applications of partial differential equations v applications of partial differential equations 5.1...

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=


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

+ + − ++


and again from (3), we have

dy

dY

Y

1= k− ⇒

dYkdy

Y= −


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

⇒ =


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


⇒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)


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


⇒ 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= ±


∴ 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+


⇒ 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


( , )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)


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


),( 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.


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 =


),( 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


∴ 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


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


0 = )sincos( 431 cptccptcc + ⇒ c1 = 0


y(x, t) = )sincos(sin 432 cptccptcpxc + ...(3)


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


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−

−

+


= 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 −=


ty

∂∂

= }cossin{sin 432 cptcccptccpxc +−

Using (iii) in above relation, we get

0 = pxccc sin)( 42 ⇒ 4 0.c =


Using the values of c and p in equation (3), we get

y = lctn

lxn

ccππ

cossin32


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


where p = π

.nl


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 =


),( 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.


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


0 = )sincos(sin 432 cptccptcplc +


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


),( 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


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

∂ ∂ ∂ ∂ ∂ ∂ ∂ + + = + + ∂ ∂ ∂ ∂ ∂ ∂∂ ∂


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.


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


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


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


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)


and its solution is



(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 −=


),( txy = .2sin2sin03.0

sinsin05.0

ctxc

ctxc

− Ans.


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


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 =

∑


⇒

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

∂ ∂ ∂ ∂+ + +

∂ ∂ ∂ ∂∂ ∂


⇒ 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


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.


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


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


∴ 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


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)


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



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)


(i) u(0, t) = 0 and (ii) u(l, t) = 0

and initial condition is (iii) u(x, 0) = f(x).


0 = 0131

22

=⇒− cecc tcp


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)


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)


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.


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)


(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)


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


= 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


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


∴ 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


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


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.


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)


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 =


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)


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


),( 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


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.


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


321 )sincos( −+ ...(2)


(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.


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)


(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


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


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.


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.


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


⇒ 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.


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)



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

⋅


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)


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 =


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


Solution: The steady state temperature u(x, y) is the solution of

2

2

2

2

y

u

x

u

∂∂

+∂∂

= 0 ...(1)


(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


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


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)


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]


u(x, y) = )(sin 432pypy ececpxc −+ ...(3)

at x = l, u = 0

0 = )(sin 432pypy ececplc −+


plπ=


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 + ⇒ + = ⇒ = −


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


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


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∫ ∫


= 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)


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)


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



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


⇒ 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,


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.


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'


∴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 ∂ ∂ ∂


= 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−+


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)


(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)


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.


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∂


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


⇒ 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).


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= +


(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


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


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


=

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.


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∂∂−=

∂∂

∂∂− ,


∂∂

=∂∂

∂∂

=∂∂

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 +


V(x, t) =

++

LC

ptc

LC

ptcpxcpxc sincos)sincos( 4321 ...(3)



(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


⇒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.


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.

applications of partial differential equations v applications of partial differential equations 5.1...

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