string y(x,t) representing the displacement of the string at any point x, it must be periodic in ‘t’. Hence the above solution which consists of periodic functions in‘t’ is the proper solution of the problems on vibration of strings.

6. A tightly stretched string with fixed end points x=0 and x=l is initially at rest in its equilibrium position. If it is set vibrating giving each point a velocity x(l-x), write down the boundary and initial conditions of the above problem.

lxforxlxt

xyandxy

tfortlyandty

≤≤−=∂

∂=

≥==

0)()0,(

0)0,(

00),(0),0(

λ

7. State Fourier law of heat conduction.

The rate at which heat flows through an area is jointly proportional to the area and to the temperature gradient normal to the area.

8. Write down the partial differential equation that r epresents one dimensional heat equation.

,2

22

x

u

t

u

∂

∂=

∂∂

α where 2α = c

k

ρ

9. What are the various possible solutions of one dimensional diffusion equation?

(i) u(x ,t) = tppxpx CeBeAe22

)( α−+

(ii) u(x ,t )= tpCepxBpxA22

)sincos( α−+

(iii) u(x ,t) = Ax + B 10. Why can’t u(x ,t) = tppxpx CeBeAe

22

)( α−+ be the correct solution in solving one dimension heat

equation?

As t→ ∞, u → ∞. It is not possible. 11. In steady state condition derive the solution of one dimension heat flow equation.




The one dimension heat flow equation is 2

22

x

u

t

u

∂∂

=∂∂

α .

In steady state condition, 0=∂∂

t

u. ∴ 02

2

=∂∂x

u ⇒ A

x

u=

∂∂

⇒ BAxu += .

12. An insulated rod of length 60 cm has its ends A and B maintained at oooo20C20C20C20C and oooo80C80C80C80C respectively. Find the steady state temperature in the rod. (Nov. 2012)

Steady state heat equation is 2

20

d u

dx= . The solution is ( )u x Ax B= + ……..(1)

Given conditions are ( ) ( )0 20, 60 80u u= = .

Substituting in (1) we get 20u x= + , which is the steady state temperature in the rod.

13. Write down the partial differential equation that r epresents variable heat flow in two dimensions? Deduce the equations of steady state heat flow in two dimensions.

The heat flow equation in two dimensional Cartesian co-ordinate system is

∂∂

+∂∂

=∂∂

2

2

2

22

y

u

x

u

t

uα When

steady state conditions prevail in the plate the temperature at any point of the plate does not depend on

‘t’, but depends on x and y only (i.e.) 0=∂∂

t

u. Thus steady state temperature distribution in a two

dimensional plate is 02

2

2

2

=∂∂

+∂∂

y

u

x

u (Laplace equation).

14. Write any two solutions of the Laplace equation obtained by the method of separation of variables involving exponential terms in x and y.

u(x ,y) = (Aepx + Be – px) (C cos py + sin py) & u(x ,y)= (A cos px + Bsin px) (C e py + e - py)

15. Given the boundary conditions on a square or rectangular plate, how will you identify the proper solution?

If the non-zero temperature is prescribed either on x=0 or on x=a (vertical edges), the proper solution will be u(x ,y) = (Aepx + Be – px) (C cos py + sin py) If the non-zero temperature is prescribed either on y=0 or y=b (horizontal edges), the proper solution will be u(x ,y)= (A cos px + Bsin px) (C e py + e - py)

16. An infinitely long plate is bounded by two parallel edges and an end at right angles to them. The breath of the edge y = 0 is ππππ and it is maintained at constant temperature u0 at all points and the other edges are kept at zero temperatures. Formulate the boundary value problem to determine the steady state temperature.

2 0u∇ = subject to π

π≤≤==∞

≥==

xforuxuandxu

yforyuyu

0)0,(0),(

00),(,0),0(

0

17. A plate is bounded by the lines x=0, y=0, x=l, and y=l. Its faces are insulated. The edge coinciding with x-axis is kept at oooo100C100C100C100C . The edge coinciding with y-axis is kept at oooo50C50C50C50C . The other two




edges are kept atoooo0C0C0C0C . Write the boundary conditions that are needed for solving two dimensional heat flow equation. (Nov. 2012)

Boundary conditions are

(i) ( ),0 100 ; 0u x C x l= < <�. (ii) ( )0, 50 ;0u y C y l= < <�

.

(ii) ( ), 0 ;0u x l C x l= < <�. (iv) ( ), 0 ; 0u l y C y l= < <� .

18. The boundary value problem governing the steady state temperature distribution in a flat thin

plate is given by 02

2

2

2

=∂

∂+

∂

∂

y

u

x

u, 0)0,( =xu , 0),0( =yu , 0),( =yau and

=a

xaxu

π3sin4),( . Find

Cn when the most general solution is ∑∞

=

=1

sinhsin),(n

n a

yn

a

xncyxu

ππ.

−=

==∑∞

= 4

3sinsin3

4sin4sinhsin),( 3

1

a

x

a

x

a

xn

a

xncaxu

nn

πππ

ππ

a

x

a

x

a

xc

a

xc

a

xc

πππ

ππ

ππ

π 3sinsin3........3sinh

3sin2sinh

2sinsinhsin 321 −=+++

1 3

1 3

2 4 5 6

sinh 3, sinh 3 1,

3 1

sinh sinh 3.......... 0

c c

c c

c c c c

π π

π π

∴ = =

⇒ = ⇒ =

= = = = =

19. Define thermally insulated ends.

The end at which the temperature gradient is zero is called thermally insulated ends. (i.e) there is no heat flow throws the ends.

20. When the heat flow is called two dimensional?

When the heat flow is along curves instead of along straight lines, all the curves lying in parallel planes, then the flow is called two dimensional.

PART B 1. A uniform string is stretched and fastened to two points ''l apart. Motion is started by displacing

the string into the form of the curve )( xlkxy −= and then releasing it from this position at

time 0=t . Find the displacement of the point of the string at a distance x from one end at timet .

Solution:

One dimensional wave equation is 2

22

2

2

x

ya

t

y

∂

∂=

∂

∂

Boundary conditions are (i) 0),0( =ty (ii) 0),( =tly (iii) 0)0,( =∂∂

xt

y (iv) )()()0,( xlkxxfxy −==




The correct solution is

)sincos)(sincos(),( patDpatCpxBpxAtxy ++= (1)

Put x = 0 and apply the boundary condition (i) in (1)

=),0( ty 0)sincos)(( =+ patDpatBA ⇒ .

Equation (1) becomes )sincos(sin),( patDpatCpxBtxy += (2)

Put x = l and apply the boundary condition (ii) in (2)

0)sincos(sin),( =+= patDpatCplBtly ⇒ 0sin =pl πnpl =⇒ ⇒

Equation (2) becomes

+=l

atnD

l

atnC

l

xnBtxy

πππsincossin),( (3)

Differentiate Partially (3) w. r. to t

=∂∂

),( txt

y

+−l

atn

l

anD

l

atn

l

anC

l

xnB

πππππcossinsin (4)

Put t = 0 and apply the boundary condition (iii) in (4)

0sin)0,( ==∂∂

l

anD

l

xnBx

t

y ππ ⇒


l

atn

l

xnB

l

atnC

l

xnBtxy

n

ππ

ππ

cossin

cossin),(

=

= where BCBn =

By the superposition principle, the most general solution is

∑∞

=

=1

cossin),(n

n l

atn

l

xnBtxy

ππ (5)

Put t = 0 and apply the boundary condition (iv) in (5)

l

np

π=

0=A

0=D




)(1.sin)0,(1

xfl

xnBxy

nn ==∑

∞

=

π (6)

By half range Fourier sine series ∑∞

=

=1

sin)(n

n l

xnbxf

π (7)

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

( ) ( ) ( )

[ ]

[ ]

n

n

n

l

l

l

n

B

oddisnifn

kl

evenisnifn

kl

n

l

l

k

n

ln

n

l

l

k

l

nl

xn

l

nl

xn

xl

l

nl

xn

xlxl

k

dxl

xnxlkx

l

dxl

xnxf

lb

=

=

−−=

−−=

+−+−+=

−+

−

−−

−−=

−=

=

∫

∫

33

2

33

2

33

3

33

3

33

3

03

33

2

222

0

0

8

0

)1(14

)1(122

1..200cos.2002

cos2

sin2

cos2

sin)(2

sin)(2

π

π

π

ππ

π

π

π

π

π

π

π

π

π

∴The equation (5) becomes ∑∞

=

=5,3,1

33

2

cossin8

),(n l

atn

l

xn

n

kltxy

πππ

2. If a string of length l is initially at rest in its equilibrium position and each of its points is given a velocity

v such that

<<−

<<=

lxl

xlc

lxcx

v

2;)(

20;

, find the displacement at any timet .

Solution:

One dimensional wave equation 2

22

2

2

x

ya

t

y

∂

∂=

∂

∂

The boundary conditions are

nn Bb =




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

(iv)

<<−

<<=

∂∂

lxl

xlc

lxcx

xt

y

2);(

20;

)0,(

The correct solution is

)sincos)(sincos(),( patDpatCpxBpxAtxy ++= (1)

Put x = 0 and apply the boundary condition (i) in (1)

=),0( ty 0)sincos)(( =+ patDpatBA ⇒ .

Equation (1) becomes )sincos(sin),( patDpatCpxBtxy += (2)

Put x = l and apply the boundary condition (ii) in (2)

0)sincos(sin),( =+= patDpatCplBtly ⇒ 0sin =pl πnpl =⇒ ⇒


+=l

atnD

l

atnC

l

xnBtxy

πππsincossin),( (3)

Put t = 0 and apply the boundary condition (iii) in (3)

=)0,(xy 0sin =Cl

xnB

π ⇒ C = 0


l

atn

l

xnB

l

atnD

l

xnBtxy

n

ππ

ππ

sinsin

sinsin),(

=

= where BDBn =

By the superposition principle, the most general solution is

∑∞

=

=1

sinsin),(n

n l

atn

l

xnBtxy

ππ (4)

Differentiate Partially (4) w.r.to t

l

np

π=

0=A

C = 0




=∂∂

),( txt

y

l

an

l

atn

l

xnB

nn

πππ∑∞

=1

cossin (5)

Apply the boundary condition (iv) in (5)

=∂∂

)0,(xt

y)(

2);(

20;

sin1

xf

lxl

xlc

lxcx

l

an

l

xnB

nn =

<<−

<<=∑

∞

=

ππ (6)

By half range Fourier sine series ∑∞

=

=1

sin)(n

n l

xnbxf

π (7)

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

lB

l

anBb

ππ

=⇒=

( )

−

−−

−−+

−

−

−

=

−+=

=

∫∫

∫

l

l

l

n

l

l

l

l

l

l

nl

xn

l

nl

xn

xl

l

nl

xn

l

nl

xn

xl

c

dxl

xnxldx

l

xnx

l

c

dxl

xnxf

lb

2

2

2

2

2

22

02

22

0

0

sin)1(

cos)(

sin1

cos2

sin)(sin2

sin)(2

π

π

π

π

π

π

π

π

ππ

π

2sin

4

2sin

2cos

22sin

2cos

2

2

22

22

22

22

22

ππ

ππ

ππ

ππ

ππ

n

n

clb

n

n

ln

n

ln

n

ln

n

l

l

c

n =

+++−=

2sin

4

2sin

433

2

22

ππ

πππ

n

an

cln

n

cl

an

lBn ==∴

∴The equation (4) becomes




∑∞

=

=1

33

2

sinsin2

sin4

),(n l

atn

l

xnn

an

cltxy

ππππ

3. The ends A and B of a rod l cm long have their temperatures kept at C�30 and C�80 , until steady state conditions prevail. The temperature of the end B is suddenly reduced to C�60 and that of A is increased to C�40 . Find the steady state temperature distribution in the rod after time t.

Solution:

One dimensional heat equation 2

22

x

ua

t

u

∂

∂=

∂∂

(1)

In the steady state condition the 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) 8030)( =+= allu l

a50

=⇒

Then the equation (3) becomes 3050

)( +=l

xxu

Now consider the unsteady state condition.

In unsteady state the correct solution is

tpaCepxBpxAtxu22

)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 correct solution is tpaCepxBpxAtxu22

)sincos(),( −+= (8)

Apply the boundary condition (vi) in (8)

0),0(22

== − tpat ACetu

here 0=A

∴the equation (8) becomes tpat CepxBtxu

22

sin),( −= (9)

Apply the boundary condition (vii) in (9)




0sin),(22

== − tpat pleBCtlu

l

np

pl

π=⇒

= 0sin

∴the equation (9) becomes t

l

na

t Cexl

xnBtxu

2

222

sin),(ππ −

=

t

l

na

n

tl

na

t

el

xnB

el

xnBCtxu

2

222

2

222

sin

sin),(

π

π

π

π

−

−

=

= where nBBC =

By the super position principle, the most general solution is ∑∞

=

−

=1

2

222

sin),(n

tl

na

nt el

xnBtxu

ππ (10)

Apply the boundary condition (viii) in (10)

∑∞

=

=−==1

)(1030

1.sin)0,(n

nt xfl

x

l

xnBxu

π

Half range Fourier sine series of )(xf is ∑∞

=

=1

sin)(n

n l

xnbxf

π (11)

From the equations (10) & (11) we get nn Bb =

−−

=

−

−

−

−=

−=

=

∫

∫

1.10

cos202

sin30

cos10

302

sin10302

sin)(2

02

22

0

0

ππ

π

π

π

π

π

π

π

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

Bn

=

−+−

= )1(2120

π

∴the equation (10) becomes

[ ]∑∞

=

−

−+−

=1

2

222

sin)1(2120

),(n

tl

nan

t el

xn

ntxu

πππ

Then the required temperature distribution function is

[ ]∑∞

=

−

−+−+=1

2

222

sin)1(2120

4020

),(n

tl

nan e

l

xn

nl

xtxu

πππ

4. A rectangular plate is bounded by the lines byaxyx ==== ,,0,0 . Its surfaces are insulated. The

temperature along 0=x and 0=y are kept at C�0 and the others at C�100 . Find the steady state temperature at any point of the plate.

Solution:

We know that two dimensional heat equation 02

2

2

2

=∂

∂+

∂

∂

y

u

x

u

Let l be the length of the square plate


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

( ) ( , ) 100

i u y ii u x iii u a y

iv u x b

= = =

=

We write the temperature function 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)((),(1 pyDpyCBeAeyxu pxpx ++= − (1)

Apply the boundary condition (v) in (1)

0)sincos)((),0(1 =++= pyDpyCBAyu

here 0=+ BA

AB −=

then the equation (2) becomes

)sincos)((

)sincos)((),(1

pyDpyCeeA

pyDpyCAeAeyxupxpx

pxpx

+−=

+−=−

−

(2)

Apply the boundary condition (vi) in (2)

0)()0,(1 =−= − CeeAxu pxpx

here 0=C


pyDeeAyxu pxpx sin)(),(1−−= (3)

Apply the boundary condition (vii) in (3)

0sin)(),(1 =−= − pbDeeAbxu pxpx

here 0sin =pb πnpb=⇒b

np

π=⇒


b

yn

b

xnB

b

yn

b

xnAD

b

ynDeeAyxu

n

b

xn

b

xn

ππ

ππ

πππ

sinsinh

sinsinh

sin)(),(1

=

=

+=−

where ADBn 2=




The most general solution is

b

yn

b

xnByxu

nn

ππsinsinh),(

11 ∑

∞

=

= (4)

Apply the boundary condition (viii) in (4)

)(100sinsinh),(1

1 yfb

yn

b

anByau

nn ===∑

∞

=

ππ (5)

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

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

∑∞

=

=1

sin)(n

n b

ynbyf

π (6)

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

b

anb

Bb

anBb n

nnn ππ

sinhsinh =⇒=

( )

[ ]

=

−+=

+−=

−=

=

=

∫

∫

oddisnifn

evenisnifn

nn

b

b

b

nb

yn

b

dyb

yn

b

dyb

ynyf

bb

n

b

b

b

n

π

π

ππ

π

π

π

π

400

0

)1(1200

1cos200

cos200

sin1002

sin)(2

0

0

0




∴ππ nn

Bn sinh

400= if n is odd


b

yn

b

xn

b

ann

yxun

πππ

πsinsinh

sinh

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(),(2pypy HeGepxFpxEyxu −++= (7)

Apply the boundary condition )(ix in (7)

0)(),0(2 =+= − pypy HeFeEyu

here 0=E


)(sin),(2pypy GeGepxFyxu −+= (8)

Apply the boundary condition )(x in (8)

0)(sin)0,(2 =+= HGpxFxu

here GHHG −=⇒=+ 0


)(sin

)(sin),(2

pypy

pypy

eepxFG

GeGepxFyxu−

−

−=

−= (9)




Apply the boundary condition )(xi in (9)

0)(sin),(2 =−= − pypy eepaFGyau

here 0sin =pa πnpa =⇒a

np

π=⇒


a

xn

a

ynC

a

xn

a

ynFG

eea

xnFGyxu

n

a

yn

a

yn

ππ

ππ

π ππ

sinsinh

sinsinh2

)(sin),(2

=

=

−=

where FGBn 2=


a

xn

a

ynCyxu

nn

ππsinsinh),(

12 ∑

∞

=

= (11)

Apply the boundary condition )(xii in (11)

)(100sinsinh),(1

2 xfa

xn

a

bnCbxu

nn ===∑

∞

=

ππ

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

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

∑∞

=

=1

sin)(n

n a

xndxf

π (12)

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




a

bnd

Ca

bnCd n

nnn ππ

sinhsinh =⇒=

a

a

a

n

a

na

xn

a

dxa

xn

a

dxa

xnxf

ad

0

0

0

cos200

sin1002

sin)(2

−=

=

=

∫

∫

π

π

π

π

( )

[ ]

=

−+=

+−=

oddisnifn

evenisnif

n

nn

a

a

n

π

π

ππ

400

0

)1(1200

1cos200

∴

a

bnn

Cn ππ sinh

400= if n is odd


a

xn

a

yn

nnyxu

n

ππππ

sinsinhsinh

400),(

5,3,12 ∑

∞

=

=

Then the equation (A) becomes

a

xn

a

yn

nnb

yn

b

xn

nn

yxuyxuyxu

nn

ππππ

ππππ

sinsinhsinh

400sinsinh

sinh

400

),(),(),(

5,3,15,3,1

21

∑∑∞

=

∞

=

+=

+=

5. A rectangular plate with insulated surface is 10cm wide and so long compared to its width that it

may be considered infinite in length without introducing appreciable error. The temperature at

short edge 0=y is given by 20 , 0 5

20(10 ), 5 10

x xu

x x

≤ ≤=

− ≤ ≤ and all the other three edges are kept




at C�0 . Find the steady state temperature at any point in the plate.

Solution:

We know that two dimensional heat equation 02

2

2

2

=∂

∂+

∂

∂

y

u

x

u


≤≤−

≤≤=

=∞==

105),10(20

50,20)0,()(

0),()(0),10()(0),0()(

xx

xxxuiv

xuiiiyuiiyui

The suitable solution which satisfying the given boundary conditions is

))(sincos(),( pypy DeCepxBpxAyxu −++= (1)

Apply the boundary condition (i) in (1)

( ) 0),0( =+= − pypy DeCeAyu

here 0=A

Then the equation (1) becomes

( )pypy DeCepxByxu −+= sin),( (2)

Apply the boundary condition (ii) in (2)

( ) 010sin),10( =+= − pypy DeCepByu

here 010sin =p πnp =⇒ 1010

πnp =⇒


+=

−

1010sin),(ynyn

DeCea

xnByxu

πππ (3)

Apply the boundary condition (iii) in (3)




( ) 010

sin),( =+=∞ ∞−∞ DeCexn

Bxuπ

here 0=C


10

10

sin

sin),(

yn

n

yn

ea

xnB

ea

xnByxu

π

π

π

π

−

−

=

=


10

1 10sin),(

yn

nn e

xnByxu

ππ −∞

=∑= (4)

Apply the boundary condition (iv) in (4)

∑∞

=

==1

1.10

sin)0,(n

n

xnBxu

π

≤≤−

≤≤

105),10(20

50,20

xx

xx

To find nB , expand )(xf in Half range Fourier sine series.

We know that Half range Fourier Cosine series

∑∞

=

=1 10

sin)(n

n

xnbxf

π (5)

From (4) and (5) we get nn bB =

−+=

=

∫∫

∫10

5

5

0

10

0

10sin)10(20

10sin20

5

1

10sin)(

10

2

dxxn

xdxxn

x

dxxn

yfbn

ππ

π




−

−−

−−+

−

−

−=

10

5

22

5

0

22

100

10sin

)1(

10

10cos

)10(

100

10sin

)1(

10

10cos

5

20

π

π

π

π

π

π

π

π

n

xn

n

xn

xn

xn

n

xn

x

2sin

800

2sin

100

2cos

50

2sin

100

2cos

504

22

2222

ππ

ππ

ππ

ππ

ππ

n

n

n

n

n

n

n

n

n

nbn

=

+++−

=


∑∞

=

−

=1

1022 10

sin2

sin800

),(n

yn xne

n

nyxu

πππ

π

UNIT IV – FOURIER TRANSFORMS PART - A

1. Write the Fourier transform pair. [Nov/Dec 2011, 2010] The Fourier transform pair is defined as

( )[ ] ( ) ( )sFdxxfisxexfF =∫∞

∞−=

π2

1 : ( ) ( ) ( )11

2isxF F f x e F s ds f x

π

∞− − = = ∫ −∞

2. State Fourier integral theorem. [Nov/Dec 2011] If ( )xf is piecewise continuous, continuously differentiable and absolutely integrable in ( )∞∞− , then

( ) ( ) ( ) dtdstxisetfxf −∫∞

∞−∫∞

∞−=

π21

3. Find the Fourier transform of

>

<=

2xfor;0

2xfor;1f(x)

The Fourier transform of the function )(xf is

[ ] ∫∞

∞−= dxisxexfxfF )(

2

1)(

π.

[ ]2 2 221 1 1

( ) 1.2 2 22 2

isx i s i se e eisxF f x e dxis is isπ π π

− ∴ = = = −∫ − −

2 21 1 1 2 sin 2

(2 sin 2 ) ( )2 2

i s i se e si s F s

is is sππ π

− − = = ∴ =

4. Find )(xf from the integral equation sf(x)cos x e

0s

∞ −=∫




( ) cos0

sgiventhat f x sxdx e∞ −=∫ ,

20

2 2cos

0

2 2 2 2 1( ) ( ) cos cos

10

sc

s se sx dx e

f x F s sx ds e sx dsx

π π

π π π π

∞−

∞ − −=∫

∞ = = =∫ + ∫

5. Define Self-reciprocal Fourier transform and give an example. [ Nov / Dec 2013]

If ( )f s is the Fourier transform of( )xf , then ( )xf is said to be self-reciprocal under Fourier

transform.2 2

2 2x s

F e e − − =

.

6. Find the Fourier transform of |x| ae− , if a 0 [ Nov / Dec 2012 ]

[ ] 1 1( ) (cos sin )

2 2

a x a xisxF f x e e dx e sx i sx dxπ π

∞ ∞− −∴ = = +∫ ∫

−∞ −∞

1 1 2cos sin 2 cos

2 22 2 ( )0

a x a x aaxe sxdx i e sxdx e sxdxs aππ π

∞ ∞ ∞ − − − = + = =∫ ∫ ∫ + −∞ −∞

7. State Convolution theorem in Fourier Transform. [Nov / Dec 2012 ] The Fourier transform of the convolution of ( )xf and ( )xg is the product of their Fourier transforms

i.e. ( ) ( ){ } ( ){ } ( ){ }xgFxfFxgxfF =∗ 8. State Parseval’s identity on Fourier Transform. [Nov/Dec 2011 ]

)()]([,2)(2)( sFxfFwheredxxfdssF =∫∞

∞−∫∞

∞−=

9. Find ( ){ }xxfF from ( )( ) 2 2

2 aF f x

π s a = +

.

By property [ ] ( )[ ]xfFds

dixxfF −=)( [ ] [ ]xaxa eF

ds

dixeF −− −=∴

( ) ( )2 22 2 2 2 2 2

2 2 .2 2 2d a a s asi i ids s a s a s aπ π π

− = − = = − == + + +

10. Find the Fourier sine transform of

a xe

x

− where 0>a .

[ ]

0

2( ) ( )sinSF f x f x sxdx

π

∞

= ∫ .

0

2sin

ax ax

S

e eF sxdx

x xπ

∞− − =

∫

Differentiate both sides with respect tos ,




2 2sin sin

0 0

2 2 2cos cos

2 20 0

ax ax axd e d e eF sxdx sxdxsds x ds x x s

axe aaxx sxdx e sxdxx s a

π π

π π π

− − − ∞ ∞ ∂ = =∫ ∫

∂ −∞ ∞ −= = =∫ ∫

+

Now integrating both sides w. r. tos ,

=+

=

−−

∫ a

sds

as

a

x

eF

ax

S1

22tan

22

ππ

11. Find the Fourier sine transform of axe− , 0>a . Hence find

− xaxesF . [ May / June 2012 ]

[ ] ∫∞

=0

sin)(2

)( sxdxxfxfsFπ

.

22

2

0sin

2

as

ssxdxaxeaxe

sF

+π=∫

∞ −π

=

−

We know that by property ( )[ ] ( )[ ]xfFds

dxxfF CS −=

+

−=

+

−+−=

+−=

−∴

222

222222

)2.(22222

2

as

as

as

ssas

as

s

ds

daxxesFπππ

12. Write the Fourier sine transform pair and Fourier Cosine transform pair. The Fourier sine transform pair is defined as

( )[ ] ( )sFsxdxxfxfF ss =π

= ∫∞

0

sin)(2

: [ ] ( ) ( )21 sin0

F F s f x sx F s dss sπ

∞− = = ∫

The Fourier cosine transform pair is defined as

( )[ ] ( )sFsxdxxfxfF cc =π

= ∫∞

0

cos)(2

: [ ] ( ) ( )1

0

2cosc cF F s f x sx F s ds

π

∞− = = ∫

13. Prove that ( )( )c c

1 sF f ax F , 0

a aa = ≠

( )[ ] ∫∞

π=

0

cos)(2

sxdxaxfaxfFc , a

dtdxdtadxtaxPut === ,,

∞=∞=== txandtxwhen ,0,0 ( )0

2( ) cosc

st dtF f ax f t

a aπ

∞ = ∫

( )0

1 2( ) cosc

sF f ax f t t dt

a aπ

∞ = ∫ =

a

sF

a c1

.

14. State Parseval’s identity in Fourier sine and cosine Transform.

∫∫∞∞

=0

2

0

2 )()( dxxfdssFc

& ∫∫∞∞

=0

2

0

2 )()( dxxfdssFs




15. Find the Fourier Sine transform of

x1

. [Nov/Dec 2011, Nov / Dec 2009]

0 0

1 2 sin 2 sin 2

2 2ssx t

F dx dtx x t

π ππ π π

∞ ∞ = = = = ∫ ∫

16. If )())(()())(( sFxfFandsFxfF sscc == , prove that [ ]1( ( )sin ) ( ) ( )

2c s sF f x ax F a s F a s= + + −

[Nov/Dec 2011, MA1201] Solution:

0 0

0 0

2 1 2( ( )sin ) ( ) sin cos ( )[sin( ) sin( ) ]

2

1 2 2( )[sin( ) ] ( )[sin( ) ]

2

cF f x ax f x ax sx dx f x s a x a s x dx

f x a s x dx f x a s x dx

π π

π π

∞ ∞

∞ ∞

= = + + −

= + + −

∫ ∫

∫ ∫

[ ]1( ( )sin ) ( ) ( )

2c s sF f x ax F a s F a s= + + − .

17. State and prove the change of scale property of Fourier Transform. [Apr/May 2011,May/June 2013] Statement:

If F [f(x)] = F(s) then 1

( ( )) , 0s

F f ax F aa a

= ≠

Proof:

( )[ ] ∫∞

∞−

= dxeaxfaxfF isx)(2

1

π ,

If a > 0 a


∞=⇒∞=−∞=⇒−∞= txandtxwhen ( ) 1( )

2

si t

a dtF f ax f t e

aπ

∞

−∞

⇒ = ∫ =

a

sF

a

1. –(1)

If a < 0 a


when x t and x t= −∞ ⇒ = ∞ = ∞⇒ = −∞

( ) 1 1( ) ( )

2 2

s si t i t

a adt dtF f ax f t e f t e

a aπ π

−∞ ∞

∞ −∞

−⇒ = = ∫ ∫ =

a

sF

a

1. ---(2)

From (1) & (2) we get 1

( ( )) , 0s

F f ax F aa a

= ≠

18. Find the Fourier cosine transform of

<<−

=otherwise,0

1x0;x1f(x)

2.




[ ] ∫∞

π=

0

cos)(2

)( sxdxxfxfFC

[ ]

−

π=

+

−π

=

−−+

−−−−

π=

−π

=∴ ∫

332

1

032

2

1

0

2

cossin22

sin2cos22

sin)2(

cos)2(

sin)1(

2

cos)1(2

)(

s

sss

s

s

s

s

s

sx

s

sxx

s

sxx

sxdxxxfFc

19. Find the Fourier Sine Transform of 0,)( >= − xexf x [Apr/May2011, MA1201] We know that,

[ ] 2( ) ( )sin

0F f x f x sxdxs π

∞= ∫ .

( ) 2

0

2 2 2sin sin cos

2 110

xe sx xF e e sxdx sx s sxs ssπ π π

∞−∞ − − = = − − =∫ + +

20 If F(s) is the Fourier transform of f (x), then show that )()}({ sFeaxfF ias=−

[ Nov/Dec 2013, 2010 ,May/June 2012]

F {f(x-a)}= ∫∞

∞−

− dxeaxf isx)(2

1

π= ∫

∞

∞−

+ dtetf stai )()(2

1

π where x - a = t

= 1

( )2

ias itse f t e dtπ

∞

−∞∫ = F(s) = iase F{f(x)}

PART - B

1 Show that the Fourier transforms of

>>

<−=

0;0

;)(

22

ax

axxaxf is 3

2 sinsa sacossa2

s−−−−

ππππ . Hence

deduce that4

cossin

03

π=

−∫∞

dtt

ttt. Using Parseval’s Identity, show that

15

cossin

0

2

3

π=

−∫∞

dtt

ttt.

The Fourier transform of the function )(xf is [ ] ∫

∞

∞−

= dxexfxfF isx)(2

1)(

π.

Given 22)( xaxf −= in axa <<− and 0 otherwise.




[ ]

[ ]

−

π=

+−−

+−

π=

−−+

−−−

−π

=

+−π

=

−+−π

=

+−π

=−π

=∴

∫

∫ ∫

∫∫

− −

−−

3

32

032

22

0

22

2222

2222

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

dxsxisxxadxexaxfF

a

a

a

a

a

a

a

a

a

a

isx

By inverse Fourier transform ∫∞

∞−

−= dsesFxf isx)(2

1)(

π

−

−

=

−

−

−

=

−

−

=

−

=

∫

∫ ∫

∫

∫

∞

∞

∞−

∞

∞−

∞

∞−

∞

∞−

−

03

33

3

3

0coscossin

22

)(

sincossin

coscossin2

)sin(coscossin2

cossin22

2

1)(

sxdss

sasasaxf

sxdss

sasasaisxds

s

sasasa

dssxisxs

sasasa

dses

sasasaxf isx

π

π

π

ππ

Put 1&0 == ax , we get

4

cossin.

cossin41

03

03

π

π

=

−

−

=

∫

∫∞

∞

dss

sssei

dss

sss

Replacesbyt , we get 4

cossin

03

π=

−∫∞

dtt

ttt.

By Parseval’s Identity,




)()]([,2)(2)( sFxfFwheredxxfdssF =∫∞

∞−∫∞

∞−=

a

a

a

a

a

a

xxaxads

s

sasasa

dxxxaadss

sasasa

dxxxaadss

sasasa

dxxadss

sasasa

0

5324

2

30

0

42242

30

42242

3

2222

3

532

cossin8

)2(2cossin8

2

)2(cossin8

)(cossin2

2

+−=

−

+−=

−

×

+−=

−

−=

−

∫

∫∫

∫∫

∫∫

∞

∞

−

∞

∞−

−

∞

∞−

π

π

π

π

.15

dtt

tcosttsin

15

a

a

dt

a

t

tcosttsin

0t,0sandt,swhen,a

dtds,

a

tstasPut

15

ads

s

sacossasasin

15

a3a10a15

85

aa

3

2a

8ds

s

sacossasasin

2

30

5

2

3

30

52

30

555555

2

30

π=

−

⇒π

=

−

==∞=∞===⇒=

π=

−

+−π=

+−

π=

−

∫∫

∫

∫

∞∞

∞

∞

2 a) Find the Fourier sine transform of

x

e xa−

where 0>a . [Nov/Dec 2011, 181301]

∫∞ −

−

=∴

=

0

sin2

)(

))((

dssxx

exf

x

exfF

ax

ax

S

π

Differentiating both sides w.r.to ‘s’

( )22

022

00

2sincos

2

cos2

)(cos2

sa

asxssxa

sa

e

ds

df

dssxedsxsxx

e

ds

df

ax

axax

+⇒

+−

+=

==

∞−

∞−

∞ −

∫∫

ππ

ππ




Integrating w.r.to ‘s’

π

=∴

=⇒==

+

π

⇒+

π

⇒

+π=

−

−−∫

a

sxf

ccfxPut

ca

sc

a

s

aa

sa

dsaxf

1

1122

tan2

)(

0)0(,0

tan2

tan1

.22

)(

b) Show that 2

2x

e−

is self-reciprocal with respect to the Fourier cosine Transform.

The Fourier cosine transform of the function )(xf is [ ] ∫

∞

=0

cos)(2

)( sxdxxfxfFC π.

∫∫∫∞

−∞

∞−

−∞

−−===

∴

0

2

2

2

2

0

2

2

2

2

)(ofR.P2

1cos

2

12cos

2dxeesxdxesxdxeeF isx

xxxx

Cπππ

∫∫∞

∞−

+−∞

∞−

−== dxedxee

isxx

isxx

2

2

2

2

2

1ofR.P

2

1ofR.P

ππ

∫∫∞

∞−

−

−−∞

∞−

−−

== dxedxe

isisxisx

x2

2

2

222

2

2

1ofR.P

2

1ofR.P

ππ

∫∞

∞−

−−

−

π= dxeeofR.P

isxs

22

22

2

1

Put y2

isx=

−, then dydx 2= , ∞=⇒∞=−∞=⇒−∞= yxandyx

( )[ ] 2

2

2

2

22

22

2

2

1ofR.P

1ofR.P2

2

1ofR.P

s

C

s

ys

ys

exfFe

dyeedyee

−−

∞

∞−

−−

∞

∞−

−−

=⇒=

== ∫∫

ππ

ππ

∴ 2

2x

e−

is self-reciprocal with respect to Fourier transform

3 a) Find the Fourier transform of

≥

<−=

1;0

1;1)(

xif

xifxxf . Hence Evaluate dt

t

t∫∞

04

4sin.


∞

∞−

= dxexfxfF isx)(2

1)(

π.

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

2cos)1(

sin)1(

2

0cos)1(22

1sin)1(cos)1(

2

1

)sin)(cos1(2

1)1(

2

1)(

s

ssF

ss

s

s

sx

s

sxx

sxdxxsxdxxisxdxx

dxsxisxxdxexxfF isx

π

ππ

ππ

ππ


∞−

−= dsesFxf isx)(2

1)(

π

−

−

=

−

−

−

=

−

−

=

−

=

∫

∫ ∫

∫∫

∞

∞

∞−

∞

∞−

∞

∞−

∞

∞−

−

02

22

22

0coscos1

21

)(

sincos1

coscos11

)sin(coscos11cos12

2

1)(

sxdss

sxf

sxdss

sisxds

s

s

dssxisxs

sdse

s

sxf isx

π

π

πππ

Put 0=x , we get

( )1.

22sin2

.

cos121

02

2

02

π

π

=

−

=

∫

∫

∞

∞

dss

sei

dss

s

Let ,2

ts= then dtds 2= ∞=⇒∞==⇒= tsandts 00

( ) 2

sin

22

2

sin2

02

2

02

2 ππ=

⇒=

∫∫∞∞

dtt

tdt

t

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

∞

−∞ −

∞

−∞

∞

−∞

− ∴ = − π

− = − π

− = − π

∫ ∫

∫ ∫

∫ ∫




( )2 1

22

0 0

2 1 coss2 ds 2 1 x 2x dx

s

∞ − = + − π ∫ ∫

12 3 2

20 0

2 1 coss x 2xds x

s 3 2

∞ − = + − π ∫

[ ]( )

∫∫∞∞

=

⇒

−+−

−+=

−

0

2

2

2

0

2

2 32sin2

200013

11

cos12 ππ

dss

sds

s

s

Let ,2

ts= then dtds 2= ∞=⇒∞==⇒= tsandts 00

( ) 3

sin

3

sin

32

2

sin22

04

4

0

2

2

2

0

2

2

2 πππ=⇒=

⇒=

∫∫∫∞∞∞

dtt

tdt

t

tdt

t

t

b) Evaluate ∫

∞

++02222 )()( bxax

dx using Fourier Cosine Transforms of , , 0a x b xe and e a b− − > .

The Fourier cosine transform of the function )(xf is [ ] ∫

∞=

0C sxdxcos)x(f

2)x(fF

π.

[ ]

22

0

2

cos2

as

a

sxdxeeF axaxC

+=

=∴ ∫∞

−−

π

π and

[ ]

22

0

2

cos2

bs

b

sxdxeeF bxbxC

+=

=∴ ∫∞

−−

π

π

Let ( ) axexf −= and ( ) bxexg −=

By Convolution theorem, ( )[ ] ( )[ ] ( ) ( )∫∫∞∞

=00

dxxgxfdsxgFxfF cc

( )( )( )∫∫

∫∫∞

+−∞

∞−−

∞

=++π

=+π+π

∴

002222

002222

2..

22

dxedsasas

abei

dxeedsas

b

as

a

xba

bxax

( )( )( )

( )

( )( )( ) ( )

( )

+−−

=++

+−=

++

+−∞+−∫∞

∞+−∫∞

ba

ee

asas

dsab2

ba

e

asas

dsab2

0baba

02222

0

xba

02222

π

π




( )( ) ( )

( )( ) ( )baab2asas

ds

ba

1

asas

dsab2

02222

02222

+=

++

+−−

=++

∫∞

∫∞

π

π

Replacesbyx , we get

( )( ) ( )baabaxax

dx

+=

++∫∞

202222

π

4 a) Find the Fourier transforms of )( xf defined by

>

<=

ax,0

ax,1)x(f and hence find the value

of ∫∞

0

dtt

tsin also prove that∫

∞ π=

02

2

2dt

t

tsin.


∞

∞−= dxe)x(f

2

1)x(fF isx

π.

Given 1)( =xf in axa <<− and 0 otherwise.

[ ]

s

sasFasi

is

is

ee

is

e

is

e

is

edxexfF

iasiasiasiasa

a

isxa

a

isx

sin2)(sin2

1

2

1

2

1

2

1

2

1.1

2

1)(

π=⇒

π=

−

π=

−

π=

π=

π=∴

−−

−−∫


∞−

−= dsesFxf isx)(2

1)(

π

−=

−=

−==∴

∫∞

∫∞

∞−∫∞

∞−

∫∞

∞−∫∞

∞−

−

0sxdscoss

sasin2

1sxdssin

s

sasinisxdscos

s

sasin1

ds)sxsinisx(coss

sasin1dse

s

sasin2

2

1)x(f

0

isx

ππ

πππ

∫∞

=0

sxdscoss

sasin2)x(f

π

Put 1&0 == ax we get

∫∫∞∞

=⇒=00 2

sinsin21

ππ

dss

sds

s

s, Replace sby t, we get ∫

∞

=0 2

sin πdt

t

t.

By Parseval’s identity ∫∫∞

∞−

∞

∞−

= dxxfdssF22

)()(




[ ]

[ ]2

sin2

sin2

2)1(1

sin2

sin21

sin2

0

2

0

22

11

21

1

2

2

πππ

ππ

∫∫∫

∫∫∫∞∞∞

∞−

−

∞

∞−−

∞

∞−

=

⇒=

⇒−−=

=⇒=∴

dss

sds

s

sds

s

s

xdss

sdxds

s

s

Replace sby t, we get

∫∫∞∞

=⇒=

02

2

0

2

2

sin

2

sin ππdt

t

tdt

t

t.

b) Find Fourier Sine and Cosine transforms of axe− and hence find the Fourier sine transforms of

22 ax

x

+ and Fourier cosine transforms of

22

1

ax +.

The Fourier sine transform of the function )(xf is [ ] ∫

∞

=0

sin)(2

)( sxdxxfxfFS π.

[ ]22

0

2sin

2

as

ssxdxeeF axax

s+π

=π

= ∫∞

−−

The Fourier cosine transform of the function )(xf is [ ] ∫∞

π=

0

cos)(2

)( sxdxxfxfFc .

[ ]22

0

2cos

2

as

asxdxeeF axax

c+π

=π

=∴ ∫∞

−−

asas

ascs

asas

c

eaa

ee

ads

d

axF

ds

d

ax

xF

a

ee

asxdx

axaxF

−−

−

−−

∞

π=−

π−⇒

π

π−=

+−=

+

π⇒

π

π⇒

+π=

+∴ ∫

2)(

22

21

.22

2cos

121

2222

02222

5

a) Find the function f(x) if its sine transform is

s

e as−

[ Nov / Dec 2013 ]

Let

s

e as−

=(f(x))Fs

Then f(x) = )1(sin2

0

−−−−−∫∞ −

dssxs

e as

π

Differentiating both side with respect to ‘x’ we get

∫∫∞

−∞ −

==00

cos2

)()(cos2

dssxedsssxs

e

dx

df asas

ππ

+

=

+−

+==

∞−

220

22

2)sincos(

2

ax

asxxsxa

xa

e

dx

df as

ππ




Integrating w.r.t ‘ x ’

f(x) = ca

x

a

x

aa

ax

dxa +

=

=+

−−∫ 1122

tan2

tan122

πππ -------(2)

put x= 0, f(0) = 0 from (1) from (2) , we get f(0) = c =0,

∴

= −

a

xxf 1tan

2)(

π

b) Express

>

<=

1||,0

1||,1)(

x

xxf as Fourier integral, hence evaluate ∫

∞

0

cossinds

s

sxs and find the values of

∫∞

0

sinds

s

s

We know that the Fourier integral formula for f(x)= ∫ ∫

∞ ∞

∞−

−0

)(cos)(1

dsdtxtstfπ

f(x)=

1

100

1

1

)(sin1)(cos

1

−

∞∞

−∫∫ ∫

−=−

s

xtsdsdtxts

ππ

∫∫∞∞

=

++−=⇒

00

cossin2)1(sin)1(sin1)( ds

s

sxsds

s

xsxsxf

ππ…….(1)

This is fourier integral representation of f(x).

From (1)

>

<==⇒

=⇒

∫

∫∞

∞

1||,0

1||,2)(

2

cossin

)(cossin2

0

0

x

xxfds

s

sxs

xfdss

sxs

ππ

π

Put x= 0, 2

sin

0

π=∫

∞

dss

s

UNIT V Z – TRANSFORMS AND DIFFERENCE EQUATIONS

PART - A 1. What is the Z- transform of discrete unit step function

Solution: Discrete Unit step function is

( )1, 0

0, 0

nu n

n

≥=

<

( ) 1 21

0

11. 1 ..

1n

n

Z u n z z zz

∞− − −

−=

= = + + + = −∑ =1

z

z−if 1z >




2. Find

1

2nZ

.

Solution: 1 2 32 3

0

1 1 1 1 11

2 2 2 2 2n

n nZ z z z z

∞− − − − = = + + + +

∑ …

2 3

1 1 11 ...

2 4 8z z z= + + + +

1

2

z

z=

−

2, 2

2 1

zz

z= >

−

3. Find ( )1Z u n−

Solution:

( ) ( ) ( )0 1

2 3 21

1 1 1

1 1 1 1 1 1.... 1 ....

n n

n n

n

n

Z u n u n z u n z

zz z z z z z

∞ ∞− −

= =

∞−

=

− = − = −

= = + + + = + + +

∑ ∑

∑

1 1

1 1 1 11

z

z z z z

− −− = − =

1

1

z

z z = −

1

1z=

− if 1z >

4. Find the Z- transform of unit impulse function

Solution: Unit impulse function is

( ) 1, 0

0, 0

n n

n

δ = =

= ≠

( ) ( ) 0

0

1n

n

Z n n z zδ δ∞

− −

=

= = = ∑ .

5. If ( ) ( )Z f n U z= then find ( )nZ a f n

Solution: ( ) ( )( ) ( )

0 0

( ) / /nn n n

n n

Z a f n a f n z f n a z U z a∞ ∞

−

= =

= = = ∑ ∑ .

6. If ( ) ( )Z f n U z= , then show that ( ) ( )kZ f n k z U z+ =

Solution:

( ) ( ) ( ) ( )

0 0

n kn kZ f n k f n k z z f n k z∞ ∞

− +− ++ = + = + ∑ ∑

( ) ( )0

k r kz f r z z U z∞

+ − += =∑




7. If ( ) ( )z f n U z= , then

( ) ( )1f nZ z U z dz

n−

= −

∫ .

Solution:

( ) ( ) 1

0 0

( )n n

n n

f n f nZ z f n z dz

n n

∞ ∞− − −

= =

= = −

∑ ∑ ∫ , since 1

nnz

z dzn

−− −= −∫

( ) ( )( )1 1

0

n n

n

f n z dz z f n z dz∞

− − − −

=

= − = −∑ ∑∫ ∫ ( )1z U z dz−= −∫ .

8. If ( ) ( ) ,Z f n U z= then find ( )Z n f n

Solution:

( ) ( ) ( ) 1n nZ nf n nf n z z nf n z− − −= = − − ∑ ∑

= ( ) ( ) ( ) ( )n nd d dz f n z z f n z z U z

dz dz dz− −− = − = −∑ ∑ .

9. Find 3nZ a + .

Solution:

( )3 3 3

0

n n n n

n

Z a a z a Z a∞

+ + −

=

= = ∑ = 3 za

z a−

10. Find

( )( )1

1 2

zZ

z z−

− −

Solution:

Let ( )( )

( )1 2

zU z

z z

=

− −

Then ( )

( )( )1

1 2 1 2

U z A B

z z z z z= = +

− − − −

2, 1

1, 1

z B

z A

= =

= = −

( ) 1 1

1 2

U z

z z z

−∴ = +

− −




( ) ( ) 1 2 2 11 2

n nz zU z u n

z z

−∴ = + ⇒∴ = − + = −

− −

11. Find ( )nZ .

Solution:

{ } ....31

321

21

11

⋅+

⋅+⋅=−∑∞

==

zzznz

nnnZ

( )

221 2 1 1 1 11 3 1 .

2 21 1

z z

z z z z z zz z

− = + + ⋅ + ⋅⋅ ⋅⋅ = − = = − −

12. Find Z-Transform of 1

n.

Solution:

2 3

1

1 1 1 1 1 1 1....

2 3n

n

Z zn n z z z

∞−

=

= = + + +

∑

1 1

1 log log1

z zlog

z z z

− = − − = − = −

13. Find Z-Transform of na .

Solution:

{ } ∑∑∞

=

∞

=

−

−=

==00 n

n

n

nnn

az

z

z

azaaZ

14. Find Z-Transform of

1

!Z

n

.

Solution:

11 1 1 1

1 ....2! ! 1! 2!1

n zZ z en n z zn

∞ −= = + + =∑ =

15. Solve 1 2nn ny y+ − = , given that ( )0 1y = .

Solution:




{ } { }1 2nn nZ y y Z+ − =

( ) ( ) ( )02

zzY z y Y z

z− − =

− , ( ) { }nY z Z y=

( )( )12

zY z z z

z− = +

− ⇒ ( ) ( )

( )( )2

2 1

z z zY z

z z

− +=

− − ⇒ ( ) ( )

( )( )1

2 1

z zY z

z z

−=

− −

( )1 2

2n

n

zy Z

z−

= = −

16. Using Convolution theorem , evaluate

−−−−−−−−−−−−

)3()1(

21

zzz

Z

Solution:

We know that ( )

( )1 11

zZ z

z−

= −

and ( )

1 33

nzZ

z−

= −

Now ( ) ( )

21 1 .

1 3 1 3

z z zZ Z

z z z z− − = − − − −

0 0

1 3 1.3 3 3n n

n n m n m

m m

− −

= =

= ∗ = =∑ ∑

( )( )

1

1 1 1 1

0

11 3 1 3 / 31 1 3 3 13

3 3 .13 1 3 / 3 2 2

13

n

n n nm n nnn n

m

+

+ + + +

=

− − − − = = = = = − − −

∑

17. Find { }2 sin 2tZ e t−

Solution:

{ } { }( ) sin 22 sin 2 sin 2 2 2 2 cos2 1 2

z TtZ e t Z t Tz ze z z T Tz ze

− = =→ − + →

2 sin 2

2 4 22 cos 2 1

Tze TT Tz e ze T

=− +




18. Find { }sinZ at

Solution:

{ } ( )sin sin0

nZ at anT zn

∞ −= ∑=

( )sinaT

Z nθ

θ=

= sin sin

2 22 cos 1 2 cos 1aT

z z aT

z z z z aTθ

θ

θ=

= = − + − +

19. State initial and final value theorem of Z - transform.

Solution: Initial value Theorem

( ) ( ) , 0If Z f n U z n= ≥ then ( ) ( ) ( )0n 0

Limit Limitf n f U z

z= =

→ →∞

Final value Theorem

( ) ( ) , 0If Z f n U z n= ≥ then ( ) ( ) ( )11

Limit Limitf n z U z

n z= −

→∞ →

20. Form the difference equation from .2 .3n nny A B= +

Solution:

Given .2 .3n nny A B= + , 1 2 .2 3 .3n n

ny A B+ = + , 2 4 .2 9 .3n nny A B+ = +

Elimination of A and B forms the difference equation

1

2

1 1

2 3 0

4 9

n

n

n

y

y

y+

+

=

2 15 6 0n n ny y y+ +− + = .

PART - B

1. a) Solve ( ) ( )2 15 6 5 , 0 0, 1 0nn n ny y y y y+ +− + = = = using Z-transform.

Solution:

2 15 6 5nn n ny y y+ +− + = .Taking Z-transform on both sides

( )( ) ( )( ) ( )2 10 0 5 0 65

zz Y z z z Y z Y z

z−− − × − − + =

−




( )2 5 65

zz z Y z

z − + = −( )

( ) ( ) ( )( )( )2

1 1

5 2 3 5 2 35 5 6

Y z A B C

z z z z z z zz z z= = = + +

− − − − − −− − +

( )( ) ( )( ) ( )( )1 2 3 5 3 5 2A z z B z z C z z= − − + − − + − −

15, 1 6

6z A= = ⇒ =

3, 1 2 1/ 2z C C= = − ⇒ = −

2, 1 3 1/ 3z B B= = ⇒ =

( ) 1 1 1 1 1 1

6 5 3 2 2 3

Y z

z z z z∴ = + −

− − −

( ) 1 1 1

6 5 3 2 2 3

z z zY z

z z z= + −

− − −

1 1 15 2 3 .

6 3 2n n n

ny∴ = + −

b) State and Prove Second Shifting Property of Z-Transform.

Statement:

Let ( ) ( )nZ u U z= and 0k > then ( ) ( ) 1 2 ( 1)0 1 2 1...k k

n k kZ u Z U z u u z u z u z− − − −+ − = − − − − −

Proof

( ) ( )

0 0

n kn kn k n k n k

n n

Z u u z z u z∞ ∞

− +−+ + +

= =

= =∑ ∑

1

0 0

kk n n

n nn n

z u z u z∞ −

− −

= =

= −

∑ ∑

( ) 1 2 ( 1)0 1 2 1...k k

kz U z u u z u z u z− − − −− = − − − − −

2. a) Solve ( ) ( ) ( )3 3 1 2 0,y n y n y n+ − + + = given that ( ) ( ) ( )0 4, 1 0, 2 8y y y= = = Using Z-transform.

Solution:

( ) ( ) ( )3 3 1 2 0y n y n y n+ − + + = . Taking z-transform on both sides

[ ] ( ) ( )3 13 2 0n n nZ y Z y Z y+ +− + =




( ) ( ) ( )3 1 20 1 2 03 2 0z y z y y z y z z y z y Y z− − − − − − − + =

( )( ) ( )( ) ( )3 24 0 8 3 4 2 0z Y z z z Y z Y z−− − − − − + =

( ) 3 3 33 2 8 4 12 4 4Y z z z z z z z z − + = + − = −

( )( ) ( )

2 2

23

4 4 4 4

3 2 1 2

Y z z z

z z z z z

− −= =

− + − +

( ) ( )23 3 2 1 2z z z z∴ − + = − +

( )( ) ( ) ( )

2

2 2

4 4

1 21 2 1

Y z z A B C

z z zz z z

−∴ = = + +

− +− + −

( )( ) ( ) ( )224 4 1 2 2 1z A z z B z C z∴ − = − + + + + −

1, 0 3 0z B B= = ⇒ =

2 12 9 12 / 9 4 / 3z c c= − = ⇒ = =

2, 4 4 4 / 3 8 / 3z A C A= + ⇒ = − =

( ) 8 1 4 10

3 1 3 2

Y z

z z z∴ = + + +

− +

( ) 8 1 4 1

3 1 3 2Y z

z z∴ = + +

− +

( )8 42 .

3 3n

ny = + −

b) If ( )

( )

2

4

2 3 12,

1

z zU z

z

+ +=

−find 2u and 3u .

Solution:

( )( )

2

0 4

lim lim 2 3 120

1

it it z zu U z

z z z

+ += = =

→∞ →∞ −

( )( )

2

1 0 4

lim lim 2 3 120 0

1

it it z zu z U z u

z z z

+ + = − = − = →∞ →∞ −




( )2 12 0 1

lim itu z U z u u z

z− = − − →∞

( )

22

4

lim 2 3 120 0 2

1

it z zz

z z

+ += − − =

→ ∞ −

( )3 1 23 0 1 2

lim itu z U z u u z u z

z− − = − − − →∞

( )

23

4 2

lim 2 3 12 20 0

1

it z zz

z zz

+ += − − −

→ ∞ −

( )( )

4 2 4 3 2

342

2 3 12 2 4 6 4 1lim

1

z z z z z z zitz

z z z

+ + − − + − + =

→∞ −

( )3

4

lim11 8 2 11.

1

it zz z

z z = + − = →∞ −

3. a) Find

−−−

)14)(12(

8 21

zz

zZ by convolution theorem.

Solution:

2 21 1

1 12 4

8

(2 1)(4 1) ( )( )

z zZ Z

z z z z− −

= − − − −

1 1

1 12 4

.( ) ( )

z zZ Z by C T

z z− −

= ∗ − −

1 1

2 4

n n = ∗

0 0

1 1 1 1 1

2 4 2 2 4

n r r n r rn n

r r

− −

= =

= =

∑ ∑

2

0 0

11 1 1 1 1 1 14 1 ....

12 2 2 2 2 2 22

r

n n r n nn n

r r= =

= = = + + + +

∑ ∑




11

11 1 1 12

2 112 2 2 2

12

n

n n n

+ − = = − −

1 1 1 1 1 12 2

2 2 2 2 2 4

n n n n n = − = −

b) Find ( )( )1 2Z n n n − −

Solution:

( )( )1 2Z n n n − − { }3 23 2Z n n n= − +

{ } { } { }3 23 2Z n Z n Z n= − +

( )( )

( ) ( )

3 2

4 3 2

3 14 2

1 1 1

z zz z z z

z z z

++ += − +

− − −

( )( ) ( )( ) ( )

23 2

4 4

4 3 1 1 2 1 6

1 1

z z z z z z z z z

z z

+ + − + − + −= =

− −

4. a) Find ( )cosnZ r nθ and hence deduce

2

nZ cos

π

Solution:

( ) ( )nin ii

zZ e Z e

z eθ θ

θ− −

− ∴ = = −

( ) ( )( )( )

i

ini i i

z z ezZ e

z e z e z e

θθ

θ θ θ−

− −

−∴ = =

− − −

( )( )

( )( )22

cos sin cos sin

2 cos 11i i

z z i z z iz

z zz z e eθ θ

θ θ θ θ

θ−

− + − − = =− +− + +

( ) ( )( ) ( )2 2

cos sincos sin

2 cos 1 2 cos 1

z z zz n i n i

z z z z

θ θθ θ

θ θ

−∴ + = −

− + − +

equating real and imaginary parts,




( ) ( )2

coscos

2 cos 1

z zZ n

z z

θθ

θ−

=− +

{ } { } ( )

2 2

coscos cos

2 cosn

zz

r

z z rZ r n Z n

z rz r

θθ θ

θ→

− = = − +

{ }[ ]11cos2

)cos(

1cos2

)cos(cos

2cos

2

2

22

2

2

22 +=

+−

−=

+θ−

θ−=θ=

π

π

π

π=θ

π=θ

z

z

zz

zz

zz

zznZ

nZ

b) Find the inverse Z-Transform of

( )( )3

1

1

z z

z

+

− by residue method.

Solution:

F (z) = 3

2

)1( −

+

z

zz

∴ Zn-1F(z) = 3

1

)1( −

++

z

zz nn

Here z = 1 is a pole of order 3

[ ] ( )

−

+×−=

+

→= 3

13

2

2

11

)1(1

!2

1

z

zzz

dz

dLtRES

nn

zz

= 2

2

1 !2

1

dz

dLt

z→[ ]nn zz ++1

dz

dLtz !2

11→

= [ ]1)1( −++ nn nzzn

12

1→

=zLt [ ]21 )1()1( −− −++ nn znnznn

2

1= [ ])1()1( −++ nnnn

2

1= [ ]nnnn −++ 22( = 2n

5. a) Form the difference equation corresponding to the family of curves 2xxy ax b= + .

Solution:

2 (1)xxy ax b= + −

( ) 11 1 2x

xy a x b ++ = + +

( )12 2 2 (2)x x xy a b a b+∆ = + − = + −




( ) ( )2 12 2 2 (3)x x xy a b a b b+∆ = + − + = −

2

(3),2x

yfrom b

∆=

2

(2) 22

x

x

ysub in y a

∆∆ = +

2a y y∴ = ∆ −∆

(1)sub in ( )2

2 2 .2

x xx x

yy y y

∆= ∆ −∆ +

( ) ( ) ( ) ( )22 1 11 1 2 0x x x x x xx y x y or x y y y x y y y+ + += − ∆ + ∆ − − + + − − =

( ) ( )2 11 3 2 2 0x x xie x y x y xy+ +− − − + =

b) Find

( ) ( )1

21 1

zZ

z z−

+ − using the method of partial fraction.

Solution:

Let ( )( )( )2

1 1

zF z

z z=

+ −

( )( )( ) ( ) ( ) ( )2 2

1

1 11 1 1

F z A B C

z z zz z z= = + +

+ −+ − −

( ) ( ) ( ) ( )( )( )

2

2

1 1 1 1

1 1

A z B z z C z

z z

− + − + + +=

+ −

( ) ( )( ) ( )21 1 1 1 1A z B z z C z= − + − + + +

Putting 1z= and 1z= − we get 1

4A= and

1

2c =

Equating the coefficient of on both side

0=A+B

1

4B A= − = −

( )( ) ( ) ( )2

1 1 14 4 2

1 1 1

F z

z z z z= − +

+ − −




( )( )2

1 1 1

4 1 4 1 2 1

z z zF z

z z z= − +

+ − −

( )( )

1 1 1 12

1 1 1

4 1 4 1 2 1

z z zZ F z Z Z Z

z z z− − − −

= − + + − −

( )1 11

4 4 2n n

= − − +



TRANSFORMS & PDE (MA6351) · 2019-08-15 · St. Joseph’s College of Engineering & Page No. 1 ISO...

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Transcript of TRANSFORMS & PDE (MA6351) · 2019-08-15 · St. Joseph’s College of Engineering & Page No. 1 ISO...