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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
Problem : 3 AUC N / D 2010
Solution :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
Problem : 5 AUC N / D 2011
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
Problem : 7 AUC N / D 2012
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
y x t A px B px C pct D pct
)..................(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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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 2cos ( ) cos cos0(0) cos 0
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 :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
1
sin 0
sin 0 since it is defined for , 0, 0
This implies 0Put 0 in (3) we get
( , ) 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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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 x t A px B px C pct D pct
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
A C pct D pctC pct D pct t A
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
y l tB pl C pct D pctB B y x t C pct D pct
pl n
npl n pl
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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 0 since it is defined for , 0, 0
This implies 0Put 0 in (3) we get
( , ) 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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
Problem : 5 AUC N / D 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) , ( , )
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
y x t A px B px C pct D pct
)..................(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
A C pct D pctC pct D pct t A
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
y l tB pl C pct D pctB B y x t C pct D pct
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
1
sin 0 since it is defined for , 0, 02 2
This implies 0Put 0 in (3) we get
( , ) 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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
( , ) ( 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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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 :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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).
The most general solution is
( , ) 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
2 2 2cos ( ) cos cos0(0) cos 0
ln x
l
k l n l l n l l ll ll n l n l n n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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 :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
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
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
S.KAVITHA M.Sc.,M.Phil., AP/MATHEMATICS MEC
20
1
0
1
The most general solution is
( , ) 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