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MAHALAKSHMI ENGINEERING COLLEGE, TRICHY

S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC

MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI – 621213

QUESTION BANK - ANSWERS

SEMESTER: III MA 2211 - TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS UNIT – IV APPLICATIONS OF PDE

PART-A

2

2

2

2

2

Classify the partial differential equation 4

Given 4

4 00

Here 4, 0, 04 0

The given equation is p

xx t

xx xy yy x y

u ux t

u ux t

u uAu Bu Cu Du Eu Fu

A B CB AC

Problem :1 AUC N / D 2009

Solution :

1 2 3 4

5 6 7 8

9 10 1

arabolic

Write down all possible solutions of one dimensional wave equation

( , )

( , ) cos sin cos sin

( , )

px px pat paty x t A e A e A e A e

y x t A px A px A pct A pct

y x t A x A A

Problem : 2 AUC N / D 2009

Solution :

2 2

1 12

1 1 1

Write down all possible solutions of one dimensional heat equation

( , ) px px p t

t A

u x t A e B e C e


Solution :



2 2

2 2 2

3 3 3

( , ) cos sin

( , )

Write all three possible solutions of steady state two dimensional heat equation.

( , ) cos

p t

px px

u x t A px B px C e

u x t A x B C

u x y Ae Be C py

Problem : 4 AUC N / D 2010, A / M 2011

Solution :

2 2

2 22 2

2

sin

( , ) E cos sin

( , )

In the wave equation what does stands for?

py py

D py

u x y px F px Ge He

u x y Ix J Ky L

y yc ct x

T TensioncM Mass per unit length of the string


Solution :

Problem : A plate is bounded by the lines 0 , 0, and . Its faces are insulated.The edge coinsiding with axis is kept at 100 .The edge coinciding wit

x y x l y lx C

6 AUC N / D 2011, N / D 2012

h axis is kept at 50 .The other two edges are kept at 0 , write the boundary conditions that are needed for solving two dimen sional heat flow equation.

( ). ( ,0

yC C

i u x

Solution :

0

) 100 ,0( ). (0, ) 50 , 0( ). ( , ) 0 , 0( ). ( , ) 0 , 0

An insulated rod of length 60 cm has its ends at and maintained at 20and 80

C x lii u y C y liii u x l C x liv u l y C y l

A B C


2

2

respectively.Find the steady state solution of the rod.

Heat equation in steady state is 0

Integrating we get ( ) .....(1)when 0 , we get

(0) (0) 20 20w

C

d udx

u x ax bx

u a b b

Solution :

hen 60, we get, (60) (60) 80

60 20 80 1

x u a b

a a



30

A tightly stretched string with fixed end points 0 and is initially in a

position given by ( ,0) sin .If it is released from rest in t

x x lxy x y

l

Problem : 8 AUC A / M 2010

30

his position

write the boundary conditions.

The boundary conditions are( ). (0, t) 0 t 0( ). ( , ) 0 t 0

( ,0)( ). 0

( ). ( ,0) ( ) sin

A rod

i yii y l t

y xiiit

xiv y x f x yl

Solution :

Problem : 9 AUC A / M 2011

0 0

is 40cm long with insulated sides has its ends and are kept at 20 and 60 respectively.Find the steady state temperature at a location15cm from .

( )

A BC C

A

b au x x al

Solution :

, 0 40

60 20 20, 0 404020, 0 40

at 15 , (15) 15 20 35

What is the basic difference between the solutions of one dimensional wave equationand one d

x

x x

x xx u

Problem :10 AUC A / M 2012

2 22

2 2

imensional heat equation with respect to the time?

One dimensional wave one dimensional heat equation equation

y y uat x t

Solution :

2 2

22

2

1. It is classified as hyperbolic pde It is classified as parabolic p.d.e

2. ( , ) cos sin Ccos sin ( , ) cos sin p t

ux

y x t A px B px pct D pct u x t A px B px e



A tightly stretched string with fixed end points 0andis initially at rest in its equilibrium posit

x x l

PART BProblem :1 AUC M / J 2013

ion.If it is vibrating by

giving to each of its points a velocity ( - ) find ( , )

From the given problem , we get the following boundary conditions( ). (0, ) 0 0( ). ( , ) 0 0( ).

x l x y x t

i y t tii y l t tiii y

Solution : , .

2

( ,0) 0

( ). ( ,0) ( )

The suitable solution is given by ( , ) ( cos sin )( cos sin )..................(1)

Use (i) in (1) we get (0, ) 0

( cos 0 sin 0)( cos sin ) 0( cos s

x

iv y x lx xt

y x t A px B px C pct D pct

y tA B C pct D pct

A C pct D

in ) 0cos sin 0 (since 0) this implies 0

Put 0 in (1) we get ( , ) sin ( cos sin ).................(2).

Use ( ) in (2) we get ( , ) 0sin ( cos sin ) 0

0 (since if

pctC pct D pct t A

Ay x t B px C pct D pct

iiy l tB pl C pct D pctB B

0 then ( , ) 0), ( cos sin ) 0This implies sin 0 sin

Put in (2) we get

( , ) sin ( cos sin ).................(3).

( ,0) 0 sin ( cos 0 sin 0) 0

sin 0

y x t C pct D pctpl n

pl nnpl

npl

n x n ct n cty x t B C Dl l l

n xy x B C Dl

n xBCl



1

1

2

1

0, sin 0 0

Put 0 in (3) we get

( , ) sin sin

( , ) sin sin (4) where

( , ) sin cos

( , ) ( ) sin cos 0

n nn

nn

nn

n xB Cl

Cn x n cty x t BD

l ln x n cty x t b b BD

l ly x t n x n ct n cb

t l l ly x t n x n clx x b

t l l

2

2

1

2

1

2

0 0

2

0

2

( )

sin ( )

sin ( ) ( ) where

2 2( )sin ( )sin

2 ( )sin

cos si2 ( ) ( 2 )

nn

n n nn

l l

n

l

lx x

n x n cb lx xl l

n x n cB lx x f x B bl l

n x n xB f x dx lx x dxl l l l

n xlx x dxl l

n xllx x l xnl

l

2 2 3 3

2 30

32

3 30

3 32 2

3 3 3 3

n cos( 2)

2 2cos ( ) cos

2 2 2cos ( ) cos cos0(0) cos 0

2 0

l

l

n x n xl l

n nl l

l n x l n xlx xl n l n l

l n l l n l l ll ll n l n l n n

l

3 3

3 3 3 3

3 3

3 3 3 3

3 3 3

3 3 3 3 3 3

2

3 3

2 2cos 0

2 2 20 cos 0

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

4 ( 1) 1

n n

n

l lnn n

l lnl n n

l l ll n n l n

ln



2

3 3

2

3 3

2 2

3 3 3 3

2 3

3 3 4 4

1

3

4 4

4 1 ( 1)

8 , if is odd ...................(5)0, if is even

8 8( . )

8 8

( , ) sin sin

8( , ) sin

n

n n

n

nn

n odd

ln

l nn

n

l n c lB i e bn l n

l l lbn n c n c

n x n cty x t bl l

ly x tn c

3

4 41

sin

8 (2 1) (2 1)sin sin(2 1)

A string is stretched and fastened to two points 0and apart.Motion is started by displacing the string int

n

n x n ctl l

l n x n ctn c l l

x x l

Problem : 2 AUC M / J 2011, N / D 2011

2

2 22

2 2

o the form( - ) from which it is released at time 0.Find the

displacement of any point on the string at a distance of fromone end at time .

The wave equation is .

From th

y k lx x tx

t

y yat x

Solution :

2

e given problem , we get the following boundary conditions( ). (0, ) 0 0( ). ( , ) 0 0

( ). ( ,0) 0

( ). ( ,0) ( )The suitable solution is given by

( , ) ( cos sin )( cos sin

i y t tii y l t t

iii y xt

iv y x k lx x


)..................(1)Use (i) in (1) we get

(0, ) 0( cos 0 sin 0)( cos sin ) 0

( cos sin ) 0

y tA B C pct D pct

A C pct D pct



cos sin 0 ( 0) 0Put 0 in (1) we get

( , ) sin ( cos sin ).................(2)Use ( ) in (2) we get

( , ) 0sin ( cos sin ) 0

0 (since if 0 then (

C pct D pct t AA

y x t B px C pct D pctii

y l tB pl C pct D pctB B y x

since this implies

, ) 0), ( cos sin ) 0This implies sin 0 sin

Put in (2) we get

( , ) sin ( cos sin ).................(3).

( , ) sin sin cos

t C pct D pctpl nnpl n pl

npl

n x n ct n cty x t B C Dl l ln x n ct n c n cty x t B C D

t l l l l

( ,0) 0 implies

sin cos 0

sin 0

sin 0 since it is defined for , 0, 0

This implies 0Put 0 in (3) we get

( , ) sin cos

n cl

y xt

n x n cB D ol ln x n cBD

l ln x n cx B

l lD

Dn x n cty x t BC

l l

1

2

2

1

0

The most general solution is ( , ) sin cos .............(4).

Use ( ) in (4) we get ( ,0) ( )

sin ( ) ( )

2 ( )sin

nn

nn

l

n

n x n cty x t bl l

ivy x k lx x

n xb k lx x f xl

n xb f x dxl l



2 2

0 0

22 2 3 3

2 30

32

3 3

2 2( )sin ( )sin .

cos sin cos2 ( ) ( 2 ) ( 2)

2 2cos ( ) cos

l l

l

n x k n xk lx x dx lx x dxl l l l

n x n x n xk l l llx x l xn n nl

l l l

k l n x l n xlx xl n l n l

0

3 32 2

3 3 3 3

3 3

3 3 3 3

3 3

3 3 3 3

3 3

3 3 3


2 2 20 cos 0

2 2 20 cos 0

2 2 2( 1)

l

n

k l n l l n l l ll ll n l n l n n

k l lnl n n

k l lnl n n

k l ll n n

3

3 3 3

2 2

3 3 3 3

2

3 3

2

3 3odd

2 2 ( 1) 1

4 4( 1) 1 1 ( 1)

8 , if is odd ...................(5)0, if is even

Use (5) in (4) we get 8( , ) sin cos .

8

n

n n

n

k ll n

kl kln nkl n

nn

kl n x n cty x tn l l

kl

2

3 31

(2 1) (2 1)sin cos(2 1)n

n x n ctn l l

A string is tightly stretched and its ends are fastened at two points 0and .The midpoint of the string is displaced transversely througha small distance ' ' and the string is

xx l

h

Problem : 3 AUC M / J 2010

released from rest in that position.Find the transverse displacement of the string at any time during thesubseqent motion.Solution :



1 1

2 1 2 1

1 1

2 1 2 1

Equation of

(0,0) , ,2

0 00 0

22

2 , in 2

Equation of

, , ( ,0)2

2 .0

22

22 .

22

OAlO A h

y y x x y xly y x x h

y xh l

hx ly o xl

ABlA h B l

lxy y x x y hly y x x h l

x ly h y h x l

lh h l

hx hly h yl

2 2 2 .

2 in 2

hx hl hl hx hll l

h ly l x x ll



The boundary conditions are( ). (0, ) 0 0( ). ( , ) 0 0

( ). ( ,0) 0

2 , in 02( ). ( ,0)

2 ( ) , in2

The suitable solution is given by ( , ) ( cos sin )( cos sin

i y t tii y l t t

iii y xt

hx lxliv y xh l x l x l

l

y x t A px B px C pct D pc

)..................(1)Use (i) in (1) we get

(0, ) 0( cos 0 sin 0)( cos sin ) 0

( cos sin ) 0cos sin 0 (since 0) this implies 0

Put 0 in (1) we get ( , ) sin ( cos sin

t

y tA B C pct D pct

A C pct D pctC pct D pct t A

Ay x t B px C pct D

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

Use ( ) in (2) we get ( , ) 0sin ( cos sin ) 0

0 (since if 0 then ( , ) 0), ( cos sin ) 0This implies sin 0 sin

Put in (2) we get

( ,

pctii

y l tB pl C pct D pctB B y x t C pct D pct

pl npl n

npl

npl

y x

) sin ( cos sin ).................(3).

( , ) sin sin cos

( ,0) 0 implies

sin cos 0

n x n ct n ctt B C Dl l ln x n ct n c n ct n cy x t B C D

t l l l l l

y xt

n x n cB D ol l



1

sin 0



( , ) sin cos

The most general solution is ( , ) sin cos .......nn

n x n cBDl l

n x n cx Bl l

DD

n x n cty x t BCl l

n x n cty x t bl l

1

2

0 02

1 22

......(4).

Use ( ) in (4) we get ( ,0) ( )

2 , in 02sin ( )

2 ( )2 , in2

2 2 2 2 ( )( )sin sin sin

4 ............

nn

ll l

nl

ivy x f x

hx lxn x lb f x

h l x ll x ll

n x hx n x hx l x n xb f x dx dx dxl l l l l l l

h I Il

2

2

1 2 20

20

2 2

2 20

2 2 2 2

2 2 2 2

.(*)

cos sinsin (1)

sin cos

sin cos 0 sin cos2 2 2 2 2 2

similarly

l

l

l

n x n xn x l lI x dx x n nl

l l

l n x l n xxn l n l

l n l n l n l nn n n n

2 2

2 2 2

2

1 2 2 2

2

1 22 2 2 2

sin cos2 2 2

2 sin2

4 4 2 sin .2n

l n l nIn n

l nI In

h h l nb I Il l n



2 2

1

2 21

8 sin2

( , ) sin cos

8( , ) sin sin cos .2

A tightly stretched string with fixed end points 0 and is initially

at rest in a position

nn

n

h nn

n x n cty x t bl l

h n n x n cty x tn l l

x x l

Problem : 4 : AUC N / D 2012

30

2 22

2 2

is given by sin .If it is released from rest

from this position , find ( , ).

The wave equation is .

From the given problem , we get the following boundary conditi

xy yl

y x t

y yat x

Solution :

30

ons( ). (0, ) 0 0( ). ( , ) 0 0

( ). ( ,0) 0

( ). ( ,0) sin

The suitable solution is given by ( , ) ( cos sin )( cos sin )..................(1)

Use (i) in (1) we get (

i y t tii y l t t

iii y xt

xiv y x yl


y

0, ) 0( cos 0 sin 0)( cos sin ) 0

( cos sin ) 0cos sin 0 (since 0) this implies 0

Put 0 in (1) we get ( , ) sin ( cos sin ).................(2).

Use ( ) in (2) we ge

tA B C pct D pct


Ay x t B px C pct D pct

ii

t ( , ) 0sin ( cos sin ) 0

0 (since if 0 then ( , ) 0), ( cos sin ) 0This implies sin 0 sin


pl n

npl n pl



Put in (2) we get

( , ) sin ( cos sin ).................(3).

( , ) sin sin cos

( ,0) 0 implies

sin cos

npl

n x n ct n cty x t B C Dl l ln x n ct n c n ct n cy x t B C D

t l l l l l

y xt

n x n cB D ol l

1

0

sin 0



( , ) sin cos .

The most general solution is ( , ) sin cos .....nn

n x n cBDl l

n x n cx Bl l

DD

n x n cty x t BCl l

n x n cty x t bl l

2

30

1

0 01 2 3

01

........(4).

Use ( ) in (4) we get ( ,0) ( )

sin sin

32 3 3sin sin sin sin sin4 4

Equating the coefficient of sin terms both sides we get 34

nn

ivy x k lx x

n x xb yl l

y yx x x x xb b bl l l l l

yb

02 3 4 5

1

1 3

0 0

, 0, , 04

( , ) sin cos

3 3sin cos sin cos

3 3 3sin cos sin cos4 4

nn

yb b b b

n x n cty x t bl l

x ct x ctb bl l l l

y yx ct x ctl l l l



A string is tightly stretched and its ends are fastened at two points 0and 2 .The midpoint of the string is displaced transversely througha small distance ' ' and the string is

xx l

b


released from rest in that position.Find the transverse displacement of the string at any time during thesubseqent motion.Solution :

1 1

2 1 2 1

1 1

2 1 2 1

Equation of (0,0) , ( , )

0 00 0

, in

Equation of ( , ), (2 ,0)

0 2

2

OAO A l hy y x x y xy y x x h ly xh l

hxy o x ll

ABA l h B ly y x x y h x ly y x x h l ly h x l hx hly h

h l lhx hl hx hl hl hx hly h y

l l l

(2 ) , in 2h l xy l x ll



The boundary conditions are( ). (0, ) 0 0( ). (2 , ) 0 0

( ). ( ,0) 0

, in ( ). ( ,0)

(2 ) , in 2

The suitable solution is given by ( , ) ( cos sin )( cos sin

i y t tii y l t t

iii y xt

hx o x lliv y x

h l x l x ll


)..................(1)Use (i) in (1) we get

(0, ) 0( cos 0 sin 0)( cos sin ) 0

( cos sin ) 0cos sin 0 ( 0) 0

Put 0 in (1) we get ( , ) sin ( cos sin

y tA B C pct D pct


Ay x t B px C pct D p

since this implies

).................(2).Use ( ) in (2) we get

(2 , ) 0sin 2 ( cos sin ) 0

0 (since if 0 then ( , ) 0), ( cos sin ) 0This implies sin 2 0 sin

22

Put in (2) we get 2

ctii


pl nnpl n p

lnp

l

( , ) sin ( cos sin ).................(3).2 2 2

( , ) sin sin cos2 2 2 2 2

( ,0) 0 implies

sin cos 02 2

sin2

n x n ct n cty x t B C Dl l ln x n ct n c n ct n cy x t B C D

t l l l l l

y xt

n x n cB D ol ln x n cBD

l

02

l



1

sin 0 since it is defined for , 0, 02 2


( , ) sin cos .2 2

The most general solution is ( , ) sin cos .............(4).2 2

Use

nn

n x n cx Bl l

DD

n x n cty x t BCl l

n x n cty x t bl l

1

2

0 0

2

0

2

( ) in (4) we get ( ,0) ( )

, in sin ( )

(2 )2 , in 2

2 2( )sin ( )sin2 2

1 (2 )sin sin2 2

sin (22

nn

l l

n

l l

l

ivy x f x

hx o x ln x lb f xh l xl l x l

ln x n xb f x dx f x dx

l l l l

hx n x h l x n xdx dxl l l l l

h n xx dx l xl l

2

0

1 22

1 2 20

20

2 2 2

2 2 2 20

)sin2

...........(*)

cos sin2 2sin (1)

22 4

4 2 4 2sin cos sin cos2 2 2

l l

l

l

l

l

n x dxl

h I Il

n x n xn x l lI x dx x n nl

l l

l n x l n x l n l nxn l n l n n

2

2

2

2 2

2

2

(2 )sin2

cos sin2 2(2 ) ( 1)

2 4

l

l

l

l

n xI l x dxl

n x n xl ll x n n

l l



22

2 2

2 22 2

2 2 2 2

2 2 2 2

1 2 2 2 2 2

2

2 2 2

4 sin2 2(2 ) cos2

2 cos 4 sin 2 42 20 cos sin2 2

4 2 2 4sin cos cos sin2 2 2 2

8(*)

l

l

n

n xll n x ll xn l n

n nl l l n l nn n n n

l n l n l n l nI In n n n

h lbl n

2 2

1

2 21

2 21

8sin sin2 2

Now ( , ) sin cos2 2

8 sin sin cos2 2 2

8 1 sin sin cos2 2 2

nn

n

n

n h nn

n x n cty x t bl l

h n n x n ctn l l

h n n x n ctn l l



1. ( , ) ( cos sin )( ) if 0 is a short edge2. ( , ) ( )( cos sin ) if 0 is

py py

px px

u x y A px B px Ce De yu x y Ae Be C py D py x

PROBLEMS UNDER INFINITE PLATESTWO DIMENSIONAL HEAT FLOW EQUATIONS The Suitable solutions are

0

a short edge

A long rectangular plate has its surfaces insulated and the two long sides and one of the short sides are maintained at 0 . Find anexpression for the steady state temp

C

Problem : 6 :

00

erature ( , ) if the short side 0 is long and is kept at C.

u x yy cm u

Solution :

00

Solution:The Boundary conditions are( ). (0, ) 0( ). ( , ) 0( ). ( , ) 0( ). ( ,0)The suitable solution is

i u yii u yiii u xiv u x u C



( , ) ( cos sin )( ) (1)Use ( ) in (1) we get

(0, ) 0( cos0 sin 0)( ) 0( . ) ( ) 0( ) 0 since it is defined for y 0Put 0 in (1) we get

( , ) sin (

py py

py py

py py

py py

u x y A px B px Ce Dei

u yA B Ce Dei e A Ce DeCe De A

Au x y B px Ce

) (2).Use ( ) in (2) we get

( , ) 0sin ( ) 0

0, ( ) 0, sin 0 sin

Put in (2) we get ( , ) sin ( ) (3).

Use ( ) in (3) we get ( , ) 0s

py py

py py

py py

ny ny

Deii

u yB p Ce De

B Ce De p np n

p nu x y B nx Ce De

iiiu xB

1

0

in ( ) 0( . ) sin ( 0) 0

0, sin , 0 0Put 0 in (3) we get

( , ) sin (3).The most general solution is

( , ) sin (4)

Use ( ) in (4) we get( ,0)

( . ).

ny

nyn

n

nx Ce Dei e B nx C D

B nx CC

u x y BD nx e

u x y b nx e

ivu x u

i e

00

1

01

00 0

sin

( . ). sin ( ).

2 2where ( ) sin sin .

nn

nn

n

b nx e u

i e b nx u f x

b f x nxdx u nxdx



0 0

00

0

0

0

0

0

1

0

2 2 cossin

2 cos cos 0

2 ( 1) 1

2 1 ( 1)

4 , if ' ' is odd

0 , if ' ' is evenNow (4) implies

4( , ) sin .

4 1 sin

n

n

n

ny

n

ny

n odd

u u nxnxdxn

u nn

unu

nu n

b nn

uu x y nx en

u nx en

0

.

An infinitely long rectangular plate with insulated surface is 10 wide. The two long sides as well as one of the short sides are maintained at 0 , while the other short side 0

cmC

x

Problem : 7 :

is kept at temperature given by20 , 0 520(10 ), 5 10

Find the steady state temperature distribution in the plate .

y yu

y y

Solution :



The boundary conditions are( ). ( ,0) 0( ). ( ,10) 0( ). ( , ) 0

20 , 0 5( ). (0, )

20(10 ), 5 10The suitable solution is

( , ) ( )( cos sin )................(1)Use ( ) in (1) we

px px

i u xii u xiii u y

y yiv u y

y y

u x y Ae Be C py D pyi

get

( ,0) 0( )( cos 0 sin 0) 0

( ) 0( ) 0 0

Put 0 in (1) we get ( , ) ( ) sin ................(2)

Use ( ) in (2) we get ( ,10) 0

( ) sin10 0(

px px

px px

px px

px px

px px

u xAe Be C D

C Ae BeAe Be C

Cu x y Ae Be D py

iiu x

Ae Be D pAe

10 10

) 0 , 0,sin10 0 sin

10Now (2) implies

( , ) ( ) sin ................(3)10

Use ( ) in (3) we get ( , ) 0

( ) sin 010

( 0) sin 0 010

( , )

px px

n x n x

n

Be Dp n

np

n yu x y Ae Be D

iiiu y

n yAe Be D

n yA D A

u x y BDe

10 sin10

x n y



10

1

0

1

The most general solution is

( , ) sin ..............(4)10

Use ( ) in (3) we get 20 , 0 5

(0, )20(10 ), 5 10

20 , 0 5( . ). sin

20(10 ), 5 1010

( . ). sin10

n x

nn

nn

nn

n yu x y b e

iiiy y

u yy y

y yn yi e b ey y

n yi e b

1

10

0 0

5 10

0 5

5 10

0 5

1 2

1

20 , 0 5( )

20(10 ), 5 10

2 2Now ( )sin . ( )sin10 10

1 20 sin 20(10 )sin5 10 10

4 sin (10 )sin10 10

4 .

Now sin

l

n

y yf y

y y

n y n yb f y dy f y dyl l

n y n yy dy y dy

n y n yy dy y dy

I I

n yI y

5

5

2 20

05

2 20

2 2

1 2 2

2

cos sin10 10(1)

1010 100

100 10sin cos10 10

100 50sin cos (0)2 2

100 50sin cos2 2

similarly

n y n y

dy y n n

n y n yyn n

n nn n

n nIn n

I

2 2

1 2 2 2

1 2 2 2

10 102 2

1 1

100 50sin cos2 2

200 sin2

8004 sin2

800( , ) sin sin sin .10 2 10

n

n x n x

nn n

n nn n

nI In

nb I In

n y n n yu x y b e en



2

A long rectangular plate with insulated surface is wide.If the temperature along oneshort edge 0 is ( ,0) ( ) degrees, for 0 , while the other two longedges 0 and as w

lcmy u x k lx x x l

x x l

Problem : 8 : AUC M / J 2012

0

2

ell as the other short edge are kept at 0 find the steady statetemperature function ( , ).

The Boundary conditions are( ). (0, ) 0( ). ( , ) 0( ). ( , ) 0( ). ( ,0) ( )

( , ) ( cos

Cu x y

i u yii u l yiii u xiv u x k lx x

u x y A

Solution :

sin )( ) (1)Use ( ) in (1) we get

(0, ) 0( cos0 sin 0)( ) 0( . ) ( ) 0( ) 0 since it is defined for y 0Put 0 in (1) we get

( , ) sin ( )

py py

py py

py py

py py

py py

px B px Ce Dei

u yA B Ce Dei e A Ce DeCe De A

Au x y B px Ce De

(2).Use ( ) in (2) we get

( , ) 0sin ( ) 0

0, ( ) 0, sin 0 sin

py py

py py

iiu l y

B pl Ce DeB Ce De pl n

npl



Put in (2) we get

( , ) sin ( ) (3).

Use ( ) in (3) we get ( , ) 0

sin ( ) 0

( . ) sin ( 0) 0

0, sin 0, 0 0

Put 0 in (3) we get

( , ) sin

n y n yl l

npl

n xu x y B Ce Del

iiiu x

n xB Ce Del

n xi e B C Dl

n xB Cl

C

n xu x y BD el

1

2

0 2

1

2

1

(3).


( , ) sin (4)

Use ( ) in (4) we get( ,0) ( )

( . ). sin ( )

( . ). sin ( ) ( ).

2where ( )

n yl

n yl

nn

nn

nn

n

n xu x y b el

ivu x k lx x

n xi e b e k lx xl

n xi e b k lx x f xl

b f xl

2

0 0

2

0

22 2 3 3

2 30

32

2sin ( )sin .

2 ( )sin .

cos sin cos2 ( ) ( 2 ) ( 2)

2 2cos ( )

l l

l

l

n x n xdx k lx x dxl l l

k n xlx x dxl l

n x n x n xk l l llx x l xn n nl

l l l

k l n x llx xl n l n

3 30

3 32 2

3 3 3 3

cos


ln x

l

k l n l l n l l ll ll n l n l n n



3 3

3 3 3 3

3 3

3 3 3 3

3 3 3

3 3 3 3 3 3

2 2

3 3 3 3

2

3 3

2 2 20 cos 0

2 2 20 cos 0

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

4 4( 1) 1 1 ( 1)

8 , i

n n

n n

k l lnl n n

k l lnl n n

k l l k ll n n l n

kl kln nkl

n

2

3 31 1

f is odd ...................(5)0, if is even

Use (5) in (4) we get 8( , ) sin sin

An infinitely long rectangular plate with insula

n y n yl l

nn n

n

n

n x kl n xu x y b e el n l

Problem : 9 : AUC M / J 2011

0

ted surface is 20 wide. The two long sides as well as one of the short sides are maintained at 0 , while the other short side 0 is kept at temperature given by

20 , 0 1020(10 ), 10 20

cmC

xy y

uy y

Find the steady state temperature distribution in the plate .

Solution :



The boundary conditions are( ). ( ,0) 0( ). ( , 20) 0( ). ( , ) 0

10 , 0 10( ). (0, )

10(20 ), 10 20The suitable solution is

( , ) ( )( cos sin )................(1)Use ( ) in (1) w

px px

i u xii u xiii u y

y yiv u y

y y

u x y Ae Be C py D pyi

e get

( ,0) 0( )( cos 0 sin 0) 0

( ) 0( ) 0 0

Put 0 in (1) we get ( , ) ( ) sin ................(2)

Use ( ) in (2) we get ( , 20) 0

( ) sin 20 0(

px px

px px

px px

px px

px px

u xAe Be C D

C Ae BeAe Be C

Cu x y Ae Be D py

iiu x

Ae Be D p

20 20

) 0 , 0,sin 20 0 sin

20Now (2) implies

( , ) ( ) sin ................(3)20

Use ( ) in (3) we get ( , ) 0

( ) sin 020

( 0) sin 0 020

( , )

px px

n x n x

Ae Be Dp n

np

n yu x y Ae Be D

iiiu y

n yAe Be D

n yA D A

u x y BDe

20 sin20

n x n y



20

1

0

1


( , ) sin ..............(4)20

Use ( ) in (3) we get 10 , 0 10

(0, )10(20 ), 10 20

10 , 0 10( . ). sin

10(20 ), 10 2020

( . ). sin

n x

nn

nn

n

n yu x y b e

iiiy y

u yy y

y yn yi e b ey y

ni e b

1

10

0 0

10 10

0 5

10 20

0 10

1 2

1

10 , 0 10( )

10(20 ), 10 2020

2 2Now ( )sin . ( )sin20 20

1 10 sin 10(20 )sin10 20 20

sin (20 )sin20 20

.

Now

n

l

n

y yy f yy y

n y n yb f y dy f y dyl l

n y n yy dy y dy

n y n yy dy y dy

I I

I

10

10

2 20

010

2 20

2 2

1 2 2

cos sin20 20sin (1)

2020 400

400 20sin cos20 20

400 200sin cos (0)2 2

400 200sin cos2 2

s

n y n yn yy dy y n n

n y n yyn n

n nn n

n nIn n

2 2 2

1 2 2 2

1 2 2 2

20

1

202 2

1

400 200imilarly sin cos2 2

800 sin2

800 sin2

( , ) sin ..............(4)20

800( , ) sin sin2 20

n

n x

nn

n x

n

n nIn n

nI In

nb I In

n yu x y b e

n n yu x y en

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