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UNIT V
Applications of Partial DifferentialEquations
5.1 INTRODUCTION
The problems related to fluid mechanics, solid mechanics, heat transfer, wave equation and other areasof physics are designed as Initial Boundary Value Problems consisting of partial differential equationsand initial conditions.
These problems can be solved by “Method of separation of variables,” in this unit we derive andsolve one dimensional heat equation, wave equation, Laplace’s equation in two dimensions etc. byseparation of variables method. The general solution of partial differential equation consists arbitraryfunctions which can be obtained by Fourier Series.
5.2 METHOD OF SEPARATION OF VARIABLES
In this method, we assume that the required solution is the product of two functions i.e.,
),( yxu = )()( yYxX ...(i)Then we substitute the value of u(x, y) from (i) and its derivatives reduces the P.D.E. to the form
),,(1 LLXXf ′ = ),,(2 LLYYf ′ ...(ii)
which is separable in X and Y. Since ),,(2 LYYf ′ is function Y only and ),,(1 LXXf ′ is function of X
only, then equation (ii) must be equal to a common constant say k. Thus (ii) reduces to
),,(1 LXXf ′ = .),,(2 kXXf =′ L
Example 1: Using, the method of separation of variables solve ,2 utu
xu +
∂∂=
∂∂
where )0,(xu = .6 3xe− (U.P.T.U. 2006)
Solution: We have
xu
∂∂
= utu +
∂∂
2 ...(1)
Let u = )()( tTxX ⋅ ...(2)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 437
then TxX
xu
∂∂=
∂∂
= dX
Tdx
, As is the
function of alone
X dXX
x dxx
∂=
∂
and tT
Xtu
∂∂=
∂∂
= dtdT
X
Putting these values in equation (1), we get
dxdX
T = XTdtdT
X +2
⇒dxdX
X1
= 12 +
dtdT
T(On dividing by XT)
⇒dxdX
X1
= kdtdT
T=+1
2
NowdxdX
X1
= k ⇒ kdxXdX = ...(3)
On integrating, we get
Xelog = 1log ckx e+
⇒1
logc
Xe = kx ⇒ 1 .kxX c e=
and again from (3), we get
12 +
dtdT
T= k
⇒dtdT
T2
= 1−k ⇒1
2dT k
dtT
−=
On integrating, we obtain
Telog = 2log2
1ct
ke+− ⇒ t
k
c
Te
2
1log
2
−=
⇒ 1
22 .
kt
T c e
− =
Putting the values of X and T is equation (2), we get
u =
1
21 2
kt
kxc e c e
− ⋅
⇒ u = ( )1
12
1 2
kx k tc c e
+ −...(4)
On putting t = 0 and u = 6e–3x in (4), we have
xe 36 − = 3 and62121 −==⇒ kccecc kx
438 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Hence the required solution of given equation is
u =1
3 ( 3 1) 3 226 6x t x te e
− + − − − −=
⇒ u = .6 23 tx ee −− Ans.
Example 2: Solve the following equation by the method of separation of variables.
xu
∂∂
y
u
∂∂
= 4 where yeyu 38),0( −= (U.P.T.U. 2008)
Solution: We have
xu
∂∂
y
u
∂∂
= 4 ...(1)
Let u = Xy ...(2)
thenxu
∂∂ ,
dX u dYY X
dx y dy
∂= =
∂
Putting these values in equation 1, we get
YdxdX
dy
dYX4=
ordxdX
X⋅1 4 dY
kY dy
= ⋅ =
NowdxdX
X1
1log loge edX
K kdx X kx cX
= ⇒ = ⇒ = +
or kxecX 1=
And dy
dY
Y
42log log
4 4e edY k k
k dy Y y cY
= ⇒ = ⇒ = +
or 42
ky
ecY =
∴ From (2), we get
uk
yx
ecc
+
= 421 ...(3)
Putting x = 0, in the equation (3), we get
ye 38 − 421
yk
ecc=
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 439
Comparing on both sides
1234
and821 −=⇒−== kk
cc
Hence
u )312(8 yxe +−= Ans.
Example 3: Solve by the method of separation of variables
y
z
x
z
x
z
∂∂
+∂∂
−∂∂
22
2
= 0. (U.P.T.U. 2005)
Solution: We have
y
z
x
z
x
z
∂∂
+∂∂
−∂∂
22
2
= 0 ...(1)
Let z = )()( yYxX ...(2)
⇒xz
∂∂
= YdxdX ⇒ Y
dx
Xd
x
z2
2
2
2
=∂∂
andy
z
∂∂
= dY
Xdy
⋅
Putting these values in equation (1), we get
dy
dYX
dx
dXY
dx
XdY ⋅+− 2
2
2
= 0
⇒dy
dY
Ydx
dX
Xdx
Xd
X
1212
2
+− = 0 (divide by XY)
⇒dy
dY
Ydx
dX
Xdx
Xd
X
1212
2
−=− = k ...(3)
Nowdx
dX
Xdx
Xd
X
212
2
− = k ⇒ 0)2( 2 =−− XkDD
The A.E. is 022 =−− kmm
⇒ m = kmk +±=⇒+±
112
442
∴ X = { } { }1 1 1 1 }
1 2
k x k xc e c e
+ + − ++
440 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
and again from (3), we have
dy
dY
Y
1= k− ⇒
dYkdy
Y= −
On integrating, we get
Yelog = 3logcky +−
⇒3
logc
Ye = 3
kyky Y c e−− ⇒ =
Hence the required solution of equation (1) is
z = { } { }1 1 1 131 2 .
k x k x kyc ec e c e+ + + − −
+
Ans.
Example 4: Solve by the method of separation of variables 2 ,u u
ux y
∂ ∂= +
∂ ∂ where
5 3( ,0) 3 2 .x xu x e e− −= −Solution: We have
xu
∂∂
= 2u
uy
∂+
∂...(1)
Let u = XY ...(2)
Putting this value in equation (1), we get
)(XYx∂∂
= XYXYy
+∂∂
)(2
⇒dxdX
Y = XYdy
dYX +2
⇒dxdX
X1
= kdy
dY
Y=+ 1
2...(3)
NowdxdX
X1
= dX
k kdxX
⇒ =
On integrating, we get
Xelog = 1log ckx e+
⇒ 1 .kxX c e=
And taking last two terms of equation (3), we have
12
+dy
dY
Y= k ⇒ 1
2−= k
dy
dY
Y
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 441
⇒YdY
= dyk
2)1( −
On integrating, 2log2
)1(log cy
kY ee +−=
⇒ ( 1)2
2 .y
kY c e
− ⋅=
From (2), we get
u = ( 1) / 2
1 2kx k yc c e e −⋅
⇒ u = ( 1) / 2
1
n nk x k yn
n
b e e∞
−
=∑ ...(4)
)( 21ccbn = and )( nkk =which is the most general solution of equation (1).
Putting y = 0 and xx eeu 35 23 −− −= in equation (4), we get
xx ee 35 23 −− − = xkxk
n
xkn ebebeb n 21
211
+=∑∞
=
Comparing the terms on both sides, we get
,31 =b ,51 −=k ,22 −=b 32 −=k
Hence the required solution of given equation is from (4), we have
u = yxyx eeee 2335 )2(3 −−−− ⋅−+⋅
⇒ u = (5 3 ) (3 2 )3 2 .x y x ye e− + − +− Ans.
5.3 ONE DIMENSIONAL WAVE EQUATION (U.P.T.U. 2007)
The one dimensional wave equation arises in the study of transverse vibrations of an elastic string.Consider an elastic string, stretched to its length ‘l’ between two points O and A fixed. Let the functiony(x, t) denote the displacement of string at any point x and at any time t > 0 from the equilibriumposition (x-axis). When the string released after stretching then it vibrates and therefore the transversevibrations formed a one dimensional wave equation.
Let the string is perfectly flexible and does not offer resistance to bending. Let T1 and T2 betensions at the end points P and Q of the portion of the string. Since there is no motion in the horizontaldirection. Thus the sum of the forces in the horizontal direction must be zero i.e.,
β+α− coscos 21 TT = 0
⇒ αcos1T = β =2 cosT T (constant) ...(i)
442 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Let m be the mass of the string per unit length then the mass of portion PQ = mδs.
Now by Newton’s second law of motion, the equation of motion in the vertical direction is
mass × acceleration = resultant of forces
2
2
t
ysm
∂
∂δ = α−β sinsin 12 TT ...(ii)
Dividing (ii) by (i), we have
2
2
t
y
T
sm
∂∂δ
= αα
−ββ
cos
sin
cos
sin
1
1
2
2
T
T
T
T
⇒ 2
2
t
y
T
sm
∂∂δ
= tan β – tan α
⇒sm
T
t
y
δ=
∂
∂2
2
)tan(tan α−β
=
δ
∂∂
−
∂∂
δ+
x
xy
xy
m
T xxxs xδ → δ
Taking 0→δx
⇒2
2
t
y
∂
∂=
0lim
→δxmT
x x x
y yx x
x+ δ
∂ ∂ − ∂ ∂
δ
O
Y
X
T1
T2
β
P
x x + x δ
δx
α
δ s
y x, t( ) y x + x, t( )δ
Q
A( , 0)l
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 443
⇒ 2
2
t
y
∂
∂= 2
2
x
y
m
T
∂
∂
Thus 2 2
22 2
.y y
ct x
∂ ∂=
∂ ∂...(iii)
where .2
mT
c = Equation (iii) is known as one dimensional wave equation.
5.3.1 Solution of One Dimensional Wave EquationThe wave equation is
2
2
t
y
∂
∂= 2
22
x
yc
∂∂
...(i)
Let y = X(x) T(t). ...(ii)
where X is the function of x only and T is the function of t only.
Then 2
2
t
y
∂
∂=
2
2
dt
TdX and 2
2
2
2
dx
XdT
x
y=
∂
∂
Putting these values in equation (i), we get
2
2
d TX
dt= 2
22
dx
XdTc
⇒ 2
21
dx
Xd
X= k
dt
Td
Tc=
2
2
2
1...(iii)
Now kdx
Xd
X=
2
21
⇒ XkD )( 2 − = 0; Ddxd =
The A.E. is 02 =− km
m = k±
∴ 1 2k x k xX c e c e−= +
and, again from (iii), we get
2
2
dt
Td= Tkc2 ⇒ 0)( 22 =− TkcD ;
dtd
D =
⇒ The A.E. is 22 kcm − = 0 ⇒ m c k= ±
444 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
∴ 3 4 .c kt c ktT c e c e−= +
Thus, from equation (ii), we get
y = 1 2 3 4( )( )k x kx c kt c ktc e c e c e c e− −+ +There are arise following cases:
Case I: If k > 0 let k = p2
then 1 2 3 4( )( ).px px cpt cpty c e c e c e c e− −= + + ...(A)
Case II: If k < 0, let k = – p2
then m2 = – p2 ⇒ m = ± pi
∴ X = pxcpxc sincos 21 +and m2 = – p2c2 ⇒ m = ± ipc
∴ T = cptccptc sincos 43 +
then 1 2 3 4( cos sin )( cos sin .y c px c px c cpt c cpt= + + ...(B)
Case III: If k = 0
then XD2 = 0 ⇒ m = 0, 0
∴ X = (c1 + c2x)
And D2 T = 0 ⇒ m = 0, 0∴ T = (c3 + c4t)
Then 1 2 3 4( )( ).y c c x c c t= + + ...(C)
Of these three solutions, we have choose the solution which is consistent with the physical natureof the problem.
Since the physical nature of one dimension wave equation is periodic so we consider the solutionwhich has periodic nature.
Here only the solution in equation (B) is periodic (as both sine and cosine are periodic).Thus, the desired solution, for one dimensional wave equation is
1 2 3 4( , ) ( cos sin )( cos sin ).y x t c px c px c cpt c cpt= + + ...(iv)
Now using the boundary conditionsAt x = 0 (origin), y = 0 from (iv), we get
0 = )sincos( 431 cptccptcc + ⇒ c1 = 0using the value of c1 in (iv), we get
),( txy = )sincos(sin 432 cptccptcpxc + ...(v)At x = l (at A), y = 0 from (v), we get
0 = 2 3 4sin ( cos sin )c pl c cpt c cpt+
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 445
⇒ sin pl = 0 = sin nπ ⇒ pl = nπ
⇒ l
np
π= , where n = 1, 2, 3, LL
Hence, the solution of wave equation satisfying the boundary conditions is, from (v)
y (x, t) =
π
+ππ
l
ctnc
l
ctnc
l
xnc sincossin 432 l
np
π=
= l
xn
l
ctnb
l
ctna nn
π
π
+π
sinsincosn
n
bcc
acc
==
42
32
∴ The general solution of wave equation is
1
( , ) cos sin sinn nn
n ct n ct n xy x t a b
l l l
∞
=
π π π = +
∑ ...(vi)
Remark: We can apply initial conditions on above equation (vi) in time domain i.e., at t = 0.
Example 5: A string of length l is fastened at both ends A and C. At a distance ‘b’ from the endA, the string is transversely displaced to a distance ‘d’ and is released from rest when it is in thisposition. Find the equation of the subsequent motion.
Solution: Let y (x, t) is the displacement of the string.
Now, by the one dimensional wave equation, we have
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂ ...(1)
The solution of equation (1) is (from equation (iv), page 444)
Y
X
B b, d ( )
A (0, 0) C l ( , 0)D b ( , 0)
d
l
446 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
( , )y x t = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Now using the boundary conditions as follows:
The boundary conditions are
(i) At x = 0(at A), y = 0 ⇒ y(0, t) = 0
and (ii) At x = l (at C), y = 0 ⇒ y(l, t) = 0
From (2), we have
0 = )sincos( 431 cptccptcc + ⇒ c1 = 0Using c1 = 0, in equation (2), we get
),( txy = )sincos(sin 432 cptccptcpxc + ...(3)Using (ii) boundary condition, from (3), we have
0 = )sincos(sin 432 cptccptcplc +
⇒ sin pl = 0 ⇒ sin pl = sin nπ ⇒ .n
plπ=
Using the value of p in (3), we obtain
),( txy =
π
+ππ
l
ctnc
l
ctnc
l
xnc sincossin 432 ...(4)
Next, the initial conditions are as follows:
(iii) velocityty
∂∂
= 0 at t = 0
and displacement at t = 0 is
(iv) y(x, 0) = , 0
( ),
( )
d xx b
bd x l
b x lb l
⋅≤ ≤
− ≤ ≤ −
Equation of is
and equation of is
( )
( )
AB
d xy BC
bd x l
yb l
⋅=
−=
−From (4),
ty
∂∂
= 3 42 sin sin cos
n cc n ccn x n ct n ctc
l l l l l
π ππ π π − +
Using (iii) in above equation, we get
0 = lxn
lcn
ccπ⋅π
sin42 ⇒ c4 = 0
Using c4 = 0 in equation (4), we get
),( txy = lctn
lxn
ccπ⋅π
cossin32
∴ The general solution of given problem is
),( txy = l
ctn
l
xnb
nn
ππ∑∞
=
·cossin1
...(5)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 447
Using initial condition (iv) in equation (5), we get
y(x, 0) = ∑∞
=
π
1
sinn
nl
xnb
which is half range Fourier sine series, so we have
bn = 0
2( ,0) sin
ln x
y x dxl l
π⋅∫
= 0
2 2sin ( )sin
( )
b l
b
d n x d n xx dx x l dx
l b l l b l l
π π ⋅ + − − ∫ ∫
=
b
l
xn
n
l
l
xn
n
lx
bl
d
0
22
2
sin·cos2
π
π−
−π
π−
l
bl
xn
n
l
l
xn
n
llx
lbl
d
π⋅
π−
−π
π−
−−
+ sincos)()(
222
2
⇒ bn = –2
2 2
2 2 2cos sin cos
d n b dl n b d n bn l l n lbln
π π π+ ⋅ + ⋅π ππ
lbn
nlbl
dl ππ−
− sin)(
222
2
⇒ nb = lbn
nblb
dl ππ−
sin)(
222
2
∴ From (5), we get
),( txy = .cossinsin1
)(
2
122
2
l
ctn
l
xn
l
bn
nblb
dl
n
π⋅
π×
ππ− ∑
∞
=
Ans.
Example 6: A string is stretched and fastened to two points l apart. Motion is started by displacing
the string in the form y = a sin lxπ
from which it is released at a time t = 0. Show that the displacement
of any point at a distance x from one end at time t is given by
),( txy = sin cos .x ct
al l
π π
(U.P.T.U. 2004)
Solution: Let y(x, t) be the displacement at any point P(x, y) at any time.
448 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Then by the wave equation, we have
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
The solution of equation (1) is of the form
),( txy = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Now using the boundary conditions
(i) At x = 0, the displacement y = 0 ⇒ y(0, t) = 0
(ii) At x = l, the displacement y = 0 ⇒ y(l, t) = 0
Using (i) boundary condition in (2), we get
),0( ty = )sincos(0 431 cptccptcc += ⇒ 1 0.c =
∴ From (2), we get
),( txy = )sincos(sin 432 cptccptcpxc + ...(3)Using (ii) boundary condition in equation (3), we get
),( tly = )sincos(sin0 432 cptccptcplc +=
⇒ sin pl = 0 = sin nπ ⇒ .n
plπ=
Now using the initial conditions
(iii) At t = 0, the velocity 0=∂∂ty ⇒
0
0t
y
t =
∂ = ∂
(iv) At t = 0, the displacementlx
ayπ= sin ⇒
lx
axyπ= sin)0,(
∴ From (3), we have
ty
∂∂
= ]cos)(sin)([sin 432 cptcpccptcpcpxc +−
Using (iii) initial condition in above equation, we get
0 = pxcpcc sin42 ⇒ 042 =cpcc
⇒ 4 0.c =≠2 0,otherwise there
is trivial solution
c
Usingl
np
π= and ,04 =c in equation (3), we get
),( txy = lctn
lxn
ccππ
cossin32
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 449
∴ The general solution is
),( txy = ∑∞
=
ππ
1
cossinn
nl
ctn
l
xnb )( 32ccbn = ...(4)
Finally using (iv) initial condition in equation (4), we get
)0,(xy = ∑∞
=
π=
π
1
sinsinn
nl
xnb
l
xa
orlx
aπ
sin = LL+π+πlx
blx
b2
sinsin 21
Equating the coefficient of sin ,lxπ
we get
,1 ab = 032 === Lbb
Hence the required solution of given problem is
),( txy = lct
lx
aππ
cossin ; n = 1. Proved.
Example 7: Find the displacement of a string stretched between two fixed points at a distance 2lapart when the string is initially at rest in equilibrium position and points of the string are given initialvelocity v where
v = , when 0
2, when 2
xx l
ll x
l x ll
< <
− < <
x being the distance measured from one end.
Solution: The displacement y(x, t) is given by wave equation
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
The solution of equation is given by
),( txy = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Now, the boundry conditions are(i) At x = 0, y = 0 ⇒ y(0, t) = 0
(ii) At x = 2l, y = 0 ⇒ y(2l, t) = 0
Using (i) boundary condition in (2), we get
0 = )sincos( 431 cptccptcc + ⇒ c1 = 0
∴ From (2), we get
y(x, t) = )sincos(sin 432 cptccptcpxc + ...(3)
450 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Using (ii) condition in (3), we get
0 = )sincos(2sin 432 cptccptcplc +
⇒ sin 2pl = 0 = sin nπ ⇒ .2n
plπ=
Now, the initial conditions are
(iii) At t = 0 the displacement y(x, 0) = 0.
(iv) At t = 0, .y
vt
∂ =∂
Making use of initial condition (iii) in (3), we get
)0,(xy = 0)(sin0 332 =⇒= ccpxc
∴ From (3), we get
y(x, t) = 2 4 sin sin2 2
n x n cc c t
l lπ π
l
np
2
π=
The general solution is
),( txy = ∑∞
=
π⋅
π
1 2sin
2sin
nn t
l
cn
l
xnb ...(4)
⇒ty
∂∂
= tl
cn
l
xnnb
l
c
nn
2cos
2sin
2 1
πππ ∑∞
=
Using initial condition (iv) in above equation, we get
v = ∑∞
=
ππ
1 2sin
2 nn
l
xnnb
l
c
which represents half range Fourier sine series
∴ nnblc
2π
= 2
0
2sin
2 2
ln x
v dxl l
π∫
= 2
0
1 1 2sin sin
2 2
l l
l
x n x l x n xdx dx
l l l l l l
π − π +
∫ ∫
=
l
l
xn
n
l
ll
xn
n
l
l
x
l0
22
2
2sin
4)1(
1
2cos
2)1(
1
ππ
−−π
π−
l
ll
xn
n
l
ll
xn
n
l
l
xl
l
2
22
2
2sin
41
2cos
221
π
π−
−
−π
⋅
π−
−
+
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 451
= 2
sin4
2cos
2
2sin
4
2cos
22222
π
π+
ππ
+π
π+
ππ
−n
n
n
n
n
n
n
n
⇒ bn = 2 2
2 8sin
2
l n
n c n
π ⋅ π π
⇒ bn = 2
sin16
33
π⋅
π
n
cn
l
Hence the displacement function is given by, from equation (4), we get
y(x, t) = 3 3
1
16 1sin sin sin .
2 2 2n
l n n x n ct
l lc n
∞
=
π π π× ⋅
π ∑ Ans.
Example 8: A string is stretched and fastened to two points l apart. Motion is started by displacingthe string into the form y = k(lx – x2) from which it is released at time t = 0. Find the displacement ofany point on the string at a distance of x from one end at time t. (U.P.T.U. 2002)
Solution: Let the displacement y(x, t) given by the wave equation
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
∴ ),( txy = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Using the boundary conditions 0),0()( =tyi 0),()( =tlyiiUsing (i) in equation (2), we get
y = )sincos(sin 432 cptccptcpxc + ...(3)
Using (ii), in equation (3), we get
0 = )sincos(sin 432 cptccptcplc + ⇒ sin pl = 0
⇒ sin pl = sin nπ ⇒ .n
plπ=
and the initial conditions are
(iii) 00
=
∂∂
=tt
y(iv) )()0,( 2xlxkxy −=
∴ From (3), we get
ty
∂∂
= }cossin{sin 432 cptcccptccpxc +−
Using (iii) in above relation, we get
0 = pxccc sin)( 42 ⇒ 4 0.c =
452 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Using the values of c and p in equation (3), we get
y = lctn
lxn
ccππ
cossin32
∴ The general solution is
y = ∑∞
=
ππ
1
cossinn
nl
ctn
l
xnb ...(4)
Making use of (iv) in (4), we get
)( 2xlxk − = ∑∞
=
π
1
sinn
nl
xnb
which represents half range Fourier sine series
∴ nb = 2
0
2( )sin
lk n x
lx x dxl l
π−∫
⇒ bn =
π
π−−−
π
π−−
22
22 sin)2(cos)(
2
n
l
l
xnxl
x
l
l
xnxlx
l
k
l
n
l
l
xn
033
3
cos)2(
π
π−+
⇒ bn = odd. is when822
)1(2
33
2
33
3
33
31 n
n
kl
n
l
n
l
l
k n
π=
π+
π− +
= 0 when n is even.Hence the required displacement is, from (4), we get
y = ∑∞
=
πππ1
33
2
,cossin8
n
tl
cn
l
xn
n
klwhen n is odd. Ans.
Example 9: A tightly stretched string with fixed end points x = 0 and x = l is initially at rest inits equilibrium position. If it is set vibrating by giving to each of its points on initial velocity λx(l – x),find the displacement of the string at any distance x from one end at any time t.
[U.P.T.U. 2002 (C.O.) 2005]
Solution: We know that the solution of one dimensional wave equation with boundary conditions
y(0, t) =y (l, t) = 0 is
y(x, t) = )sincos(sin 432 cptccptcpxc + ...(1)
where p = π
.nl
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 453
Now the initial conditions are
(a) 0)0,( =xy (b) )(0
xlxt
y
t
−λ=
∂∂
=
Making use of (a) in (1), we get
0 = pxcc sin23 ⇒ 3 0.c =
∴ From (1), we get
),( txy = lctn
lxn
ccππ
sinsin42
The general solution of wave equation is
),( txy = ∑∞
=
π⋅
π
1
sinsinn
nl
ctn
l
xnb ...(2)
From (2),ty
∂∂
= ∑∞
=
ππ⋅π
1
cossin)/(n
nl
ctn
l
xnblcn
⇒0=
∂∂
tty
= ∑∞
=
π
π=−λ
1
sin)(n
nl
xnb
l
cnxlx
∴ nblcnπ
= 0
2( )sin
ln x
x l x dxl l
πλ −∫
=
−−
π
π−−
λ)2(cos)(
2xl
l
xn
n
lxlx
l
l
l
xn
n
l
l
xn
n
l
033
3
22
2
cos)2(sin
ππ
−+
ππ
−
= [ ]n
n
ln
n
l)1(1
4)cos1(
433
2
33
2
−−π
λ=π−
πλ
⇒ bn =
πλ
even is when 0,
odd is when ,8
44
3
n
ncn
l
∴ From (2) the required solution is
),( txy = ∑∞
=
πππλ
144
3
,sinsin18
n l
ctn
l
xn
nc
ln is odd. Ans.
454 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Example 10: If a string of length l is initially at rest in equilibrium position and each of its
points is given the velocity 3
0
sin ,t
y xb
t l=
∂ π = ∂ find the displace y(x, t). (U.P.T.U. 2001, 2006)
Solution: The equation for vibrations of the string is
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂
The solution of above equation is
),( txy = 1 2 3 4( cos sin )( cos sin )c px c px c cpt c cpt+ + ...(1)
At x = 0, the displacement y = 0
⇒ 0 = )sincos( 431 cptccptcc + ⇒ 1 0.c =
From (1), we get ),( txy = )sincos(sin 432 cptccptcpxc + ...(2)
Again at x = l, the displacement y = 0
∴ From (2), we get
0 = )sincos(sin 432 cptccptcplc +
⇒ sin pl = 0 = sin nπ ⇒ .n
plπ=
Next making the use of initial conditions i.e., at t = 0, the displacement y(x, 0) = 0.
Again from (2), we get
0 = pxcc sin32 ⇒ 03 =c
Using c3 = 0 and p = ,l
nπ equation (2) takes the form
y(x, t) = 2 4 sin sinn x n ct
c cl lπ π
∴ The general solution is
),( txy = 1
sin sinnn
n x n ctb
l l
∞
=
π π⋅∑ ...(3)
⇒ty
∂∂
= 1
sin cosnn
n c n x n ctb
l l l
∞
=
π π π⋅∑
At t = 0, ,sin3
0 l
xb
t
y
t
π=
∂∂
= we get
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 455
lx
bπ3sin =
1
sinnn
n c n xb
l l
∞
=
π π⋅∑
= 1 2 32 2 3 3
sin sin sinc x c x c x
b b bl l l l l l
π π π π π π ……+ ⋅ + +
lx
bπ3sin =
31 3 2 3
2 2 12( 9 ) sin sin sin
c x c x c xb b b b
l l l l l lπ π π π π π ……+ + ⋅ − +
Equating the coefficient of like powers of sin ,lxπ
we get
31 9bb + = 0 ...(i)
and 312
bl
cπ−= b ⇒ b3 =
cblπ
−12
from (i), 13
,4
blb
c=
π0542 ==== Lbbb
Hence from (3), the required displacement is
y(x, t) = 3 3 3
sin sin sin sin .4 12
bl x ct bl x ct
c l l c l l
π π π π + − π π Ans.
Example 11: Solve the equation 022
22
2
2
=∂∂−
∂∂∂+
∂∂
y
uyx
u
x
u using the transformation v = x + y,
z = 2x – y. [U.P.T.U. (C.O.) 2005]
Solution: We have
2
22
2
2
2y
uyx
u
x
u
∂∂−
∂∂∂+
∂∂
= 0 ...(1)
Given thatv = x + y ...(2)
and z = 2x – y ...(3)
Nowx
z
z
u
x
v
v
u
x
u
∂∂
∂∂
+∂∂
⋅∂∂
=∂∂
· = z
u
v
u
∂∂
+∂∂
2 (3) and (2) using,
⇒x∂∂
= 2v z
∂ ∂+
∂ ∂
∴ 2
2
x
u
∂
∂=
2 2 2
2 22 2 4 4
u u u u u
v z v z z vv z
∂ ∂ ∂ ∂ ∂ ∂ ∂ + + = + + ∂ ∂ ∂ ∂ ∂ ∂∂ ∂
456 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
andy
u
∂∂
= z
u
v
u
y
z
z
u
y
v
v
u
∂∂
−∂∂
=∂∂
⋅∂∂
+∂∂
⋅∂∂
⇒y∂
∂=
zv ∂∂
−∂∂
∴2
2
y
u
∂∂
= 2
22
2
2
2z
u
vz
u
v
u
z
u
v
u
zv ∂∂
+∂∂
∂−
∂∂
=
∂∂
−∂∂
∂∂
−∂∂
Alsoyx
u
∂∂∂2
= 2u u
v z v z
∂ ∂ ∂ ∂ + − ∂ ∂ ∂ ∂
= 2
222
2
2
22z
u
vz
u
zv
u
v
u
∂∂
−∂∂
∂+
∂∂∂
−∂∂
= 2
22
2
2
2z
u
zv
u
v
u
∂∂
−∂∂
∂+
∂∂
`
Putting the values of yx
u
x
u
∂∂∂
∂∂ 2
2
2
, and 2
2
y
u
∂∂
in equation (1), we get
∂ ∂ ∂ ∂ ∂ ∂+ + + + −
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
2 2 2 2 2 2
2 2 2 24 4 2
u u u u u u
z v v zv z v z– 0242
2
22
2
2
=∂∂
−∂∂
∂+
∂∂
z
u
vz
u
v
u
⇒ 0449222
2
2
22
2
2
2
2
=∂∂−
∂∂+
∂∂∂+
∂∂−
∂∂
z
u
z
u
vz
u
v
u
v
u
⇒ 092
=∂∂
∂vz
u
⇒ 02
=∂∂
∂vz
u
Integrating the above equation w.r.t. ‘z’ partially, we get
vu
∂∂
= )(1 vf
Again integrate w.r.t. ‘v’ partially, we get
u = ∫ +∂ )()( 21 zFvvf
= )()( 21 zFvF +
⇒ u = ).2()( 21 yxFyxF −++ Ans.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 457
Example 12: The points of trisection of a string are pulled aside through the same distance onopposite sides of the position of equilibrium and the string is released from rest. Derive an expressionfor the displacement of the string at subsequent time and show that the mid-point of the string alwaysremains at rest. (A.M.I.E.T.E., 1999)
Solution: Let the string OA be trisected at B and C.
Let the equation of vibrating string is
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
∴ The solution of equation (1) is
y(x, t) = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Now using the boundary conditions
(i) y(0, t) = 0 (ii) y(l, t) = 0
Making use of (i) and (ii) in equation (2), we get
y(x, t) = )sincos(sin 432 cptccptcpxc + ...(3)
whereπ= .
np
l
Next equation of OB' is y = 3/l
ax ⇒ y = xla3
Equation of B'C' is y – a =
−
−
+3
3
2
3
lx
llaa 2 1
1 12 1
( )y y
y y x xx x
−− = −
−
⇒ y = )2(3
xlla −
Y
XAO Bl/3
( /3, )l aB'
– a
a( 0 )l, C
−′ al
C ,3
223l
458 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
and equation of C'A is y – 0 = )(
3
20
lxl
la
−−
−−
⇒ y = )(3
lxla −
∴ The initial conditions of given problem are
(iii) 0=
∂∂
tt
y = 0
(iv) y(x, 0) =
≤≤−
≤≤−3
≤≤
lxl
lxl
a
lx
lxl
l
a
lxxl
a
3
2),(
33
2
3),2(
3/0,3
From equation (3), we get
ty
∂∂
= )cossin(sin 432 cptcpccptcpcpxc +−
Using (iii) initial condition in above, we obtain
0 = )(sin 42 cpcpxc ⇒ 4 0.c =
Again from (3), we have
y(x, t) = lctn
lxn
ccπ⋅π
cossin32
∴ The general solution of equation (1) is
y(x, t) = ∑∞
=
ππ
1
cossinn
nl
ctn
l
xnb ...(4)
Using (iv) condition in equation (4), we get
.y(x, 0) = ∑∞
=
π
1
sinn
nl
xnb
∴ bn = ∫π⋅
ldx
lxn
xyl 0
sin)0,(2
π π π = + − + − ∫ ∫ ∫
2
332
033
2 3 3 3sin ( 2 )sin ( )sin
lll
ll
ax n x a n x a n xdx l x dx x l dx
l l l l l l l
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 459
2 3
2 2 2 20
6 6cos (1) sin ( 2 ) cos
l
a l n x l n x a l n xx l x
n l l n ll n l
π π − π = − − − + − π ππ
22 2 23
2 2 2 2 22
3 3
6( 2) sin ( ) cos (1) sin
ll
l l
l n x a l n x l n xx l
l n l ln l n
− π − π − π− − + − − ππ π
= 2 2 2 2
2 2 2 2 2
6 2 2 2cos sin cos sin
3 3 3 3 3
a l n l l n l nn
n nl n n
π π π− + π + − π ππ π
2 2 2 2
2 2 2 2
2 2 2cos sin cos sin
3 3 3 3 3 3
l n l n l n l n
n nn n
π π π π+ + − + π ππ π
=
π2−
ππ 3
sin3
sin3
·6
22
2
2
nn
n
l
l
a
= 2 2
18sin [1 ( 1) ]
3na n
n
π+ −
π 3sin)1(
3sin
3
2sin
π−−=
π
−π=π nn
nn n
⇒ bn=
ππ
even is when,3
sin36
odd is when ,0
22 nn
n
a
n
Putting the value of bn in equation (4), we get
y(x, t) = 2 22(even)
36sin sin cos .
3n
a n n x n ct
l ln
∞
=
π π ππ∑ Ans.
Example 13: A tightly stretched string with fixed end points x = 0 and x = π is initially at rest inits equilibrium position. If it is set vibrating by giving to each of its points an initial velocity
0=
∂∂
tt
y= xx 2sin06.0sin05.0 −
then find the displacement y (x, t) at any point of string at any time t.
Solution: We know that
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
460 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
and its solution is
),( txy = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
The boundary conditions are
(a) 0),0( =ty (b) 0),( =π ty
Using (a) in equation (2), we get c1 = 0
⇒ y(x, t) = )sincos(sin 432 cptccptcpxc + ...(3)
Using (b) in equation (3), we get
0 = )sincos(sin 432 cptccptcpnc +
⇒ sin pn = 0 = sin nπ ⇒ .p n=
Now, the initial conditions are
(c) 0)0,( =xy (d) xxt
y
t
2sin06.0sin05.00
−=
∂∂
=
At t = 0, from (3), we have
0 = pxcc sin32 ⇒ 3 0.c =
∴ Again from (3), we obtain
),( txy = 2 4 sin sinc c nx nct np = As
The general solution is
),( txy = ∑∞
= 1
sinsinn
n nctnxb ...(4)
⇒ty
∂∂
= ∑∞
=
⋅1
cos)(sinn
n nctncnxb
At t = 0
0=
∂∂
tt
y= ∑
∞
=
⋅=−1
sin2sin06.0sin05.0n
n nxbncxx
⇒ xx 2sin06.0sin05.0 − = L++ xcbxcb 2sin2sin 21
⇒ 1cb = 05.0 ⇒c
b05.0
1 = and 06.02 2 −=cb ⇒c
b03.0
2 −=
∴ From (4), we get
),( txy = .2sin2sin03.0
sinsin05.0
ctxc
ctxc
− Ans.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 461
Example 14: If the string of length l is initially at rest in equilibrium position and each of itspoints is given the velocity.
π
π
l
x
l
xv
2cos
3sin0 where lx <<0 at t = 0 determine the displacement function
).,( txy
Solution: The displacement ),( txy given by wave equation
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
we know that the solution of (1) is
),( txy = )sincos)(sincos( 4321 cptccptcpxcpxc ++ ...(2)
Using the boundary conditions
(i) 0),0( =ty (ii) 0),( =tly
We get from (2)
),( txy = )sincos(sin 432 cptccptcpxc + ...(3)
where p =π
.nl
and the initial conditions are
(iii) 0)0,( =xy (iv) 00
3 2sin cos
t
y x xv
t l l=
∂ π π = ∂
Using (iii) in equation (3), we get
0 = cptpxcc cossin32 ⇒ 3 0.c =
Making use 03 =c in equation (3), we get
),( txy = 2 4 sin sinc c px cpt
where p =π
.nl
or the general form of solution is
),( txy = 1
sin sinnn
n x n ctb
l l
∞
=
π π∑ ...(4)
Differentiating partially w.r.t. ‘t’ equation (4), we get
ty
∂∂
= 1
sin cosnn
n c n x n ctb
l l l
∞
=
π π π ⋅
∑
∴0t
y
t =
∂ ∂
= 01
3 2sin cos sinn
n
x x n c n xv b
l l l l
∞
=
π π π π =
∑
462 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
⇒
π
+π
l
x
l
xvsin
5sin
20 = ∑
∞
=
π
π
1
sinn
nl
xn
l
cnb
Equating the coefficient of like terms, we have
20v
=
π
l
cb1 ⇒
π=
c
lvb
20
1
20v
= π
=⇒
π
c
lvb
l
cb
5
5 055
and 065432 ====== Lbbbbb
Using these values in equation (4), we get the required solution
),( txy = .5
sin5
sin5
sinsin2
00
l
ct
l
x
c
lv
l
ct
l
x
c
lv π
π
π
+π
π
π
Ans.
5.4 D’ALEMBERT’S SOLUTION OF WAVE EQUATION
Transform the equation 2
22
2
2
x
yc
t
y
∂
∂=
∂
∂ to its normal form using the transformation ,ctxu += ctxv −=
and hence solve it. Show that the solution may be put in the form 1
( ) ( ) .2
y f x ct f x ct= + + −
Assume initial conditions )(xfy = and 0=∂∂ty at t = 0. [U.P.T.U., 2003]
Proof: Consider one dimensional wave equation
2
2
t
y
∂
∂= 2
22
x
yc
∂
∂...(1)
Let ctxu += and ,v x ct= − be a transformation of x and t into u and v.
thenxy
∂∂
= y u y v y yu x v x u v
∂ ∂ ∂ ∂ ∂ ∂+ = +∂ ∂ ∂ ∂ ∂ ∂
1,1 =∂∂
=∂∂
x
v
x
u
orx∂∂
= u v∂ ∂+
∂ ∂
⇒ 2
2
x
y
∂
∂=
y y y
x x u v u v
∂ ∂ ∂ ∂ ∂ ∂ = + + ∂ ∂ ∂ ∂ ∂ ∂
= 2 2 2 2
2 2
y y y y
u v v uu v
∂ ∂ ∂ ∂+ + +
∂ ∂ ∂ ∂∂ ∂
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 463
⇒ 2
2
x
y
∂
∂=
2 2 2
2 22
y y y
u vu v
∂ ∂ ∂+ +
∂ ∂∂ ∂...(2)
andty
∂∂
= y u y v y y
c cu t v t u v
∂ ∂ ∂ ∂ ∂ ∂+ = −∂ ∂ ∂ ∂ ∂ ∂
ct
vc
t
u−=
∂∂
=∂∂
,
ort∂
∂=
∂∂
−∂∂
vuc
∴ 2
2
t
y
∂
∂=
∂∂
−∂∂
∂∂
−∂∂
=
∂∂
∂∂
v
yc
u
yc
vuc
t
y
t
=
∂
∂+
∂∂∂
−∂∂
∂−
∂
∂2
222
2
22
v
y
uv
y
vu
y
u
yc
⇒ 2
2
t
y
∂
∂=
∂
∂+
∂∂∂
−∂
∂2
22
2
22 2
v
y
vu
y
u
yc ...(3)
Making use of equations (2) and (3) in equation (1), we get
∂
∂+
∂∂∂
−∂
∂2
22
2
22 2
v
y
vu
y
u
yc =
2 2 22
2 22
y y yc
u vu v
∂ ∂ ∂+ + ∂ ∂∂ ∂
⇒vuy
c∂∂
∂224 = 0 ⇒ 0
2
=∂∂
∂vuy
...(4)
Integrating equation (4), w.r.t. ‘v’, we get
uy
∂∂2
= ( )uφ ...(5)
where φ(u) is a constant in respect of v.
Again integrate equation (5) w.r.t. ‘u’, we get
y = 2( ) ( )u du vφ + φ∫⇒ y = )()( 21 vu φ+φ
⇒ 1 2( , ) ( ) ( ).y x t x ct x ct= φ + + φ − ...(6)
The solution (6) is D’Alembert’s solution of wave equation
Now, we applying initial conditions )(xfy = and 0=∂∂ty
at t = 0
464 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
From (6), we get at t = 0
f(x) = )()( 21 xx φ+φ ...(7)
andty
∂∂
= 1 2( ) ( )c x ct c x ct′ ′φ + − φ −
⇒0=
∂∂
tt
y= 0 = 1 2( 0) ( 0)c x c x′ ′φ + − φ −
⇒ 1 2( ) ( )x x′ ′φ − φ = 0
⇒ 1( )x′φ = 2( )x′φOn integrating, we get
)(1 xφ = 12 )( cx +φ ...(8)
Using equation (8) in equation (7), we get
f(x) = 12212 )(2)()( cxxcx +φ=φ++φ
⇒ )(2 xφ = 1 2 11 1
( ) ( ) ( )2 2
f x c x ct f x ct c− ⇒ φ − = − −
and )(1 xφ = 1 1 11 1
( ) ( ) ( )2 2
f x c x ct f x ct c+ ⇒ φ + = + +
Putting the values of )(1 ctx +φ and )(2 ctx −φ in equation (6), we get
y(x, t) = 1
( ) ( ) .2
f x ct f x ct+ + − Proved.
EXERCISE 5.1
1. Solve completely the equation ,2
22
2
2
x
yc
t
y
∂∂
=∂∂
representing the vibrations of a string of
length l, fixed at both ends, given that y(0, t) = 0, y(l, t) = 0; y(x, 0) = f(x) and 00
=
∂∂
=tt
y,
0 < x < l.1 0
2 ( , ) sin cos ; ( )sin
l
n nn
n x n ct n xy x t b b f x
l l l l
∞
=
π π π = =
∑ ∫Ans. (U.P.T.U. 2005)
2. A taut string of length 2l is fastened at both ends. The midpoint of the string is taken to aheight h and then released from rest in that position. Find the displacement of the string.
1
2 21
8 ( 1) (2 1) (2 1) ( , ) sin .cos
2 2(2 1)
n
n
h n x n cty x t
l ln
∞ −
=
− − π − π = π −
∑Ans.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 465
3. A uniform elastic string of length 60 cm is subjected to a constant tension of 2 kg. If theends are fixed and the initial displacement is y(x, 0) = 60x – x2 for 0 < x < 60 while theinitial velocity is zero, find y(x, t).
∞
=
− π× − π = × π − ∑
2
3 31
(2 1)8 60 1 (2 1) ( , ) sin cos
60 60(2 1)n
n ctn xy x t
nAns.
4. A taut string of length 20 cms fastened at both ends is displaced from its position of equilibriumby imparting to each of its points an initial velocity is given by
v =
<<−<<
2010 in20
100 in
xx
xx
x being the distance from one end. Determine the displacement at any subsequent time.
3 31
1600 1 ( , ) sin sin ·sin
2 20 20n
n n x n cty x t
c n
∞
=
π π π = π ∑Ans.
5. A tightly stretched string with fixed end points x = 0 and x = π is initially at rest in the
equilibrium position. If it is set vibrating by giving each point a velocity 0=
∂∂
tt
y= 0.03
sin x – 0.04 sin 3x then find the displacement y(x, t) at any point of the string at any time t.
1 ( , ) [0.03sin sin 0.133sin3 sin3 ]y x t x ct x ct
c = − Ans.
6. A tightly stretched string with fixed end points x = 0 and x = l is initially in a position given
by y = y0 sin3
π
l
x. If it is released from rest from this position, find the displacement y(x, t).
0 3 3 ( , ) 3sin cos sin cos
4
y x c t x c ty x t
l l l l
π π π π = − Ans.
7. A tightly stretched flexible string has its ends fixed at x = 0 and x = l. At time t = 0 the stringis given a shape defined by y = µx(l – x), µ is a constant and then released. Find thedisplacement y(x, t) of any point x of the string at any time t > 0.
2
3 31
4 1 ( 1) ( , ) ·cos sin
n
n
l n ct n xy x t
l ln
∞
=
µ − − π π = π
∑Ans.
8. A uniform string of length l is struck in such a way that an initial velocity of v0 is imparted
to the portion of the string 43
and 4
ll while the string is in equilibrium position. Find the
466 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
subsequent displacement of the string as a function of x and t.
π π π π π π = − + − π
02 2 2 2
4 1 1 3 3 1 5 5 ( , ) sin .sin sin sin sin sin ·····
2 1 3 5
lv x ct x ct x cty x t
l l l l l lcAns.
9. Find the displacement if a string of length a is vibrating between fixed end points with
initial velocity zero and initial displacement given by y(x, 0) =
2for 0
22
2 for .2
px ax
apx a
p x aa
< <
− < <
1
2 21
8 ( 1) (2 1) (2 1) ( , ) sin ·cos
(2 1)
n
n
p n x n cty x t
a an
∞ +
=
− − π − π = π −
∑Ans.
10. A taut string of length l has its ends x = 0 and x = l fixed. The mid-point is taken to a smallheight h and released from rest at time t = 0. Find the displacement y(x, t).
ππππ
= ∑∞
=122
·cos·sin2
sin8
),( n
l
ctn
l
xnn
n
htxyAns.
11. If a string of length l is released from rest in the position y = 2
)(4
l
xlkx − show that the
motion is described by the equation.
π+π++π
= ∑∞
= 132
)12(cos
)12(sin
)12(
132),(
nl
ctn
l
xn
n
ktxyAns.
5.5 ONE DIMENSIONAL FLOW
In this section we set up the mathematical model for one dimensional heat flow and derive thecorresponding partial differential equation.
Consider a bar or a rod of equal thickness at every point.
Let the area of cross-sectional = A cm2.
and density of material of rod = 3/cmgrρ
Here we consider a small element PQ of length .xδ
X
P' Q'
O Px δx Q
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 467
∴ The mass of the element PQ = xAρδLet u(x, t) is the temperature of the rod at a distance x at time t.
We know that the amount of heat in a body is always proportional to the mass of the body and tothe temperature change.
Thus the rate of increase of heat in element
= u
sA xt
∂ρδ∂
(s is specific heat) ...(i)
Since the direction of heat flow in a body becomes always toward decreasing temperature. Physicalexperiment shows that the rate of flow is proportional to the area and to the temperature gradientnormal to the area. If we suppose Q1 and Q2 are the quantities of heat flowing at the points P and Qrespectively,
then Q1 = secondper xx
ukA
∂∂
−
and Q2 = per secondx x
ukA
x + δ
∂ − ∂
where k is a constant known as thermal conductivity.
∴ Total amount of heat in the element = Q1 – Q2 = kA
∂∂
−
∂∂
δ+ xxx x
u
x
u per second ...(ii)
From (i) and (ii)
tu
xsA∂∂δρ =
∂∂
−
∂∂
δ+ xxx x
u
x
ukA
⇒tu
∂∂
= x x x
u ux xk
s x+ δ
∂ ∂ − ∂ ∂ ρ δ
Taking the limit as 0→δx i.e., when xxx →δ+
∴tu
∂∂
= 0
lim x x x
x
u ux xk
s x+ δ
δ →
∂ ∂ − ∂ ∂ ρ δ
= 2
2
k u k u
s x x s x
∂ ∂ ∂ = ρ ∂ ∂ ρ ∂
The negative sign shows
the direction of heat flow
towards lower temperature
468 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Let 2kc
s=
ρ is called diffusivity of the substance
Thus 2
22
.u u
ct x
∂ ∂=
∂ ∂
Alternate: Since the rate of flow is proportional to the gradient of the temperature
∴ Vr
= uk∇− ...(i)
where Vr
is the velocity of the heat flow and u(x, y, z, t) is temperature and k is thermal conductivity..
Let R be the region in the body and let s be its boundary surface.
Then amount of heat leaving R per unit time = ∫∫S
dAnV ˆ·r
where n̂ is the outward unit normal of s.
From (i) and by Gauss divergence theorem, we obtain
ˆ·S
V ndA∫∫r
= ·( )R
k u dxdydz− ∇ ∇∫∫∫
= 2
R
k udxdydz− ∇∫∫∫ ...(ii)
And the amount of heat in R is given by
Q = R
s udxdydzρ∫∫∫where s is the specific heat of the material of the body. Here the rate of decrease of Q is given by
tQ
∂∂− =
∂− ρ
∂∫∫∫R
us dxdydz
t...(iii)
Since the rate of decrease of Q must be equal to the amount of heat leaving R , from (ii) and (iii),we get
R
us dxdydz
t
∂− ρ
∂∫∫∫ = 2
R
k udxdydz− ∇∫∫∫
⇒ 2
R
us k u dxdydz
t
∂ ρ − ∇ ∂ ∫∫∫ = 0
Since this holds for any region R in the body the integrand must be zero everywhere.
∴tu
s∂∂ρ = uk 2∇
⇒tu
∂∂
= us
k 2∇ρ
...(iv)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 469
Letρs
k= c2, then equation (iv) reduces in the form
tu
∂∂
=2
2 2 22
.u u
c u ct x
∂ ∂∇ ⇒ =
∂ ∂
2
22
2
2
2
2
2
22
. along.,.dimension one in
·
x
uu
Xei
z
u
y
u
x
uuu
∂∂
=∇
∂∂
+∂∂
+∂∂
=∇∇=∇
5.5.1 Solution of One Dimensional Heat Equation (U.P.T.U. 2007)
We know that the heat equation
tu
∂∂
= 2
22
x
uc
∂∂
...(1)
Let u(x, t) = X(x) T(t) ...(2)
⇒xu
∂∂
= 2
2
2
2
or dx
XdT
x
u
dx
dXT =
∂∂
andtu
∂∂
= dtdT
X
Using these values in equation (1), we get
dtdT
X = 2
22
dx
XdTc
⇒dt
dT
Tc2
1= k
dx
Xd
X=
2
21...(3)
Taking 2nd and 3rd terms
∴2
21
dx
Xd
X= k ⇒ 0
2
2
=− kXdx
Xd
⇒ XkD )( 2 − = 0
Hence X = xkxk ecec −+ 21
anddt
dT
Tc2
1= k ⇒ dtkc
TdT 2=
On integrating, loge T = 32 logctkc +
⇒ T = tkcec2
3
470 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
∴ From (2), we get
tkcxkxk ececectxu2
321 )(),( −+=
There are arise following cases:
Case I: If k > 0, let k = p2
then 2 2
1 2 3( , ) ( ) .px px p c tu x t c e c e c e−= + ...(A)
Case II: If k < 0, Let k = –p2
then A.E. is m2 = –p2 ⇒ m = ± pi
∴ X = pxcpxc sincos 21 +
then 2 2
1 2 3( , ) ( cos sin ) .p c tu x t c px c px c e−= + ...(B)
Case III: If k = 0, then
1 2 3( , ) ( ) .u x t c c x c= + ...(C)
Since the physical nature of the problem is periodic so the suitable solution of the heat equationis
2 2
1 2 3( , ) ( cos sin ) .p c tu x t c px c px c e−= + ...(4)
The boundary conditions are
(i) u(0, t) = 0 and (ii) u(l, t) = 0
and initial condition is (iii) u(x, 0) = f(x).
Using (i) boundary condition in (4), we get
0 = 0131
22
=⇒− cecc tcp
∴ From (4), we get
u(x, t) = c2c3 sin px· tcpe22− ...(5)
Using (ii) boundary condition in (5), we get
0 = l
npnplplcc
π=⇒π==⇒ sin0sinsin32
∴ u(x, t) = 2
222
sin32l
tcn
elxn
ccπ
−π
The general form of above solution is
u(x, t) = 2 2 2 /
1
sin · n c t ln
n
n xb e
l
∞− π
=
π∑ (bn = c2c3) ...(6)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 471
Again using initial condition (iii) in equation (6), we get
f (x) = ∑∞
=
π
1
sinn
nl
xnb
which represents Fourier half range sine series so, we have
bn = ∫π
l
dxl
xnxf
l0
sin)(2
Thus the required solution is
2 2 2
1
0
( , ) sin
2where ( )sin .
n c t
ln
n
l
n
n xu x t b e
l
n xb f x dx
l l
π∞ −
=
π=
π=
∑
∫
Remark: In steady state 2
20, so 0.
u u
t x
∂ ∂= =
∂ ∂
Example 15: Determine the solution of one dimension heat equation
tu
∂∂ = 2
22
x
uc
∂∂
Under the conditions u(0, t) = u(l, t) = 0 and )0,(xuif 0 /2
if .2
x x l
ll x x l
≤ ≤=
− ≤ ≤
Solution: We have
tu
∂∂
= 2
22
x
uc
∂∂
...(1)
We know that the solution of equation (1) (on page 470, equation 4) is
u(x, t) = tcpecpxcpxc22
321 )sincos( −+ ...(2)
At x = 0, we get
0 = 2 2
1 3 1 0.p c tc c e c− ⇒ =∴ From equation (2), we get
u(x,t) = tcpepxcc22
·sin32− ...(3)
472 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Again at x = l from (3), we get
0 = π==⇒− npleplcc tcp sin0sin·sin22
32
⇒ p = l
nπ
From (3), we get
u(x, t) = t
l
cn
elxn
cc2
222
sin32
π−π
⇒ Therefore the general form of solution can be written as
u(x, t) = ∑∞
=
π−π
1
2
222
·sinn
l
tcn
n el
xnb ...(4)
At t = 0, from equation (4), we get
u(x, 0) = ∑∞
=
π
1
sinn
nl
xnb
∴ bn = 0
2( ,0)sin
ln x
u x dxl l
π∫
= π π + − ∫ ∫/ 2
0 /2
2sin ( )sin
l l
l
n x n xx dx l x dx
l l l
= π π
− + π π
/22
2 20
2cos sin
ll n x l n x
xl n l ln
π π + − − −
π π
2
2 2/2
( ) cos sin
l
l
l n x l n xl x
n l ln
=
ππ
+π
π+
ππ
+π
π−
2sin
2cos
22sin
2cos
2
222
22
22
22 n
n
ln
n
ln
n
ln
n
l
l
= 2 2
4sin .
2
l n
n
π π
∴ From equation (4), we get
u(x, t) = ∑∞
=
π−ππ
π 122 .··sin
2sin
14 2
222
n
tl
cn
el
xnn
n
l Ans.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 473
Example 16: An insulated rod of length l has its ends A and B maintained 0ºC and 100ºCrespectively until steady state conditions prevail. If B is suddenly reduced to 0ºC and maintained at 0ºCfind the temperature at a distance x from A at time t. [U.P.T.U., 2004, 2005]
Solution: From one dimensional and equation, we have
tu
∂∂
= 2
22
x
uc
∂∂
...(1)
The boundary conditions are
(i) u(0, t) = 0ºC and (ii) u(l, t) = 100ºC
In steady state condition 0=∂∂
tu here from (1), we get
2
2
x
u
∂∂
= 0
On integrating, we get u(x) = c1x + c2 ...(2)
where c1 and c2 are constants to be determined
At x = 0, from equation (2), we have
20 .c=
and at x = l, 100 = c1l + 0 ⇒ .100
1l
c =
∴ From (2)
u(x) = xl
100...(3)
Now the temperature at B is suddenly changed we have again transient state. If u(x, t) is thesubsequent temperature function, the boundary conditions are
(iii) u(0, t) = 0ºC, (iv) u(l, t) = 0ºC and the initial condition (v) u(x, 0) = xl
100
Since the subsequent steady state function us(x) satisfies the equation
2
2
x
us
∂∂
= 0
or2
2sd u
dx= 0 ⇒ 43)( cxcxus +=
at x = 0, we get 0 = c4
and at x = l, we get 0 = c3l + 0 ⇒ c3 = 0
Thus us(x) = 0 ...(4)
474 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
If uT (x, t) is the temperature in transient state then the temperature distribution in the rod u(x, t)can be expressed in the form
u(x, t) = ),()( txuxu Ts +
⇒ u(x, t) = uT(x, t) 0)( As =xus ...(5)
Again from heat equation, we have
tuT
∂∂
= 2
22
x
uc T
∂∂
...(6)
The solution of equation (6) is
uT(x, t) = tpcecpxcpxc22
321 )sincos( −+
⇒ u(x, t) = tpcecpxcpxc22
321 )sincos( −+ ...(7)
At x = 0, u(0, t) = 0
⇒ 0 = 2 2
1 1 0.c p tc e c− ⇒ =From (7), we get
u(x, t) = tpcecpxc22
32 ·sin − ...(8)
Again at x = l, u (l, t) = 0
⇒ 0 = c2c3 sin pl· π==⇒− nple tpc sin0sin22
⇒ p = l
nπ
From (8), we get
u(x, t) = 2
222
·sin32l
tcn
elxn
ccπ
−π
⇒ u(x, t) =
π∞ −
=
π∑2 2 2
2
1
sin ·
n c t
ln
n
n xb e
l...(9)
Using initial condition i.e., at t = 0, u = ,100
xl
we get
u(x, 0) = ∑∞
=
π=
1
sin100
nn
l
xnbx
l
∴ bn = ∫π
l
dxl
xn
l
x
l0
·sin1002
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 475
= 2
2 2 2 200
200 200·sin cos sin
lln x xl n x l n x
x dxl n l ll l n
π π π= − +
π π ∫
⇒ bn = 2
12
200 200cos ( 1)nl
nn nl
+ −π = −
π π Hence from equation (9), we get
u(x, t) = ∑∞
=
π−+ π−
π 1
1
.sin)1(200 2
222
n
tl
cnn
el
xn
nAns.
Example 17: Determine the solution of one dimensional heat equation 2
22
x
uc
t
u
∂∂
=∂∂
subject to
the boundary conditions u(0, t) = 0, u(l, t) = 0 (t > 0) and the initial condition u(x, 0) = x, l being thelength of the bar. (U.P.T.U. 2006)
Solution: We have
tu
∂∂
= 2
22
x
uc
∂∂
...(1)
We know that the solution of equation (1) is given by
u(x, t) = tcpecpxcpxc22
321 )sincos( −+ ...(2)
At x = 0, u = 0
⇒ 0 = 0131
22
=⇒− cecc tcp
From (2), we get
u(x, t) = tcpepxcc22
sin32− ...(3)
Again at x = l, u = 0
⇒ 0 = tcpeplcc22
·sin32−
⇒ sin pl = 0 = sin nπ
⇒ p = .l
nπ
From (3) the general solution of equation (1) is
u(x, t) = ∑∞
=
π−π
1
2
222
·sinn
l
tcn
n el
xnb ...(4)
At t = 0, u = x
⇒ x = ∑∞
=
π
1
sinn
nl
xnb
476 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
∴ bn = ∫
ππ
−−
π−
π=
πl l
l
xn
n
l
l
xn
n
lx
ldx
l
xnx
l0 0
22
2
sincos·2
sin2
=
−
π
π+π−
π0sin)cos(·
222
2
nn
ln
n
ll
l
⇒ bn = 2
12 2( 1) ( 1) ·n nl l
l n n+ −
− = − π π
Putting the value of bn in equation (4), we get
u(x, t) =
2 2 2
21
1
2 ( 1)sin .
n c tnl
n
l n xe
n l
π∞ + −
=
− ππ ∑ Ans.
Example 18: A homogeneous rod of conducting material of length l has its ends kept at zerotemperature. The temperature at the centre is T and falls uniformly to zero at the two ends. Find thetemperature distribution.
Solution: From heat equation in one dimension, we have
tu
∂∂
= 2
22
.u
cx
∂
∂...(1)
The boundary conditions are (i) u (0, t) = 0 (ii) u(l, t) = 0.Since the temperature at the centre is T and falls uniformly to zero at the two ends, its distribution
at t = 0 is given in the figure.
The equation of line segment OB is u = 2
0;l
xxlT ≤≤2
and the equation of line segment BA is
given by
u
XD
B l T( /2, )
O (0, 0) A l ( , 0)
0,2
l
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 477
u = lxl
lxlT ≤≤−
2;
)(2 )( line of equation 112
121 xx
xxyy
yy −−−=−
Thus initial condition i.e., at t = 0 is given by
u(x, 0) =
≤≤−
≤≤
lxl
l
xlT
lx
l
Tx
2when,
)(22
0when,2
We know that the solution of equation (1) satisfying the boundary conditions (i) and (ii) is givenby
u(x, t) = ∑∞
=
π−π
1
2
222
sinn
l
tcn
n el
xnb ...(2)
At t = 0, from (2), we get
u(x, 0) = ∑∞
=
π
1
sinn
nl
xnb
∴ bn = 0
2( ,0) sin
ln x
u x dxl l
π⋅∫
bn = / 2
0 2
2 2 2 ( )sin sin
l l
l
Tx n x T l x n xdx dx
l l l l l
π − π + ⋅ ∫ ∫
=
/ 22 2
2
0 /2
4cos sin ( ) cos sin
l l
l
T l n x l n x l n x l n xx l x
n l n l n l n ll
π π π π − + + − − − π π π π
= 2 2 2 2
2 2 2 2 2
4cos sin cos sin
2 2 2 2 2 2
T l n l n l n l n
n nl n n
π π π π− + + +
π ππ π
⇒ bn = 2 2
8sin
2
T n
n
π
πPutting the value of bn in equation (2), we get
),( txu = ∞
−
=
π π ππ ∑
2 2 2
2 2 21
8 1sin sin .
2n
T n n x n c te
ln lAns.
478 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
60)10(
90)0(1
=
=⇒→
s
s
s
u
u
uu
Example 19: A rod of length 10 cm has its ends A and B kept at 50ºC and 100ºC until steady stateconditions prevail. The temperature at A suddenly raised to 90ºC and that at B is lowered to 60ºC andthey are maintained. Find the temperature at a distance x from one end at time t.
Solution: From one dimensional heat equation
tu
∂∂
= 2
22
uc
x
∂
∂...(1)
The boundary conditions are (i) u (0, t) = 50º and (ii) u(10, t) = 100ºC
In steady state condition 0=∂∂
tu
, so from (1), we get
2
2
x
u
∂∂
= 0
On integrating, we get u(x) = C1x + C2 ...(2)
Making use of boundary conditions in above equation, we get
50 = C2
and 100 = 2110 CC + ⇒ C1 = 5
∴ From equation (2)u(x) = 505 +x ...(3)
Now, the temperatures at A and B are suddenly changed we have again transient state.If u1(x, t) is subsequent temperature function then the boundary conditions are
),0(1 tu = 90ºC, u1(10, t) = 60ºC
and the initial condition
)0,(1 xu = 5x + 50
If us(x) is the subsequent steady state function then
2
2su
x
∂
∂= 0 ⇒
2
20sd u
dx=
⇒ us(x) = C3x + C4 ...(4)At x = 0, we get
90 = C4 and at x = 10, we get
60 = 10 C3 + 90 ⇒ C3 = –3
∴ us(x) = –3x + 90
Thus the total temperature distribution in the rod at a distance x at time t is given by
u (x, t) = ),()( txuxu Ts +
where ),( txuT is the temperature in transient state
⇒ u(x, t) = ),()903( txux T++− ...(5)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 479
Now again by heat equation, we have
tuT
∂∂
= 2
22Tu
cx
∂
∂We know that the solution of above equation uT (x, t) is given by
),( txuT = 2 2
1 2 3( cos sin ) c p tC px C px C e−+ ...(6)
Now the boundary conditions which satisfied by ),( txuT are as follows:
),0( tuT = 09090)0(),0(1 =−=− sutu
),( tluT = 1(10, ) (10) 60 60 0su t u− = − =
and )0,(xuT = )()0,(1 xuxu s−
⇒ )0,(xuT = 408)903(505 −=+−−+ xxx
The general solution for ),( txuT is given by
),( txuT = ∞
−
=
π π∑2 2 2
1
sin10 100
nn
n x n c tb e ...(7)
At t = 0, uT = 8x – 40
⇒ 8x – 40 = ∑∞
=
π
1 10sin
nn
xnb
∴ bn = 10
0
2(8 40)sin
10 10
n xx dx
π−∫
= 10
022 10
sin800
10cos)408(
10
5
1
π
π+
π−
π−
xn
n
xnx
n
=
π−π
π+π
π−
nn
nn
n
400sin
800cos
400
5
122
⇒ bn = { }1)1(80400
)1(400
5
1 1 −−π
=
π−−
π− +nn
nnn
⇒ bn = 0,when is odd
160,when is even
n
nn
− πFrom (7), we get
),( txuT =
2 2 24
100
1
160 2sin
2 10
m c t
m
m xe
m
π∞ −
=
π−
π∑ (even)2mn =
480 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
or uT(x, t) =
2 2 2
25
1
80 1sin
5
m c t
m
m xe
m
π∞ −
=
π−
π ∑ ...(8)
Hence from equation (5), we get
u(x, t) =
2 2 2
25
1
80 13 90 sin · .
5
m c t
m
m xx e
m
π∞ −
=
π− + −
π ∑ Ans.
Example 20: The temperature of a bar 50 cm long with insulated sides is kept at 0ºC at one endand 100ºC at the other end until steady conditions prevail. The two ends are then suddenly insulated sothat the temperature gradient is zero at each end thereafter. Find the temperature distribution.
Solution: The temperature function u(x, t) is the solution of the one dimensional heat equation
tu
∂∂ = 2
22
x
uc
∂∂
...(1)
When the steady state condition prevails 0=∂∂
tu and hence from (1), we get
2
2
x
u
∂∂
= 0
On integrating, we getu(x) = c1x + c2 ...(2)
At x = 0, u = 0∴ 0 = c2
and at x = 50, u = 100, from (2), we get
100 = 50 c1 + 0 ⇒ c1 = 2
Hence u(x) = 2x ⇒ u(x, 0) = 2x ...(3)
and the subsequent temperature function u1 (x, t) satisfy the boundary conditions
u1 (0, t) = 0, u1 (50, t) = 0
Under these conditions, we find the steady state function us(x) vanishes
i.e., us(x) = 0 suu →1
⇒ u(x, t) = ),(0),()( txutxuxu TTs +=+
⇒ u(x, t) = ),( txuT ...(4)
where uT (x, t) is the temperature in transient state which satisfied the boundary conditions
uT (0, t) = 0 = uT (50, t)
∴ The temperature uT(x, t) can be obtained by the solution of one dimensional heat equation
⇒ ),( txuT = tpcecpxcpxc22
321 )sincos( −+ ...(5)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 481
At x = 0, uT = 0
⇒ 0 = tpcecc22
31− ⇒ c1 = 0 (otherwise uT (x, t) = 0)
From (5), we get
),( txuT = tpcepxcc22
·sin32− ...(6)
And at x = 50, 0=Tu
⇒ 0 = tpcepcc22
·50sin32−
⇒ sin 50p = 0 = sinnπ ⇒ p = 50πn
∴ From (6), we get
),( txuT =
2 2 2
25002 3 sin
50
n c tn x
c c eπ−π
The general form of solution is
),( txuT =
2 2 2
2500
1
sin50
n c t
nn
n xb e
π∞ −
=
π∑ ...(7)
At t = 0, uT = 2x (from equation 3)
⇒ 2x = ∑∞
=
π
1 50sin
nn
xnb
∴ bn = 50 50
2 200
2 2 50 25002 sin cos sin
50 50 25 50 50
n x n x n xx dx x
n n
π π π = − + π π ∫
bn = 12 2
2 2500 2500 200cos sin 0 ( 1)
25nn n
n nn+ − π + π + = − π ππ
Putting the value of bn in equation (7), we get
uT (x, t) =
2 2 2
1 2500
1
200( 1) sin
50
n c tn
n
n xe
n
π∞ −+
=
π− ⋅ ⋅
π∑ ...(8)
Hence from (4) and (8), we get
u(x, t) =
2 2 21
2500
1
200 ( 1)sin .
50
n c tn
n
n xe
n
π∞ + −
=
− π⋅
π ∑ Ans.
482 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Example 21: A rod of length l with insulated sides is initially at a uniform temperature u. Its endsare suddenly cooled 0ºC and are kept that temperature. Prove that the temperature function u(x, t) isgiven by
u(x, t) =
2 2 2
2
1
sin
n c t
ln
n
n xb e
l
π∞ −
=
π⋅∑
where bn = 0
0
2( )sin .
ln x
u x dxl l
π∫
Solution: Since the temperature distribution u(x, t) is the solution of heat equation
∴tu
∂∂
= 2
22
x
uc
∂∂
...(1)
The solution of equation (1) is as follows
u(x, t) = tcpecpxcpxc22
321 )sincos( −+ ...(2)
The boundary conditions are
(i) u(0, t) = 0 (ii) u(l, t) = 0
Making use of these boundary conditions the equation (2) takes the form
u(x, t) =
2 2 2
2
2 3 sinn c t
ln xc c e
l
π−π
The general form of the solution is
u(x, t) =
2 2 2
2
1
sin
n c t
ln
n
n xb e
l
π∞ −
=
π⋅∑ ...(3)
and initial condition i.e., at t = 0, u = u0(x) (say)
u0(x) = 1
sinnn
n xb
l
∞
=
π⋅∑
which is half range Fourier sine series
bn = 0
0
2( ) sin
ln x
u x dxl l
π⋅∫
Hence
u(x, t) =
2 2 2
2
1
sin
n c t
ln
n
n xb e
l
π∞ −
=
π⋅∑
where bn = 0
0
2( ) sin .
ln x
u x dxl l
π⋅∫ Proved.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 483
Example 22: Two ends A and B of a rod 20 cm long have the temperature at 30ºC and 80ºCrespectively until steady state prevails. The temperature at the end are changed to 40ºC and 60ºCrespectively find the temperature distribution in the rod.
Solution: The heat equation in one dimensional is
tu
∂∂
= 2
22
x
uc
∂∂
...(1)
The boundary conditions are
(i) u(0, t) = 30ºC (ii) u(20, t) = 80ºC
In steady condition 0=∂∂
tu
∴ From (1), we get ,02
2
=∂∂x
u on integrating, we get
u(x) = c1x + c2 ...(2)at x = 0, u = 30, so 30 = 0 + c2 ⇒ c2 = 30
and at x = 20 u = 80 so 80 = c1 × 20 + 30 ⇒ c1 = 25
From equation (2), we get
u(x) = 3025 +x
...(3)
Now the temperatures at A and B are suddenly changed we have again gain transient state.
If u1(x, t) is subsequent temperature function then the boundary conditions are
1(0, )u t = 40ºC and u1 (20, t) = 60ºC
and the initial condition i.e., at t = 0, is given by equation (3)
Since the subsequent steady state function us(x) satisfies the equation
2
2
x
us
∂∂
= 0 or2
20sd u
dx=
The solution of above equation is
us(x) = c3x + c4 ...(4)
At x = 0, us = 40 ⇒ 40 = 0 + c4 ⇒ c4 = 40
and at x = 20, us = 60 ⇒ 60 = 20 c3 + 40 ⇒ c3 = 1∴ From (4), we get
us(x) = x + 40 ...(5)
Thus the temperature distribution in the rod at time t is given by
u(x, t) = ),()( txuxu Ts +
⇒ u(x, t) = ),()40( txux T++ ...(6)
==
(0) 40ºC
(20) 60ºCs
s
u
u
484 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
where uT (x, t) is the transient state function which satisfying the conditions
and
−=−−+=−=
=−=−==−=−=
102
34030
2
5)()0,()0,(
06060)20(),20(),20(
04040)0(),0(),0(
1
1
1
xx
xxuxuxu
ututu
ututu
sT
sT
sT
The general solution for uT (x, t) is given by
uT (x, t) =
2 2 2
400
1
sin20
n c t
nn
n xb e
π∞ −
=
π∑ ...(7)
At t = 0, from (7), we get
1023 −x
= ∑∞
=
π
1 20sin
nn
xnb
∴ bn = 20
0
2 310 sin
20 2 20
x n xdx
π −
∫
=
20
2 20
1 3 20 3 40010 cos sin
10 2 20 2 20
x n x n x
n n
π π − − − − π π
= 1 20 20 20
20 ( 1) ( 10) 2( 1) 110
n n
n n n
− − − − = − − + π π π Putting the value of bn in equation (7), we get
uT (x, t) =
2 2 2
400
1
20 2( 1) 1sin
20
n c tn
n
n xe
n
π∞ −
=
− + π− ⋅
π ∑ ...(8)
From (6) and (8), we get
u(x, t) =
2 2 2
400
1
20 2( 1) 1( 40) sin .
20
n c tn
n
n xx e
n
π∞ −
=
− + π+ − ⋅ ⋅
π ∑ Ans.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 485
EXERCISE 5.2
1. Solve the boundary value problem 2
22
x
uc
t
u
∂∂
=∂∂
with the conditions 0),( =tlu for all t ≥ 0
0),0( =∂∂
ttu and u(x, 0) = 20x for 0 < x < l.
2 2 2
21
1
40 ( 1) ( , ) sin
n c tnl
n
l n xu x t e
n l
π∞ + −
=
− π = π
∑Ans.
2. Solve 2
22
x
uc
t
u
∂∂
=∂∂
for 0 < x < π, t > 0.
0),(),0( =π= tutu xx and ( , 0) sin .u x x=∞
−
=
= − ⋅
π π − ∑
2 242
1
2 4 1 ( , ) cos2
4 1n c t
n
u x t nx en
Ans.
3. Solve 2
22
x
uc
t
u
∂∂
=∂∂
with the conditions 0),0(,0),0( ==∂∂
tuttu , 0),( =tlu and
2( , 0) , 0 .u x lx x x l= − ≤ ≤
2 2 2
2(2 1)2
3 31
8 1 (2 1) ( , ) sin
(2 1)
c n t
l
n
l n xu x t e
ln
− π∞ −
=
− π = ⋅ π −
∑Ans.
4. Solve 2
2
x
uk
t
u
∂∂
=∂∂
under the conditions (i) ∞→∞≠ tu if
(ii) 0 for 0 and u
x x lx
∂ = = =∂
(iii) 2 for 0; 0 .u lx x t x l= − = ≤ ≤
2 2
242 2
2 21
1 2 ( , ) cos
6
m kt
l
m
l l m xu x t e
lm
π∞ −
=
π = − π
∑Ans.
5. The temperature distribution in a bar of length π, which is prefectly insulated at the ends
x = 0 and x = π is governed by the partial differential equation .2
2
x
u
t
u
∂∂
=∂∂
Assuming the
initial temperature as u(x, 0) = cos 2x, find the temperature distribution at any instant of
time.− = ⋅ 4 ( , ) cos2tu x t e xAns.
6. The ends A and B of a rod 30 cms long have their temperatures kept at 20ºC and 80ºCrespectively until steady state conditions prevail. The temperature at the end B is thensuddenly reduced to 60ºC and at the end A is raised to 40ºC and maintained so. Find u(x, t)
at any time t.
π∞ −
=
π = + − ⋅ π
∑2 2 2
225
1
2 40 ( , ) 40 sin
3 15
n c t
n
x n xu x t eAns.
486 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
7. A bar 40 cm long has originally a temperature of 0ºC along all its length. At t = 0, thetemperature at the end x = 0 is raised to 50ºC while that at other end is raised to 100ºC.Determine the resulting temperature function.
2 2 21
1600
1
5 2 50 100( 1) ( , ) 50 sin
4 40
c n tn
n
x n xu x t e
n
π∞ + −
=
+ − π = + − ⋅ ⋅ π
∑Ans.
8. An insulated rod of length l has its ends A and B maintained at 0ºC and 100ºC respectivelyuntil steady state conditions prevails. If B is suddenly reduced to 0ºC and maintained 0ºC ,find the temperature at a distance x from A at time t. (U.P.T.U. 2004)
π∞ + −
=
− π = ⋅ π
∑
2 2 2
21
1
200 ( 1) ( , ) sin
n c tnl
n
n xu x t e
n lAns.
9. Solve 2
22
x
uc
t
u
∂∂
=∂∂
under the conditions ( i) ,0),0( =∂∂
txu
( ii) 0),5( =∂∂
txu and
(iii) u(x, 0) = x.∞
− − π
=
− π= − ⋅
π − ∑
2 2(2 1) /252 2
1
5 20 1 (2 1) ( , ) cos
2 5(2 1)n t
n
n xu x t e
nAns.
10. A rod of length l has its ends A and B kept 0ºC and 120ºC until steady state conditionsprevail. If the temperature at B is reduced to 0ºC and kept so while that of A is maintained
find u(x, t).
2 2 2
21
1
240 ( 1) ( , ) sin
n cnl
n
n xu x t e
n l
π∞ + −
=
− π = π
∑Ans.
11. Solve 2
22
x
uc
t
u
∂∂
=∂∂
under the boundary conditions
(i) u is finite when t → ∞ (ii) u = 0 when x = 0 and x = π for all values of t, and (iii) u = x
from x = 0 to x = π when t = 0.+∞
−
=
−= ⋅
∑
2 21
1
( 1) ( , ) 2 sin
nn c t
n
u x t nx en
Ans.
5.6 TWO DIMENSIONAL HEAT EQUATION
We have
tu
∂∂
= )( 22 uc ∇ ...(i) In Section (5.5) (Alternate method)
where c2 = ,k
sρ k is the thermal conductivity of the body, s is the specific heat of the material of the
body and ρ is the density.In case of two dimensional, we may suppose that z-coordinate is constant.
∴ u2∇ = 02
2
2
2
+∂∂
+∂∂
y
u
x
u
0
constant)(As
2
2
=∂∂
∴
=
z
u
zu
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 487
⇒ u2∇ = 2
2
2
2
y
u
x
u
∂∂
+∂∂
...(ii)
From (i) and (ii), we get
2 22
2 2.
u u uc
t x y
∂ ∂ ∂= + ∂ ∂ ∂
...(A)
In steady state u always independent of t so that .0=∂∂
tu
Hence from equation (A), we get
2 2
2 20.
u u
x y
∂ ∂+ =
∂ ∂ ...(B)
The equation (B) is known as Laplace’s equation.
5.6.1 Solution of Two Dimensional Heat EquationWe have
tu
∂∂
= c2
∂∂
+∂∂
2
2
2
2
y
u
x
u...(i)
Let u(x, y, t) = XYT ...(ii)
Putting the value of u(x, y, t) from (ii) in equation (i), we get
dtdT
XY =
+
2
2
2
22
dy
YdX
dx
XdYTc
⇒dt
dT
TC 2
1= 2
2
2
2 11
dy
Yd
Ydx
Xd
X+
There are three possibilities
(a) 01
,01
,01
22
2
2
2
===dt
dT
Tcdy
yd
Ydx
Xd
X
(b)2 2
2 2 21 22 2 2
1 1 1, , .
d X d Y dTp p p
X Y dtdx dy c T= = =
(c)2 2
2 2 21 22 2 2
1 1 1, , .
d X d Y dTp p p
X Y dtdx dy c T= − = − = −
where = +2 2 21 2 .p p p
Out of these three possibilities, we have to select that solution which suits the physical nature ofthe problem and the given boundary conditions.
488 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Here ),,( tyxu = 54321 )()( ccyccxc ++ (For a)
),,( tyxu = tcpypypxpxp ececececec22
221154321 )()( −− ++ (For b)
and ),,( tyxu = tcpecypcypcxpcxpc22
524231211 )sincos()sincos( −++ (For c)
Example 23: A thin rectangular plate whose surface is impervious to heat flow has t = 0 anarbitrary distribution of temperature f (x, y). Its four edges x = 0, x = a, y = 0 and y = b are kept at zerotemperature. Determine the temperature at a point of the plate as t increases. (U.P.T.U. 2002)
Solution: The heat equation in two dimensional is
tu
∂∂
=
∂∂
+∂∂
2
2
2
22
y
u
x
uc ...(1)
Let u = XYT ...(2)
From (1)dtdT
XY =
+
2
2
2
22
dy
YdX
dx
XdYTc
⇒dt
dT
Tc2
1=
2 2
2 2
1 1d X d YX Ydx dy
+ ...(3)
Since the physical nature of problem is periodic so we choosen the constant as follows:
From (3), we get
2
21
dx
Xd
X= 0)(0 2
122
12
221 =+⇒=+⇒− XpDXp
dx
Xdp
∴ The A.E. is 21
2 pm + = 0 ⇒ m2 = 2 21 1i p m ip⇒ = ±
⇒ X = xpcxpc 1211 sincos +Again from (3), we get
2
21
dy
Yd
Y= 0)( 2
222
2 =+⇒− XpDp
so Y = )sincos( 2423 ypcypc +
anddt
dT
Tc2
1= ,2p− where p2 = +2 2
1 2 .p p
⇒TdT
= dtcp 22− ⇒ logeT = 522 log ctcp e+−
⇒ T = tcpec
22
5−
Y
XO
C B
A
(0, )b( , )a b
( , 0)a(0, 0)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 489
Hence from equation (2), we get
u = tcpecypcypcxpcxpc22
524231211 )sincos()sincos( −++ ...(4)
Now the boundary conditions are:
(i) u(0, y, t) = 0, (ii) u (a, y, t) = 0, (iii) u(x, 0, t) = 0; and (iv) u(x, b, t) = 0
Using first boundary condition in (4), we get
0 = 2 2
1 3 2 4 2 5( cos sin ) p c tc c p y c p y c e−+
⇒ c1 = 0 (otherwise u(x, y, t) = 0)From (4), we get
u = tcpeypcypcxpcc22
)sincos(sin 2423152−+ ...(5)
Using second boundary condition in (5), we get
0 = tcpeypcypcapcc22
)sincos(sin 2423152−+
⇒ sin p1a = 0 = a
mpm
π=⇒π 1sin
Next using 3rd boundary condition in (5), we get
0 = 2 2
3 2 5 1 3sin 0p c tc c c p x e c−⋅ ⇒ = (otherwise u(x, y, t) = 0)
From (5), we get
u = tcpeypxpccc22
··sinsin 21542− ...(6)
And using 4th boundary condition in (6), we get
0 = tcpebpxpccc22
··sinsin 21542−
⇒ bp2sin = 0 = 2sin .n
n pbππ ⇒ =
Since p2 = 22
21 pp +
so p2 =
+π
2
2
2
22
b
n
a
m
Putting the values of p1, p2 and p in equation (6), we get
),,( tyxu =
2 22 2
2 2
2 4 5 sin sin
m nc t
a bm x n yc c c e
a b
− π + π π
⋅
The general form of solution is
u(x, y, t) =
2 22 2
2 2
1 1
sin sin
m nc t
a bmn
m n
m x n yA e
a b
− π + ∞ ∞
= =
π π⋅∑∑ ...(7)
490 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
and the initial condition is
)0,,( yxu = ƒ(x, y)From (7), we get
f (x, y) = 1 1
sin sinnmm n
m x n xA
a b
∞ ∞
= =
π π⋅ ⋅∑ ∑
which is the double Fourier since series of f (x, y).
∴ Amn = 0 0
2 2( , )sin sin
a b
x y
m x n xf x y dxdy
a b a b= =
π π⋅ ∫ ∫ ...(8)
Hence the equation (7) is required temperature distribution with the equation (8). Ans.
Example 24: Solve the following Laplace equation: ...(i)
2
2
2
2
y
u
x
u
∂∂+
∂∂
= 0
in a rectangle with 0),(,0),(,0),0( === bxuyauyu and ( ,0) ( )u x f x= along x-axis
[U.P.T.U. 2008]
Solution: Let ).()(),( yYxXyxu = ...(i)
∴ 2
2
x
u
∂
∂ 2 2 2
2 2 2and
d X u d yY X
dx y y
∂= =∂ ∂
Putting these values in given equation, we get
2
2
2
2
dy
YdX
dx
XdY + = 0
⇒ 2
2
dx
XdY
22
2
2
2
2
2 11p
dy
YdYdx
XdXdy
YdX −=−=⇒−= (say) ...(ii)
Taking first two terms, we get2
22
d Xp X
dx+ = 0
⇒ X pxcpxc sincos 21 +=And taking last two terms, we get
ypdy
Yd 22
2
− = 0
⇒ Y pypy ecec −+= 43
From (i)
),( yxu ))(sincos( 4321pypy ececpxcpxc −++= ...(iii)
Putting x = 0, we get
0 00)( 1431 =⇒=+= − cececc pypyotherwise
0u =
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 491
From (iii) ),( yxu )(sin 432pypy ececpxc −+= ...(iv)
Putting x = a, in (iv), we get
0 0sin)(sin 432 =⇒+= − paececpac pypy
or sin paa
npn
π=⇒π= sin
From (iv), ),( yxu 2 3 4sin ( )n n
y ya an x
c c e c ea
π π−π
= + ...(v)
Putting y = b in (v), we get
0
+
π=
π−πa
bn
a
bn
ececa
xnc 432 sin
⇒ a
bn
a
bn
ececπ−π
+ 43a
bn
eccπ
−=⇒=2
340 ...(ii)
From (v), we get
),( yxu( 2 )
2 3 sinn n
y y ba an x
c c e ea
π π− − π = ⋅ ⋅ −
or ),( yxu( 2 )
1
sinn y n
y ba a
nn
n xb e e
a
π π∞ − −
=
π = ⋅ ⋅ −
∑ ...(vi)
Putting y = 0, we get
)(xf
−⋅
π⋅=
π+∞
=∑ a
bn
nn e
a
xnb
2
1
1sin ...(vii)
which is Fourier sine series
∴
−
πa
bn
n eb2
1
0
2( ) sin
an x
f x dxa a
π= ⋅∫
⇒ nb ∫π
⋅
−
=π2
a
a
bndx
a
xnxf
ea0
sin)(
1
2...(viii)
Hence equation (vi) is required solution with coefficient bn (equations (viii)).
Example 25: An infinitely long uniform plate is bounded by two parallel edges and an end atright angles to them. The breadth is π, the end is maintained at a temperature 100ºC at all points andother edges are at 0ºC. Show that the steady state temperature is given by
u(x, y) = 3 5400 1 1sin sin3 sin5 ···· .
3 5y y ye x e x e x− − − + + + π
492 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Solution: The steady state temperature u(x, y) is the solution of
2
2
2
2
y
u
x
u
∂∂
+∂∂
= 0 ...(1)
The boundary conditions are
(i) ),(0),0( yuyu π== for all y ≥ 0
(ii) 0),(lim),( ==∞∞→
yxxuy
for 0 ≤ x ≤ π
and initial condition is (iii) 100)0,( =xu
Let u(x, y) = XY ...(2)
Putting the value of u in equation (1), we get
2 2
2 2
d X d YY X
dx dy+ = 0
⇒2
2
d XY
dx=
2 2 22
2 2 2
1 1(say)
d Y d X d YX p
X Ydy dx dy− ⇒ = − = − ...(3)
Taking first two member, we get
Xpdx
Xd 22
2
+ = 0
⇒ X = pxcpxc sincos 21 +And now taking last two member of equation (3), we get
22
20
d Yp Y Y
dy− = ⇒ = pypy ecec −+ 43
Putting the values of X and Y in equation (2), we get
),( yxu = )()sincos( 4321pypy ececpxcpxc −++ ...(4)
At x = 0, u = 0
⇒ 0 = 1 3 4 1( ) 0.py pyc c e c e c−+ ⇒ =
From equation (4), we have
),( yxu = )(sin 432pypy ececpxc −+ ...(5)
Again at x = π, u = 0
⇒ 0 = )(sin 432pypy ececpc −+π
⇒ πpsin = 0 = nn
pn =ππ=⇒πsin
0),(
otherwise
=yxu
O A ( , 0)πX
Y
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 493
Putting the value of p in equation (5), we get
u(x, y) = )(sin 432nyny ececnxc −+ ...(6)
Now using (ii) boundary condition in equation (6) i.e., when y → ∞, u = 0, we get
0 = )(lim·sin 432nyny
yececnxc −
∞→+
0 =
+
∞→0)(lim·sin 32
ny
yecnxc As lim 0ny
ye−
→∞→
⇒ 03 =c (otherwise u(x, y) = 0)
Putting the value of c3 in equation (6), we get
u(x, y) = c2c4 sin nx·e–ny
∴ The general form of solution may be written as
u(x, y) = ∑∞
=
−
1
·sinn
nyn enxb ...(7)
Using (iii) condition (i.e., at y = 0) in equation (7), we get
100 = ∑∞
= 1
sinn
n nxb
which is Fourier half range sine series in π≤≤ x0
∴ bn = 0
0
2 200 200100 sin cos cos cos0nxdx nx n
n n
ππ
⋅ = − = − π + π π π∫
bn = − − + = π π
400if is odd200
( 1) 1
if is even0
n nn
nn
Hence from equation (7), we get
u(x, y) = .····5sin5
13sin
3
1sin
400 53
+++
π−−− xexexe yyy Proved.
Example 26: Solve 02
2
2
2
=∂∂
+∂∂
y
u
x
u which satisfies the conditions
u(0, y) = 0)0,(),( == xuylu and u(x, a) = sin .n x
l
π
(U.P.T.U. 2004)
Solution: We have
2
2
2
2
y
u
x
u
∂∂
+∂∂
= 0 ...(1)
494 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
The solution of equation (1) can be obtained as in (Example 26, equation (4) on page no. 494) isgiven below:
u(x, y) = (c1cos px + c2 sin px) (c3epy + c4e–py) ...(2)
At x = 0, 0=u
0 = 0)( 1431 =⇒+ − cececc pypy [otherwise u(x, y) = 0]
Putting the value of c1 in equation (2), we get
u(x, y) = )(sin 432pypy ececpxc −+ ...(3)
at x = l, u = 0
0 = )(sin 432pypy ececplc −+
⇒ sin pl = 0 = sin nπ ⇒ .n
plπ=
From equation (3), we get
u(x, y) =
+
π π−
πl
yn
l
yn
ececl
xnc 432 sin ...(4)
At y = 0, u = 0, we get
0 = 2 3 4 3 4 4 32 sin ( ) 0n x
c c c c c c clπ + ⇒ + = ⇒ = −
Putting the value of c4 in equation (4), we get
u(x, y) = π π− π π π − =
2 3 2 3sin 2 sin ·sinn y n y
l ln x n x n yc c e e c c h
l l l 2sin
θ−θ −=θ
eeh
The general form of above solution is given as
u(x, y) = 1
sin sinnn
n x n yb h
l l
∞
=
π π∑ (2c1c2 = bn) ...(5)
Putting y = a and u = lxnπ
sin in equation (5), we get
lxnπ
sin = 1
sin sinnn
n x n ab h
l l
∞
=
π π∑
Equating the coefficient of lxnπ
sin on both sides, we get
1 =
l
anh
bl
anhb nn π
=⇒π
sin
1sin
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 495
and b1 = b2 = b3 = ....... = bn – 1 = 0
Hence, from equation (5), we get
u(x, y) = .·sinsin
)/(sin
l
ynh
lan
h
lxn π
π
πAns.
Example 27: A thin rectangular plate whose surface is impervious to heat flow has t = 0 anarbitrary distribution of temperature f(x, y). Its four edges x = 0, x = a, y = 0, y = b are kept at 0°Ctemperature. Determine the temperature at a point of a plate as t increases. Discuss when
(U.P.T.U. 2002)
f (x, y) = .sinsin
π
π
βb
y
a
x
Solution: Two dimensional heat flow equation is
2
2
2
2
y
u
x
u
∂∂
+∂∂
= 2
1 u
tc
∂∂
...(1)
Boundary conditions are
u(0, y, t) = 0
u(a, y, t) = 0
u(x, 0, t) = 0
u(x, b, t) = 0
and the initial condition is
u(x, y, t) = f (x, y) at t = 0.
The solution of equation (1) is given below (Example 23, equation (4) on page 489)
u = −+ +2 2
1 1 2 1 3 2 4 2 5( cos sin ) ( cos sin ) p c tc p x c p x c p y c p y c e ...(2)
At x = 0, u = 0, then we get
u(0, y, t) = 0 = tpcecypcypcc22
524231 )sincos( −+
⇒ c1 = 0
∴ from (2), u(x, y, t) = c2 c5 sin p1x(c3 cos p2y + c4 sin p2y) )(22 tpce− ...(3)
from (3), u(a, y, t) = 0 = tpceypcypcapcc22
)sincos(sin 2423152−+
⇒ sin p1a = 0 = sin mπ (m ∈ I)
∴ p1 = .maπ
From (3), u(x, y, t) = 2 2
2 5 3 2 4 2sin ( cos sin )( )c p tm xc c c p y c p y e
a−π + ...(4)
y = b
y = 0
x =
0
x =
a
OX
Y
C B
A
496 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
u(x, 0, t) = 0 = 2 2
2 5 3sin c p tm xc c c e
a−π ⋅
⇒ c3 = 0.
∴ from (4), u(x, y, t) = 2 2
2 5 4 2sin sin c p tm xc c c p y e
a−π ⋅ ...(5)
u(x, b, t) = 0 = 2 2
2 5 4 2sin sin c p tm xc c c p b e
a−π ⋅
⇒ sin p2b = 0 = sin nπ (n ∈ I)
p2b = nπ
⇒ p2 = bnπ
∴ from (5), u(x, y, t) = 2 2
2 5 4 sin sin c p tm x n yc c c e
a b−π π
= 2 2
sin sin c p tmn
m x n yA e
a b−π π
...(6) mnAccc =452 where
But, p2 = 2
22
2
2222
21
b
n
a
mpp
π+
π=+
or 2mnp =
+π
2
2
2
22
b
n
a
m
By using pmn, equation (6) becomes,
u(x, y, t) = ∑∑∞
=
∞
=
−ππ
1 1
22
sinsinm n
tpcmn
mnea
xn
a
xmA ...(7)
which is the most general solution.
∴ u(x, y, 0) = f (x, y) = 1 1
sin sinmnm n
m x n yA
a b
∞ ∞
= =
π π∑ ∑which is the double Fourier half range sine series for ƒ(x, y).
Amn = ∫ ∫= =
ππa
x
b
y
dydxyxfb
yn
a
xm
ba0 0
),(sinsin2
·2
Ans.
when f (x, y) = sin sinx y
a b
π π β
∴ Amn =0 0
2 2sin sin sin sin
a b
x y
m x n y x ydxdy
a b a b a b= =
π π π π⋅ β∫ ∫
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 497
= 0 0
4sin sin sin sin
a bm x x n y y
dx dyab a a b b
β π π π π∫ ∫
=0 0
cos( 1) cos( 1) cos( 1) cos( 1)a b
x x y ym m dx n n dy
ab a a b b
β π π π π − − + × − − + ∫ ∫
=
0 0
sin( 1) sin( 1) sin( 1) sin( 1)
( 1) ( 1) ( 1) ( 1)
a bx x y y
m m n na a b b
ab m m n na a b b
π π π π − + − + β− × − π π π π − + − +
= 0abβ ⋅ = 0 (∵ sin mπ = 0 and sin nπ = 0)
From (7), v (x, y, t) = 0 when f (x, y) = β sin .sinby
ax ππ
Proved.
Example 28: The edges of a thin plate of side π cm are kept at temperature zero and facesinsulated. The initial temperature is u(x, y, 0) = f (x, y) = xy(π – x) (π – y). Determine temperature in theplate at time t.
Solution: Referred to solution of Example 4, where, a = π and b = π and f (x, y) = xy(π – x)(π – y)
2mnp = 22
2
2
2
22 nm
nm+=
π+
ππ
u(x, y, t) = 2 2 2( )
1 1
sin sin c m n tmn
m n
A mx n y e∞ ∞
− +
= =∑ ∑ ...(1)
u(x, y, 0) = 1 1
sin sinmnm n
A mx ny∞ ∞
= =∑ ∑ ...(2)
Amn = 20 0
4( )( )sin sinxy x y mx nydxdy
π π
π − π −π ∫ ∫
= 2 22
0 0
4( )sin ( )sinx x mxdx y y nydy
π π
π − π −π ∫ ∫ ...(3)
498 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Now 2
0
( )sinx x mxdxπ
π −∫ = π
−+
−
−π−
−
−π0
322 cos
)2(sin
)2(cos
)(m
mx
m
mxx
m
mxxx
= 3
2[1 ( 1) ]m
m− −
and similarly
2
0
( )siny y nydyπ
π −∫ = ])1(1[2
3n
n−−
∴ Amn = ])1(1[])1(1[16
332
nm
nm−−−−
π
or Amn = 2 3 3
0, when 2 and 2
64, when 2 1 and 2 1
(2 1) (2 1)
m p n q
m p n qp q
= = = − = − π − −
Hence from (2)
u(x, y, t) = 2
1 1
sin(2 1) sin(2 1) .pqtpq
p q
A p x q ye∞ ∞
−µ
= =
− −∑ ∑ Ans.
where 2pqµ = ])12()12[( 222 −+− qpc
and Apq = .)12()12(
64332 −−π qp
Example 29: A square plate is bounded by the lines x = 0, y = 0, x = 10 and y = 10. Its faces areinsulated. The temperature along the upper horizontal edge is given by u(x, 10) = x(10 – x), while theother three edges are kept at 0ºC. Find the steady state temperature in the plate.
Solution: The steady state temperature u(x, y) is the solution of the equation
2
2
2
2
y
u
x
u
∂∂
+∂∂
= 0 ...(1)
The solution of equation (1) is given below
u(x, y) = )()sincos( 4321pypy ececpxcpxc −++ ...(2)
[See eqn. (4) on page no. 491]The boundary conditions are
(0, ) (10, ) 0, 0 10
( , 0) 0 ; 0 10
and ( , 10) (10 ) ; 0 10
u y u y y
u x x
u x x x x
= = ≤ ≤ = ≤ ≤ = − ≤ ≤
...(A)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 499
Putting x = 10 and u = 0 in (2), we get
0 = 1 3 4 1( ) 0py pyc c e c e c−+ ⇒ =∴ From (2), we have
u(x, y) = )(sin 432pypy ececpxc −+ ...(3)
Next putting x = 10 in (3), we get
0 = 2 3 4sin10 ( ) sin10 0 sin .10
py py nc p c e c e p n p− π+ ⇒ = = π ⇒ =
Again putting y = 0 and u = 0 in equation (3), we get
0 = 2 3 4 3 4 4 3sin ( ) 0c px c c c c c c+ ⇒ + = ⇒ = −
Putting the values of c4 and p in equation (3), we get
u(x, y) = 10 102 3 2 3sin 2 sin sin
10 10 10
n y n yn x n x n y
c c e e c c hπ π− π π π − =
The general solution is given by
u(x, y) = 1
sin sin10 10
nn
n x n yb h
∞
=
π π⋅∑ ...(4)
Putting y = 10 and u = x (10 – x) in equation (4), we get
x(10 – x) = 10
1
sin sin10
nn
n xb hn
=
ππ∑
∴ bnsin hnπ = 10
0
2(10 )sin
10 10
n xx x
π−∫ constanta is sin πhn
= 21 10(10 )· cos (10 2 )
5 10
n xx x x
n
π− − + − π 2 2
100sin
10
n x
n
π
π
10
3 30
10002 cos
10
n x
n
π − π
=
π×+π
π×−
3333
10002cos
10002
5
1
nn
n
=
π=+−−π even is if0,
odd is if,800
]1)1([5
200033
33n
nn
nn
∴ bn =
ππeven is if0,
odd is if,sin
80033
n
nhnn
500 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Putting the value of bn in equation (4), we get
u(x, y) = 3 31
(2 1) (2 1)sin sin
800 10 10 ; where 2 1.(2 1) sin (2 1)m
m x m yh
n mm h m
∞
=
− π − π⋅= −
π − − π∑ Ans.
Example 30: A rectangular metal plate is bounded by x = 0, x = a, y = 0 and y = b. The three
sides x = 0, x = a and y = b are insulated and the edge y = 0 is kept at u0 cos .axπ
Show that the
temperature in the steady state is
u(x, y) = .cos)(
cos)(
sec0
π
−π
π−
a
x
a
ybh
a
abhu
Solution: The steady state temperature u(x, y) is the solution of the equation
2
2
2
2
y
u
x
u
∂∂
+∂∂
= 0 ...(1)
Since sides x = 0, x = a and y = b are insulated
0=
∂∂
xx
u= 0 ...(2)
axx
u
=
∂∂
= 0 ...(3)
byx
u
=
∂∂
= 0 ...(4)
Also u(x, a) = u0 cos
π
a
x...(5)
The solution of (1) is
u(x, y) = )()sincos( 4321pypy ececpxcpxc −++ ...(6)
Nowxu
∂∂
= )()cossin( 4321pypy ececpxpcpxpc −++−
0=
∂∂
xx
u= 0 ⇒ 0)( 432 =+ − pypy ececpc
⇒ c2 = 0 3 4( 0)py pyc e c e−+ ≠∵
axx
u
=
∂∂
= 0 ⇒ 0)(sin 431 =+− −pypy ececpapc
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 501
⇒ sin pa = 0 1( 0 otherwise ( , )
would be a trivial solution)
c u x y≠∵
⇒ pa = np (n is an integer)
⇒ p = a
nπ
Nowy
u
∂∂
= ).(cos 431pypy pecpecpxc −−
byy
u
=
∂∂
= 0 ⇒ 0)(cos 431 =− − pbpb pecpecpxc
⇒ pbpb pecpec −− 43 = 0
⇒ 3 4( )pb pbp c e c e−− = 0
⇒ pbec3 = pbec −4
⇒ c4 = pbec 23
∴ From (6) u(x, y) = an
peececpxc pypbpy π=+ − where)(cos 2331
=
+
−
−
pb
pypbpy
e
eeepxcc cos31
= 1 3 cos ( )pb py pb pypb
c cpx e e e e
e− −
− +
= ][cos )()(31
pybpybppb eepxecc −−− +
= pybhpxecc pb )(coscos2 31 −
= pbeccKa
nybh
a
xnK 312 where
)(coscos =
π−
π
∴ The most general solution of (1) is
u(x, y) = 1
( )cos cosn
n
b y nn xA h
a a
∞
=
− ππ
∑
where An’s a are to be determined.
Since u(x, a) = u0 cos
π
a
x for 0 < x < a,
502 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
u(x, a) = 01
( )cos cos cosn
n
n x b a n xA h u
a a a
∞
=
π − π π =
∑
∴
π
a
xu cos0 = ∑
∞
=
π−
π
1
)(coscos
nn
a
nabh
a
xnA
= 1 2( ) 2 ( )2
cos cos cos cos ····x b a x b a
A h A ha a a a
π − π π − π + + Comparing both sides, we get
A1 = u0 sec h
π−
a
ab )(
and A2 = A3 = ····=0
∴ u(x, y) =
π−
π
π−
a
ybh
a
x
a
abhu
)(coscos
)(sec0
or u(x, y) = .cos)(
cos)(
sec0
π
π−
π−
a
x
a
ybh
a
abhu Proved.
EXERCISE 5.31. An infinite long plane uniform plate is bounded by two parallel edges and an end at right
angles to them. The breadth is π. This end is maintained at temperature u0 at all points andother are at zero temperature. Determine the temperature at any point of the plate in thesteady state. [U.P.T.U. (C.O.) 2003]
−
−π= ∑
∞
=
−−
1
)12(0 )12sin(12
14),(
n
ynxenn
uyxuAns.
2. A square plate is bounded by the lines x = 0, y = 0, x = 20 and y = 20. Its faces are insulated.The temperature along the upper horizontal edge is given by u(x, 20) = x(20 – x) when0 < x < 20, while the other three edges are kept at 0ºC. Find the steady temperature in the rod.
31
(2 1) (2 1)sin sin
3200 20 20 ( , ) ;2 1sin (2 1)m
m x m yh
u x y m nA m
∞
=
− π − π
= − = − ππ
∑Ans.
3. A rectangular plate with insulated surfaces is 10 cm wide and so long compared to its widththat it may be considered infinite in length without introducing an appreciable error. If thetemperature along the short edge y = 0 is given by
u(x, 0) = 20x, 0 < x ≤ 5
= 20(10 – x), 5 < x < 10.
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 503
while the two long edges x = 0 and x = 10 as well as other short edges are kept at 0ºC. Findthe steady state temperature at any point (x, y) of the plate.
(2 1)110
2 21
800 ( 1) (2 1) ( , ) sin
10(2 1)
m ym
m
m xu x y e
m
− π∞ + −
=
− − π = π −
∑Ans.
4. A long rectangular plate of width lam with insulated surfaces has its temperature u equal tozero on both the long sides and one of the shorter sides so that u(0, y) = u(l, y)
= u(x, ∞) = 0 and u(x, 0) = kx. Find u(x, y).1
1
2 ( 1) ( , ) sin
n ynl
n
kl n xu x y e
n l
π∞ + −
=
− π =π
∑Ans.
5. A long rectangular plate has its surface insulated and the two long sides as well one of theshort sides are kept at 0ºC while the other sides is kept at u(x, 0) = 3x and the length being
5 cm. Find u(x, y).
15
1
30 ( 1) ( , ) sin
5
n yn
n
n xu x y e
n
π∞ + −
=
− π =π
∑Ans.
6. A square plate has its faces and the edge y = 0 insulated. Its edges x = 0 and x = π are keptat 0ºC and the edge y = π is kept at temperature f (x). Find the steady state temperature.
1 0
sin cos 2 ( , ) , where ( )sin
cosn nn
nx hnyu x y b b f x nxdx
h n
π∞
=
= =
π π ∑ ∫Ans.
7. A long rectangular plate has its surface insulated and the two long sides as well as one of theshort sides are maintained at 0ºC. Find an expression for the steady state temperature u(x, y)if the short side y = 0 is 30 cm long and kept at 40ºC.
(2 1)
30
1
160 1 (2 1) ( , ) sin ·
2 1 30
m y
m
m xu x y e
m
− π∞ −
=
− π =π −
∑Ans.
5.6.2 Two Dimensional Heat Flow in Polar CoordinateLet c1 and c2 be two arcs of circles of radius r and r + δr respectively. Here we consider an elementbetween the circles c1 and c2 and the lines OP and OQ i.e., (θ and θ + δθ) of any sheet of conductingmaterial of uniform density ρ.
We know that the area of portion QQPP ′′ = δθδrr
XO
θ
δθ
θ + δθ
θ
P
Q
r r + rδ
c1
c2
Q'
P'
504 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
∴Mass of elementary portion = ( )h r rρ δ δθ sheet. of thickness
uniform the is h
Suppose uδ is the temperature rise in the elementary portion during .tδ∴ Rate of increase of heat content in the elementary portion
= 0t
uLt hr r s
tδ →
δρ δ δθδ
heat specific the iss
= u
h s r rt
∂ρ δ δθ∂ ...(i)
If Q1, Q2, Q3 and Q4 are the heat flow across the sides of the elementary portion through the edgePP', Q'Q, PQ and P'Q' respectively then
Q1 = rhu
rk δ
θ∂∂−
θ
1k is thermal conductivity
Q2 = 1 u
k h rr θ+δθ
∂ − δ ∂θ
Q3 = r
uk hr
r
∂ − δθ ∂
Q4 = δθδ+
∂∂
−δ+
)( rrhr
uk
rr
The rate of increase of heat (Q1 – Q2 + Q3 – Q4) in the elementary portion is equal to the equation (i).
∴ uh s r r
t∂ρ δ δθ∂
= 1 1u u
kh rr rθ+δθ θ
∂ ∂ − δ ∂θ ∂θ
r r r r r
u u ur r
r r r+δ +δ
∂ ∂ ∂ + − δθ+δ δθ ∂ ∂ ∂
⇒ tu
∂∂
= 2
1 1r r r
r r
u u u u
r rk us r r rr
θ + δ θ θ + δ
+ δ
∂ ∂ ∂ ∂ − − ∂θ ∂θ ∂ ∂ ∂ + + ρ δθ δ ∂
Now taking limit 0→δr and 0→δθ , we get
tu
∂∂
= 2
1 1k u u u
s r r r rr
∂ ∂ ∂ ∂ ∂ + + ρ ∂θ ∂θ ∂ ∂ ∂
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 505
= 2 2
2 22 2 2
1 1, where
u u u k
r r sr r
∂ ∂ ∂α + + α =
∂ ρ∂θ ∂
⇒ 2 2
2 2 2 2
1 1 1,
u u u u
t r rr r
∂ ∂ ∂ ∂= + +
∂ ∂α ∂ ∂θ
This the heat equation in
polar coordinates
In steady statetu
∂∂
= 0
⇒ 011
2
2
22
2
=∂∂
+θ∂
∂+
∂∂
r
u
r
u
rr
u This is called Laplace equation in polar coordinates.
5.6.3 Solution of Two Dimensional Heat Equation (Laplace Equation) in Polar Coordinates (Steady State)
We have
r
u
r
u
rr
u
∂∂
+θ∂
∂+
∂∂ 11
2
2
22
2
= 0
⇒2
2
2
22
θ∂∂
+∂∂
+∂∂ u
r
ur
r
ur = 0 ...(i)
Let u(r, θ) = )()·( θφrR ...(ii)
Putting the value of u in equation (i), we get
2
2
2
22
θ
φ+φ+φ
d
dR
dr
dRr
dr
Rdr = 0
⇒2
2
2
22 ·
θφ
+φ
+
d
dR
dr
dRr
dr
Rdr = 0
⇒R
dr
dRr
dr
Rdr +2
22
= 2
2
1, is a constant.
dk k
d
φ− =
φ θTaking first and last member, we get
Rkdr
dRr
dr
Rdr −+
2
22 = 0.
which is linear homogeneous differential equation of second order so put r = ez, we get the solution ofabove equation as
R = 1 2 .k kc r c e−+
506 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Next, taking second and last member, we get
φ+θφ
kd
d2
2
= 0 ⇒ φ = 3 4cos sinc k c kθ + θ
Hence from equation (ii), we get
),( θru = )sincos()( 4321 θ+θ+ − kckcecrc kk ...(iii)
Case I: If k = p2, then from (iii), we have
u(r, θ) = )sincos()( 4321 θ+θ+ − pcpcrcrc pp
Case II: If k = –p2, then
u(r, θ) = { } { }θ−θ ++ pp ececrpcrpc 4321 ·)logsin()logcos(
Case III: If k = 0, then
u(r, θ) = {c1log r + c2} {c3θ + c4}.
The choice of most suitable solution among these solutions depends upon the physical nature ofthe problem and the boundary conditions.
Example 31: A thin semi circular plate of radius a has its boundary diameter kept at 0ºC and its
circumference at 100ºC. If u(r, θ) is the steady state temperature distribution, find .2
,4
πa
u
Solution: The steady state temperature distribution u(r, θ) is a solution of
2
2
22
2 11
θ∂∂
+∂∂
+∂∂ u
rr
u
rr
u= 0 ...(1)
The boundary conditions are
(i) arruru <<=π= 0,0),()0,(
and (ii) ( , ) ( ) 100, 0 2 .u a fθ = θ = < θ < π
The solution of equation (1) which suits above boundary conditions is (in Case I)
so u(r, θ) = )sincos()( 4321 θ+θ+ − pcpcrcrc pp ...(2)
At θ = 0, u = 0, from (2), we get
0 = )0(0)( 213321 ≠+=⇒+ −− pppp rcrcccrcrc
∴ From equation (2), we get
u(r, θ) = θ+ − pcrcrc pp sin)( 421 ...(3)
At θ = π, u = 0
⇒ 0 = )(sin0sinsin)( 421 npnppcrcrc pp =⇒π==π⇒π+∴ (3) becomes
),( θru = θ+ − ncrcrc nn sin)( 421 ...(4)
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 507
Now as ,0→r u(r, θ) must be finite and hence c2 = 0
∴ ),( θru = θnrcc n sin41
or u(r, θ) = ∑∞
=
θ1
sin·n
nn nrb )( 41 nbcc = ...(5)
Putting r = a and u = 100
100 = ∑∞
=
θ1
sin·n
nn nab
∴ nn ab · = 0
0
2 200 200100sin cos [1 cos ]n d n n
n n
ππ
θ θ = − θ = − π π π π∫
= ])1(1[200 n
n−−
π
⇒ bn = { }400
, if is odd2001 ( 1)
· 0, if is even
n nn
nn a
n a n
− − = π
π
From (5),
u(r, θ) = ∑ ∑∞
=
∞
=
−
θ−
−π=θ
π ,..)3,1( 1
12
)12sin()12(
1400sin
1400
n m
mn
ma
r
mn
a
r
n
(putting, n = 2m–1)
∴ ,4 2
au
π
= .2
)12(sin
4
1
)12(
1400
1
12
∑∞
=
− π−
−π m
mm
mAns.
EXERCISE 5.41. The boundaring diameter of a semi circular plate of radius 10 cm is kept 0ºC and the
temperature along the semi circular boundary given by
u(10, θ) = 50 , when 0
2
50( ), when .2
πθ ≤ θ ≤
π π − θ ≤ θ < π
Find the steady state temperature u(r, θ) at any point.
2 11
21
200 ( 1) u( , ) sin(2 1)
10(2 1)
mm
m
rr m
m
−∞ −
=
− θ = − θ π − ∑Ans.
508 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
2. Determine the steady state temperature at the points in the sector given by ar ≤≤π≤θ≤ 0,4
0
of a circular plate if the temperature is maintained at 0ºC along the side edges and at aconstant temperature kºC along the curved edge.
θ−
−π=θ ∑
∞
=
−
1
)12(4
)]12(4sin[12
14),(
m
m
mar
mk
ruAns.
3. A thin semi circular plate of radius a has its boundary diameter kept at 0ºC and itscircumference at kºC where k is a constant. Find the temperature distribution in the steady
state.
θ−
−π=θ ∑
∞
=
−
1
)12(
)12(sin12
14),(
m
m
mar
mk
ruAns.
4. A plate with insulated surfaces has the shape of quardant of a circle of radius 10 cm. The
boundary radii θ = 0 and 2π=θ are kept 0ºC and the temperature along the circular quardant
is kept at 100 (πθ – 2θ2). Find the steady state temperature distribution.
2(2 1)
31
400 1 ( , ) sin{2(2 1) }
10(2 1)
m
m
ru r m
m
−∞
=
θ = − θ π − ∑Ans.
5. Find the steady state temperature in a circular plate of radius a which has one half of itscircumference at 0ºC and the other half at 100ºC.
θ−
−π+=θ ∑
∞
=
−
1
12
})12sin{()12(
120050),(
m
m
mar
mruAns.
5.7 WAVE EQUATION IN TWO DIMENSION
Consider a tightly stretched uniform membrane in xy-plane having the same tension T at every point.Since the deflection u(x, y, t) is in the perpendicular direction to xy-plane.
The vertical component of T δy = αδ−βδ sinsin yTyT
= )tan(tan α−βδyT approximately, since and are very small
so sin tan ,sin tan
α β β ≈ β α ≈ α
= xx
uyT
x
u
x
uyT
xxx
δ∂
∂δ=
∂∂
−
∂∂
δδ+
2
2
Similarly, the vertical component due to the force T δx
= 2
2
uT x y
y
∂δ δ∂
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 509
Hence the equation of motion of the element ABCD is
2
2
t
uyxm
∂∂
δδ = 2 2
2 2
u uT x y
x y
∂ ∂+ δ δ ∂ ∂ area.
unitper membrane
of mass the is m
or2
2
t
u
∂∂
=
∂∂
+∂∂
2
2
2
2
y
u
x
u
m
T
⇒
2 2 22 2
2 2 2; .
u u u Tc c
mt x y
∂ ∂ ∂= + = ∂ ∂ ∂
This is two dimensional wave equation.
5.7.1 Solution of Two Dimensional Wave EquationWe have
2
2
t
u
∂∂
=
∂∂
+∂∂
2
2
2
22
y
u
x
uc ...(i)
Let u(x, y, t) = X(t) Y(y) T(t) ...(ii)
⇒2
2
t
u
∂∂
= 2 2 2 2 2
2 2 2 2 2, and
d T u d X u d yXY YT XT
dt x dx y dy
∂ ∂= =∂ ∂
Putting these values in equation (i), we get
2
2
dt
TdXY =
+
2
2
2
22
dy
YdXT
dx
XdYTc
u
x x + x δ
y + y δY
X
T xδ
T yδA B
CD
T xδ
T yδα
β
O
y
510 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
⇒ 2
2
2
1
dt
Td
Tc= 2
2
2
2 11
dy
Yd
Ydx
Xd
X+ (dividing by XYT)
⇒ 2
2
2
1
dt
Td
Tc= k
dy
YdYdx
XdX
=+2
2
2
2 11...(iii)
Let 2
21
dx
Xd
X= 012
2
1 =−⇒ Xkdx
Xdk
∴ X = 1 11 2
k x k xc e c e−+
and again let 2
21
dy
Yd
Y= k2 ⇒
2
220
d Yk Y
dy− =
∴ Y = ykyk ecec 2243
−+
Finally 2
2
2
1
dt
Td
Tc= k ⇒ 02
2
22
2
2
=−⇒= Tkcdt
TdTkc
dt
Td
∴ T = 5 6c kt c ktc e c e−+
From equation (ii), we get
u(x, y, t) = 1
1 1 2 21 2 3 4 5 6( )( )( )k x k x k y k y c kt c k tc e c e c e c e c e c e− − −+ + +
where k = k1 + k2.There arise the following cases:Case I: If k = 0, then
u(x, y, t) = )()()( 654321 ctccyccxc +++Case II: If k = p2, then
u(x, y, t) = )()()( 6543212211 cptcptypypxpxp ecececececec −−− +++
where p2 = 2 21 2 .p p+
Case III: If k = –p2, then
u(x, y, t) = )sincos()sincos()sincos( 6524231211 cptccptcypcypcxpcxpc +++
where p2 = 2 21 2 .p p+
Since the physical nature of the problem for a rectangular membranes (0, 0), (a, 0), (a, b), (0, b)is periodic so the solution in case III, will be consistent with the physical nature of the problem. Herewe will consider case III solution.
Example 32: Find the deflection u(x, y, t) of a rectangular membranes (0 ≤ x ≤ a, 0 ≤ y ≤ b)whose boundary is fixed given that it starts from rest and u(x, y, 0) = xy (a – x) (b – y).
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 511
Solution: The vibrations of the membrane are governed by the wave equation
2
2
t
u
∂∂
= 2 2
22 2
u uc
x y
∂ ∂+ ∂ ∂
...(1)
The physical nature of the problem is periodic so, we consider solution of equation (1) in case III.
∴ u(x, y, t) = )sincos()sincos( 24231211 ypcypcxpcxpc ++ )sincos( 65 cptccptc + ...(2)
where 2 2 21 2 .p p p= +
The boundary conditions are
(i) u(0, y, t) = u(a, y, t) = 0 (ii) u(x, 0, t) = u(x, b, t) = 0
and the initial condition are (iii) u(x, y, 0) = xy (a – x) (b – y) (iv) ,0=∂∂
tu at t = 0
Using boundary conditions in (2), we get
At x = 0, the temperature the defelection u = 0
⇒ 0 = )sincos()sincos( 6524231 cptccptcypcypcc ++⇒ c1 = 0 [(otherwise u(x, y, t) = 0) which is impossible]From (2), we obtain
u(x, y, t) = )sincos()sincos(sin 65242312 cptccptcypcypcxpc ++ ...(3)
At x = a, u = 0
⇒ 0 = )sincos()sincos(sin 65242312 cptccptcypcypcapc ++
⇒ sin p1a = 0 ⇒ 1 1sin sin .m
p a m paπ= π ⇒ =
At y = 0, u = 0
(3) gives 0 = )sincos()0(sin 65312 cptccptccxpc ++⇒ c3 = 0 [(otherwise, u = 0) which is impossible]Putting c3 = 0 in equation (3), we get
u(x, y, t) = )sincos(sinsin 652142 cptccptcypxpcc + ...(4)
At y = b, u = 0
A
BC
O
(0, )b
(0, 0)
( , )a b
( , 0)a
512 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
Equation (4) gives
0 = )sincos(·sinsin 652142 cptccptcbpxpcc +
⇒ sin p2b = 0 [otherwise u(x, y, t) = 0, which is impossible]
⇒ sin p2b = 2sin .n
n pbππ ⇒ =
Putting the values of p1 and p2 in equation (4), we get
u(x, y, t) = 2 4 5 6sin sin ( cos sin )m x n y
c c c cpt c cpta bπ π + ...(5)
Putting ∂u/∂t = 0 and u = 0, we get
⇒0=
∂∂
tt
u= 0 2 4 5 6sin sin ( sin (0) cos (0))
m x n yc c c cp cp c c p cp
a bπ π= − +
⇒ 0 = 2 4 6 6sin sin 0m x n y
c c c cp ca bπ π ⇒ = 0 otherwise =u
∴ Equation (5), gives
u(x, y, t) = 2 4 5 sin sin cosm x n y
c c c cpta bπ π
The general form of solution is
u(x, y, t) = 1 1
sin sin cosmn mnm n
m x n yA cp t
a b
∞ ∞
= =
π π∑∑ ...(6)
where Amn= 2 2
2 4 5 2 2and mn
m nc c c p
a b= π +
At t = 0, from (6), we get ∑∑∞
=
∞
=
ππ⋅=−−
1 1
sinsin))((m n
mnb
yn
a
xmAybxaxy
which is double half range Fourier sine series.
∴ Amn = 0 0
2 2( )( )sin sin
a b
x y
m x n yxy a x b y dxdy
a b a b= =
π π⋅ − −∫ ∫
= ∫ ∫π
−π
−a b
dyb
ynybydx
a
xmxax
ab0 0
sin)(sin)(4
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 513
=
2 3
2 2 3 3
0
cos4 2( ) ( 2 ) sin cos
am x
a m x a m xax a x a xmab a am ma
π π π− − + − − π π π
2 3
2 2 3 30
2( ) cos ( 2 ) sin cos
bb n y b n y b n y
y b y b yn b b bn n
π π π× − − + − −
π π π
= 3 3 3 3
3 3 3 3 3 3 3 3
4 2 2 2 2cos cos
a a b bm n
ab m m n n
− −π + π + π π π π
= { }{ }3 3
1 13 3 3 3
4 2 2( 1) 1 ( 1) 1m na b
ab m n+ +⋅ ⋅ − + − +
π π
= ;64
633
22
πnm
ba m and n both are odd.
Putting the value of Amn in equation (6), we get
u(x, y, t) = ∑ ∑∞
=
∞
=
πππ..),5,3,1( ,..)5,3,1(
633
22
cos·sin·sin64
m nmntcp
b
yn
a
xm
nm
ba
= ∑ ∑∞
=
∞
=
πππ ..),5,3,1( ,..)5,3,1(
336
22
cossinsin164
m nmntcp
b
yn
a
xm
nm
ba Ans.
where pmn = 2
2
2
2
b
n
a
m+π .
EXERCISE 5.51. The initial displacement of the rectangular membrane (0 < x < 1, 0 < y < 2) where boundary
is fixed given that it starts from rest and u(x, y, 0) = xy (1 – x) (2 – y). Find u(x, y, t).
+π=
πππ
= ∑∑
4 and odd are both and where
2sinsincos
256),,(
22
633
nmpnm
ynxmtcp
nmtyxu
mn
mnAns.
514 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
2. The initial displacement of a rectangular membrane (0 < x < a, 0 < y < b) whose boundaryis fixed is given by f (x, y) = λ (ax – x2) (by – y2) show that
1 1
( , , ) cos sin sinmn mnm n
m x n yu x y t A cp t
a b
∞ ∞
= =
π π= ∑ ∑
2 2
3 3 6
16where (1 cos )(1 cos );mn
a bA m n
m n
λ= − π − π
π
2 22 2
2 2and .mn
m np
a b
= π +
3. Find the deflection u(x, y, t) of the square membrane with a = b = c = 1, if the initial
velocity is zero and the initial deflection f (x, y) = A sinπx sin2πy.
]2sinsin.5cos),,( [ yxtAtyxu πππ=Ans.
4. Find the deflection u(x, y, t) of a rectangular membrane (0 ≤ x ≤ a, 0 ≤ y ≤ b) and
0
0.t
u
t =
∂ = ∂ The deflection at t = 0 is xy (a2 – x2) (b2 – y2).
2 2
3 3 61 1
2 22 2
2 2
144 ( , , ) sin sin cos
where .
mnm n
mn
a b m x n yu x y t cp t
a bm n
m np
a b
∞ ∞
= =
π π=
π
= π +
∑ ∑Ans.
5. Find the deflection u(x, y, t) of a membrane (0 ≤ x ≤ a, 0 ≤ y ≤ b) when the initial velocity
is zero and the initial deflection is given by 2
sin sin .x y
a bπ π
2 2
2 4 1 ( , , ) sin sin cos
x yu x y t ct
a b a b
π π= ⋅ π +
Ans.
5.8 EQUATIONS OF TRANSMISSION LINES
Suppose the resistance, inductance and capacitance vary linearly with x in transmission line, where x isthe distance from the source of electricity, measured along the line. If V(x, t) denote the potential at apoint on the cable, i(x, t) is current there R is the resistance of the cable per unit length and L isinductance per unit length.
Then the equations
xV
∂∂
= tV
Cxi
ti
L∂∂−=
∂∂
∂∂− ,
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 515
∂∂
=∂∂
∂∂
=∂∂
t
iRC
x
i
t
VRC
x
V
2
2
2
2
...(A)
and
∂∂
=∂∂
∂∂
=∂∂
2
2
2
2
2
2
2
2
t
iLC
x
i
t
VLC
x
V
...(B)
The set (A) is known as ‘Telegraph equations’, and (B) is known as ‘radio equations’.
Example 33: Neglecting R and G obtain potential V(x, t) in a line l km. long, after t seconds the
ends are suddenly grounded, if initially i(x, 0) = I0 and V(x, 0) = V0 sin .lxπ
Also current,
(U.P.T.U. 2008)
Solution: We have
2
2
x
V
∂∂
= 2
2
t
VLC
∂∂
...(1)
Let V(x, t) = )()( tTxX ...(2)
∴ 2
2
x
V
∂∂
= 2
2
2
2
2
2
anddt
TdX
t
V
dx
XdT =
∂∂
Putting these values in equation (1), we obtain
2
2
dx
XdT = 2
2
2
2
2
2
2 1p
dt
Td
T
LC
dx
Xd
Xdt
TdLCX −==⇒
∴2
21
dx
Xd
X= 02
2
22 =+⇒− Xp
dx
Xdp
⇒ X = pxcpxc sincos 21 +
and2
2
dt
Td
T
LC= 0
2
2
22 =+⇒− T
LC
p
dt
Tdp
⇒ T = LC
ptc
LC
ptc sincos 43 +
From equation (2), we get
V(x, t) =
++
LC
ptc
LC
ptcpxcpxc sincos)sincos( 4321 ...(3)
516 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
The boundary conditions are
(i) V(0, t) = 0 (ii) V(l, t) = 0 grounded. suddenly are ends the As
The initial conditions are
(iii) 0)0,(
sin)0,( 0 =∂
∂⇒π=txV
lx
VxV
(iv) 0)0,(
)0,( 0 =∂
∂⇒=t
xiIxi
At x = 0, V = 0, equation (3) gives
0 =
+
LC
ptc
LC
ptcc sincos 431
⇒ c1 = 0 [otherwise V(x, t) = 0, which is impossible]∴ From (3), we have
V(x, t) =
+
LC
ptc
LC
ptcpxc sincossin 432 ...(4)
At x = l, V = 0, equation (4) gives .n
plπ=
∴ From equation (4), we have
V(x, t) =
π+
ππ
LCl
tncc
LCl
tncc
l
xnsincossin 4232
The general form of above solution is
V(x, t) = 1
sin cos sinn nn
n x n t n ta b
l l LC l LC
∞
=
π π π+
∑ ...(5)
tV
∂∂
= 1
sin sin cosn nn
n x l LC n t l LC n ta b
l n nl LC l LC
∞
=
π π π− ⋅ + ⋅ π π
∑
At t = 0, 0=∂∂
tV
∴ Above equation gives
0 = 1
sin 0.n nn
n x l LCb b
l n
∞
=
π⋅ ⇒ =
π∑From (5), we obtain
V(x, t) = 1
sin cosnn
n x n ta
l l LC
∞
=
π π∑ ...(6)
and at t = 0, V = lx
Vπ
sin0
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 517
⇒lx
Vπ
sin0 = 1
sinnn
n xa
l
∞
=
π∑
orlx
Vπ
sin0 = ·····2
sinsin 21 +π+πlx
alx
a
⇒ 01 Va = and a2 = a3 = a4 =........ = an= 0.
Putting the value of a1 in equation (6), we get
V(x, t) = .cossin0LCl
t
l
xV
ππ Ans.
Second part: We know that
xV
∂∂
= ti
L∂∂−
So
ππ∂∂
LCl
t
l
xV
xcossin0 =
ti
L∂∂−
⇒LCl
t
l
xV
l
πππcoscos0 =
ti
L∂∂− ...(7)
Integrating (7), we obtain
i = 0 cos sin ( )C x t
V xL l l LC
π π− + φ ...(8)
But At t = 0, i = I0, equation (8), gives
I0 = φ(x)
Putting the value of φ(x) in equation (8), we get
i = 0 0 cos sin .C x t
I VL l l LC
π π− Ans.
EXERCISE 5.61. Find the potential V(x, t) in a line of length l, after t seconds the ends were suddenly
grounded, given that i(x, 0) = i0 and V(x, 0) = .5
sinsin 51 lx
Vlx
Vπ+π
1 55 5
( , ) sin cos sin cosx t x t
V x t V Vl ll LC l LC
π π π π= +
Ans.
518 A TEXTBOOK OF ENGINEERING MATHEMATICS–II
2. Neglecting R and G find the e.m.f. e(x, t) in a line one km long, t seconds after the ends
were suddenly grounded, if initially i(x, 0) = I so that 0=∂∂te
at t = 0 and e(x, 0)
= 1 22
sin sin .x x
E Ea aπ π+ 1 2
2 2 ( , ) sin cos sin cos
x t x te x t E E
a aa LC a LC
π π π π= +
Ans.
3. Obtain the solution of the radio equation 2
2
2
2
t
VLC
x
V
∂∂
=∂∂
when a periodic e.m.f. V0 cos pt is
applied at the end x = 0 of the line. }]cos{),( [ 0 LCpxptVtxV −=Ans.4. A transmission line 1000 km. long is initially under steady state conditions with potential
1300 volts at the sending end (x = 0) and 1200 volts at the receiving end (x = 1000). Theterminal end of the line is suddenly grounded, but the potential at the source is kept at 1300volts. Assuming the inductance and leakance to be negligible find the potential V(x, t).
2
2:Apply ; in steady state 0
V E ERE
t tx
∂ ∂ ∂= =
∂ ∂∂ Hint
2 2
21
1
2400 ( 1) ( , ) 1300 1 3 sin
n tnl RC
n
n xV x t x e
n l
π∞ + −
=
− π = − ⋅ + π
∑Ans.
5. A steady voltage distribution of 20 volts at the sending end and 12 volts at the receiving endis maintained in a telephone wire of length l. At time t = 0, the receiving end is grounded.Find the voltage and current t second later. Neglect leakage and inductance.
2 2
2
2 2
2
120( ) 24 ( 1) ( , ) sin
20 24( , ) ( 1) cos
n tnRcl
n tn Rcl
l x n xV x t e
l n l
n xi x t e
lR lR l
π+ −
π−
− − π = + π
π = + −
∑
∑
Ans.