APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONSchettinadtech.ac.in/storage/12-07-17/12-07-17-09-27-49-1577-velu.pdf · 1 APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS • Classification

1

APPLICATIONS OF PARTIAL DIFFERENTIAL

EQUATIONS

• Classification of second order quasi linear partial

differential equations

• Solutions of one dimensional wave equation

• One dimensional heat equation

• Steady state solution of two-dimensional heat equation

(Insulated edges excluded)

• Fourier series solutions in Cartesian coordinates.

In mathematics, partial differential equations (PDE) are a type of differential equation, i.e.,

a relation involving an unknown function (or functions) of several independent variables and

its (or their) partial derivatives with respect to those variables. Partial differential equations

are used to formulate, and thus aid the solution of, problems involving functions of several

variables; such as the propagation of sound or heat, electrostatics, electrodynamics, fluid

flow, and elasticity. Seemingly distinct physical phenomena may have identical mathematical

formulations, and thus be governed by the same underlying dynamic.

2

In mathematics, in the field of differential equations, a boundary value problem is a

differential equation together with a set of additional restraints, called the boundary

conditions. A solution to a boundary value problem is a solution to the differential equation

which also satisfies the boundary conditions.

Boundary value problems arise in several branches of physics as any physical differential

equation will have them. Problems involving the wave equation, such as the determination of

normal modes, are often stated as boundary value problems. A large class of important

boundary value problems are the Sturm-Liouville problems. The analysis of these problems

involves the eigenfunctions of a differential operator.

To be useful in applications, a boundary value problem should be well posed. This means

that given the input to the problem there exists a unique solution, which depends

continuously on the input. Much theoretical work in the field of partial differential equations

is devoted to proving that boundary value problems arising from scientific and engineering

applications are in fact well-posed.

Among the earliest boundary value problems to be studied is the Dirichlet problem, of

finding the harmonic functions (solutions to Laplace's equation); the solution was given by

the Dirichlet's principle.

3

Initial value problem

A more mathematical way to picture the difference between an initial value problem and a

boundary value problem is that an initial value problem has all of the conditions specified at

the same value of the independent variable in the equation (and that value is at the lower

boundary of the domain, thus the term "initial" value). On the other hand, a boundary value

problem has conditions specified at the extremes of the independent variable. For example, if

the independent variable is time over the domain [0,1], an initial value problem would

specify a value of y(t) and y'(t) at time t = 0, while a boundary value problem would

specify values for y(t) at both t = 0 and t = 1.

If the problem is dependent on both space and time, then instead of specifying the value of

the problem at a given point for all time the data could be given at a given time for all space.

For example, the temperature of an iron bar with one end kept at absolute zero and the other

end at the freezing point of water would be a boundary value problem.

Types of boundary value problems

The boundary value problem for an idealised 2D rod

If the boundary gives a value to the normal derivative of the problem then it is a Neumann

boundary condition. For example, if there is a heater at one end of an iron rod, then energy

would be added at a constant rate but the actual temperature would not be known.

If the boundary gives a value to the problem then it is a Dirichlet boundary condition. For

example, if one end of an iron rod is held at absolute zero, then the value of the problem

would be known at that point in space.

If the boundary has the form of a curve or surface that gives a value to the normal derivative

and the problem itself then it is a Cauchy boundary condition.

Aside from the boundary condition, boundary value problems are also classified according to

the type of differential operator involved. For an elliptic operator, one discusses elliptic

boundary value problems. For an hyperbolic operator, one discusses hyperbolic boundary

4

value problems. These categories are further subdivided into linear and various nonlinear

types.

Related mathematics:

• initial value problem

• differential equations

• Green's functions

• Stochastic processes and

boundary value problems

• Examples of boundary value

problems

• Sturm-Liouville theory

• Dirichlet boundary condition

• Neumann boundary condition

• Sommerfeld radiation

condition

• Cauchy boundary condition

Physical applications:

• waves

• normal modes

• electrostatics

• Laplace's equation

• potential theory

Numerical algorithms:

• shooting method

• direct multiple

shooting method

CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND

ORDER

Let a second order partial differential equation in the function u of the two independent

variable x,y be of the form.

0,,,,2

22

2

2

=

∂

∂

∂

∂+

∂

∂+

∂∂

∂+

∂

∂

y

u

x

uuyxf

y

uC

yx

uB

x

uA

1

5

Eqn (1) is classified as elliptic, parabolic , (or) hyperbolic at the points of a given region R

depending on whether

B2 -4AC < 0 [ elliptic equation]

B2-4AC < 0 [ Parabolic equation]

B2 -4AC < 0 [ hyperbolic equation]

CLASSIFY THE FOLLOWING EQUATIONS :-

1) x fxx + yfyy =0, x>0 , y>0.

Soln : -

Here A=x ; C=y

∴ B

2 - 4Ac = - 4xy

= - ve when x>0 , y>0

∴ The equation is elliptic.

2) fxx -2 fxy =0 , x>0 , y>0

Soln :-

Here A=1; B = -2, c=0

∴ B2 – 4Ac = 4-0 =4 =+ ve

∴ The equation is hyperbolic

3) uxx -2uxy + uyy =0

Here A=1; B = -2, c=1

∴ B2 - 4AC = 4 -4 =0

∴ The equation is parabolic

4) fxx + 2fxy + 4fyy =0 , x>0, y>0

Here A=1,B=2, C=4

6

∴ B2 -4AC = 4-16

= - 12

= -ve

∴ The equation is elliptic

5) x2fxx + (1-y

2) fyy =0 for -1 <y<1

−∞ <x< ∝

Here A=x2 ; B=0 ; C=1-y

2

∴B2-4AC = -4x

2(1-y

2)

= -4x2(y

2-1)

x2 is always +ve in -∞<x<∞

y2-1 is negative in -1<y<1

∴ B2 -4AC =- ve (x≠0)

∴ The equation is elliptic

If x=0 , B2 -4AC =0

∴ The equation is parabolic.

ONE DIMENSIONAL WAVE EQUATION

One dimensional wave equation is

(i) Problems on Vibrating string

with zero velocity

ii) Problems on vibrating string

with non –zero velocity

7

2

22

2

2

x

ya

t

y

∂

∂=

∂

∂

The various solutions of the wave equation is

y( x,t) = [ C1epx

+ C2e-px

] [C3epat

+ C4e-pat

]

y(x,t) = [ C5cospx + C6 sinpx] [ C7 cospat + C8 sinpat]

y( x,t) =[ C9x + C10] [ C11t + C12]

Note:

In problems of vibration of strings we always take the following as the correct

solution;

y(x,t) = [ C1 cospx + C2 sinpx] [C3 cospat + C4 sinpat ]

i) Problems on Vibrating string with zero initial velocity

1) A string is stretched and fastened to two points x=0 and x=l asprt. Motion is

started by displacing the string into the form y=k(lx-x2) from which it is released at time

t=o. Find the displacement of any point on the string at a distance of x from one end at time

t.

Soln:-

Here, we have to solve the wave equation

with given boundary conditions

i) y(0,t) =0 These two conditions are

fixed conditions in wave Eqn

ii) y(l,t) =0

iii) ∂ y(x,0) =0 (initial velocity is not given in problem. So we

∂t take it as 0).

iv) y(x,0) = K (lx=x2) ( initial displacement is given in problem it self. )

The correct solution of (1) is

2

22

2

2

x

ya

t

y

∂

∂=

∂

∂ 1

8

y(x,t) = [C1 cospx+ C2 sinpx] [C3 cospat + C4 sinpat]

Apply condition (i) in (2) , we get

y( 0,t) = C1 [C3 cospat + C4 sinpat]=0

=>C3 cospat + C4 sinpat ≠ 0

C1=0

Sub C1=0 in (2) , we get the solution

Apply condition (ii) in (3), we get

y( l,t) = C2 sinpl [C3 cospat + C4 sinpat]=0

=> C2 ≠ 0 and C3 cospat + C4 sinpat ≠ 0

∴ sinpl =0

ie., pl =sin-1

0

ie., pl =nπ

ie.,

Sub p= nπ in (3) , we get the solution

l

pl = nπ

l

+

=

l

atnC

l

atnC

l

xnCtxy

πππsincos[sin),( 432

4

2

y( x,t) =C2 sinpx [ C3 cospat + C4 sinpat]

3

9

`P.Diff (4) w.r.to ‘t’, we get

Apply condition (iii) in (I) , we get

Sub C4 =0 in (4) we get the solution

Bu superposition principle , we get the solution

Apply condition (iv) in (5) we get

This show that , this is the half range F.S.series of K(lx-x2). Using Fourier coefficient formula

+

−=

∂

∂

l

atn

l

anC

l

atn

l

anC

l

xnC

t

txy πππππcossinsin

),(432

0

0sin&0&0

0sin

0sin)0,(

4

2

42

42

=∴

≠≠≠⇒

=

=

=

∂

∂

C

l

xnalso

l

anC

l

xn

l

anCC

l

anC

l

xnC

t

xy

ππ

ππ

ππ

32

32

cossin)],(

cossin)],(

CCwhereCl

atn

l

xnCtxy

l

atn

l

xnCCtxy

nn =

=

=

ππ

ππ

= ∑

∞

= l

atn

l

xnCtxy

n

n

ππcossin),(

1

5

=−=

= ∑∑

∞

=

∞

= l

xnSinbxlxK

l

xnSinCxy

n

n

n

n

ππ

1

2

1

)()0,(

I

10

cn=bn=

=

V= l

xnπsin

2xlxu −= V1 = -cos(π

π

n

l

l

xn×)

xlu 21 −= V2 = - )sin(l

xnπ

22

2

πn

l×

22 −=u V3 = cos(

l

xnπ33

3

πn

l×

∴ Cn =

l

n

l

l

lxl

n

l

l

xnxlx

l

K

0

33

3

22

22 xn

cos2- n

l

xn-sin)2(cos)(

2

×

××

×−+×−−

π

π

π

π

π

π

=

=

Sub Cn= in (5) , we get

which is the required soln

dxl

xnxlx

l

kl π

sin)(2 2

∫ −

Cn= n

n

Kl)1(1(

433

2

−−π

)

dxl

xnxlxK

l

lπ

sin)(2 2

0

−∫

+×−

33

3

33

3 2cos2

2

πππ

n

l

n

ln

l

K

[ ]n

n

l

l

K)1(1

2233

3

−−π

))1(1(4

33

2n

n

Kl−−

π

( )

−−=∑

∞

= l

atn

l

xn

n

Kltxy

n

n

ππ

πcossin)1(1

4),(

133

2

11

2) A tightly stretched string with fixed end points x=0 and x=l is initially in a position given by

y(x,0) = y0 sin3(px/l).if it is released from rest from this position . Find the displacement ‘y’ at

any distance ‘x’from end at any time ‘t’.

Soln : -

Here, we have to solve the wave equation

2

22

2

2

x

ya

t

y

∂

∂=

∂

∂

With boundary conditions

i) y(0,t)=0

ii) y(l,t)=0

iii)

iv)



y(0,t) = c1[ (c3 cospat + c4sinpat] =0

=> c3 cospat + c4sinpat ≠0

sub c1=0 in (2) , we get

y(x,t) = c2 sinpx (c3 cospat + c4sinpat (3)

Apply condition (ii) in (3) , we get

y(x,t) = c2 sinpl (c3 cospat + c4sinpat) =0

=> c2 ≠ o and c3 cospat + c4 sinpat ≠0

∴sinpl =0

ie. pl =nπ

y(x,t) = (c1 cospx + c2 sinpx) (c3 cospat + c4sinpat)

c1 =0

0)0,(

=∂

∂

t

xy

−

=

=

l

x

l

xy

l

xyxy

πππ 3sin

4

1sin

4

3sin)0,( 0

3

0

1

2

12

sub l

np

π= in (3) , we get

Patially diff (4) w.r.’t’ ,we get

+

×−

=

∂

∂

l

atn

l

anc

l

atn

l

anc

l

xnc

t

txy πππππcossinsin

),(432

Apply condition (iii) in (4), we get

By superposition principle, we get

Apply condition (iv) in (5) , we get

l

np

π=

y(x,t) = c2 sin

+

l

atnc

l

atnc

l

xn πππsincos

43

( )

32

32

4

4

2

42

42

cossin),(

cossin),(

),4(0

0

0sin0,0

0sin

0sin)0,(

ccwherecl

atn

l

xnctxy

l

atn

l

xncctxy

wegetinsubc

c

l

xnand

l

anc

l

an

l

ancc

l

anc

l

xnc

t

xy

nn =

=

=

=

=∴

≠≠≠=>

=

=

=

∂

∂

ππ

ππ

ππ

ππ

ππ

= ∑

∞

= l

atn

l

xnctxy

n

n

ππcossin),(

1

−

== ∑

∞

= l

x

l

xy

l

xncoxy

n

n

πππ 3sin

4

1sin

4

3sin),( 0

1

4

5

13

c1 sin

Equating the coeffient, we get

Substitute these values in (5) , we get


3) A string is tightly stretched and its ends are fastened at two points x=0 and x=l. The mid

point of the string is displaced transversely through a small distance ‘b’ and the string is

released from rest in that position. Find an expression for the transverse displacement of the

string at any time during the subsequent motion.

Soln

Here , first we have to find initial displacement ie[y(x,o) ] of the string in given problem.

D (l/2,b)

b

l/2 l/2

A(0,0) C B(l,0)

−

=∑

∞

= l

xy

l

xy

l

xnc o

n

n

πππ 3sin

4sin

4

3sin 0

1

−

=

l

at

l

xy

l

at

l

xytxy oo ππππ 3

cos3

sin4

cossin4

3),(

−

=+

+

+

l

xy

l

xy

l

xc

l

xc

l

x o πππππ 3sin

4sin

4

3..........

3sin

2sin 0

32

0..........;4

;0;4

354

032

01 ===−=== cc

ycc

yc

14

Now, we find

Equation of the string in its initial position ADB

The equation of the string(or line) AD is.

The equation of the string (or line) DB is

2/0,2

2/

0

0

2/0

0

lxl

bxy

ylbx

b

y

l

x

<<=

−=−

−

−=

−

−

[ ] lxlxll

by

l

bxbly

l

blbxbly

bl

bxbly

l

bxblby

bl

bxby

lxbl

by

byllxb

b

by

ll

lx

<<−=

−=

+−=

+−

=

−=−

+−

=−

−−=−

−−=−

−

−=

−

−

2,

2

22

2

2

2

2

)2

(2

)(2

)2

(

02

2

15

Hence, initially the displacement of the string is in the form.

Now, we have to solve the wave equation

with boundary conditions

i) y(0,t) =0

ii) y(l,t) =0

iii) 0)0,(

=∂

∂

t

xy

iv) y(x,0)= 2/0,2

lxl

bx<<


Apply condition (i) in(2) we, get

y(0,t) =c1[c3 cospat+c4 sinpat]=0

=> c3 cospat +c4 sinpat ≠ 0

Sub c1=0 in (2), we get

lxlXll

b

lxol

bxxy

<<−

<<=

2/),(2

2/,2

)0,(

2

22

2

2

x

ya

t

y

∂

∂=

∂

∂

)sincos)(sincos(),( 4321 patcpatcpxcpxctxy ++=

∴c1=0

]sincos[sin),( 432 patcpatcpxctxy +=

+=

lxlxll

b<<− 2/),(

2

3

2

1

16

Apply condition (ii) in (3) we, get

Y(l,t)=c2sinpl[c3cospat + c4cospat]=0

=> c2≠0 and c3 cospat +c4sinpat≠0

∴ sinpl=0

ie, pl=nπ

Sub

Partially diff (4) w.r.to ‘t’, we get

=∂

∂

t

txy ),( c2 sin

Apply condition (iii) in (4) , we get

0sin

0,0

0sin

0sin)0,(

2

42

42

≠

≠≠=>

=

=

=

∂

∂

l

xnand

l

anc

l

xn

l

ancc

l

anc

l

xnc

t

xy

π

π

ππ

ππ

l

np

π=

wegetinl

np ),3(

π=

y(x,t) = c2 sin

+

l

atnc

l

atnc

l

xn πππsincos

43

c4=0

( )

+

−

l

atn

l

anc

l

atn

l

anc

l

xn πππππcossin 43

4

17

Sub c4=0 in (4) we, get

y(x,t) = C2C3 sin

l

atn

l

xn ππcos

=Cn sin 32cos ccwherecl

atn

l

xnn =

ππ

By super position principle, we get

Apply condition (iv) in (5) , we get

where cn=bn= 2

l

= ∑

∞

= l

atn

l

xnctxy

n

n

ππcossin),(

1

=

<<−

<<

== ∑∑∞

=

∞

= l

xnb

lxl

xll

b

lxl

bx

l

xncxy

n

n

n

n

ππsin

2),(

2

2/0,2

sin)0,(11

−+

∫ ∫

2/1

0 2/

2sin)(sin

4l

l

dxl

xnxldx

l

xnx

l

b ππ

−

+∫ ∫2

0 2

sin)(2

sin2

l

l

l dxl

xn

l

xlbdx

l

xn

l

bx ππ

5

18

( ) 2/22

2

2/

22

2

2

sincos

sincos4

l

l

l

n

lX

l

xn

n

lX

l

xnxl

n

lX

l

xn

n

lx

l

xnx

l

b

−−−+

+

−

π

π

π

π

π

π

π

π

u=x v=sin

l

xnπ u=l-x

u1=1 v1=-cos

l

n

l

xn ππ

u

1=-1

v2=-sin

l

n

l

xn ππ

=

=

=

Sub cn in (5) , we get


(4) A taut string of length l has its ends x=0,x=l fixed . The point where x=l/3 is drawn aside a

small distance h, the displacement y(x,t) satisfies 2

22

2

2

x

ya

t

y

∂

∂=

∂

∂ .Determine y(x,t) at any time t.

+

+

+

−

2sin

2cos

22sin

2cos

2

422

22

22

22

2

π

π

π

π

π

π

π

π

n

n

ln

n

ln

n

ln

n

l

l

b

2sin

2422

2

2

π

π

n

n

l

l

b

cn =

2

sin8

22π

πn

n

b

∑∞

=

=

12

cossin2

sin8

),(2

n l

atn

l

xnn

n

btxy

πππ

π

19

Soln

Here, first we have to find B(l/3,h)

the initial displacement of

the string ie, y(x,0) h

ie., to find, eqn of OBA l/3

Eqn. of the string (or line) OB is 0(0,0) A(l,0)

x-0 = y-0

0-l/3 0-h

ie. -3x = - y/h

l

ie., y = 3xh in 0<x<l/3

l

Eqn of the string (or line) BA is

x-l/3 = y-h

l/3 –l h-0

y-h = 3x-l

h -2l

y-h = h(3x-l)

-2l

∴y = h(3x-l) + h

-2l

= h(l-3x) + 2lh

2l

= h(l-3x+2l)

2l

∴y = 3h (l-x) in l/3 <x<l

2l

∴ The initial displacement of the string is

y(x,t) = 3xh

, 0<x<l/3

l

= 3h(l-x)

2l ,l/3 <x<l

20

Now, we have to solve the wave eqn

∂2y = a

2 ∂2y

∂t2 ∂x

2


i) y(0,t) =0

ii) y(l,t) =0

iii) ∂y(x,0) = 0

∂t

iv) y(x,0) = 3xh , 0<x<l/3

l

3h(l-x) , l/3<x<l

2l

then proceed this problem solve as previous problem.

5) The points of trisection of a string are palled aside through a distance ‘b’ on

opposite sides of the position of equilibrium and the string is released from rest. Find

the displacement of the string at any subsequent time.

Soln : -

Here, first we have to find an initial

displacement y(x,0) of the string.

21

B1(l/3,a)

a

c (l,0)

0 (0,0) B A

-a

C1(2l/3,-a)

Here, B & C be the points of the trisection of the string OA(=l).

Initially the string is held in the form OB1C

1A where BB

1 = CC

1 = a

Eqn of OB1 is x-0

= y-0

0-l/3 0-a

=> y = 3ax in (0,l/3)

l

Eqn of B1c

1 is x-l/3

= y-a

l/3-2l/3 a+a

=> x-l/3 =

y-a

-l/3 2a

=> 2a(3x-l) = y-a

-l

=> y = 2a(l-3x) +a

l

=> y = 2al-6ax+al

l

22

y = 3al-6ax

l

y = 3a(l-2x) in (l/3,2l/3)

l

Eqn of C

1A is x-2l/3

= y+a

2l/3 –l -a

3x-2l = y+a

2l—3l -a

-a(3x-2l) = y+a

-l

y = a(3x-2l) - a

l

= 3ax-2al-al

l

= 3ax-3al

l

Hence there initial displacement of the string is

y(x,0) = 3a x, 0≤x<l/3

l

=3a(l-2x),l/3 0≤x<2l/3

l

=3a(x-l), 2l/30≤x<l/3

l

Now, we have to solve the wave eqn

∂ 2y = a

2 ∂

2y

∂t2

∂x2

y=3a(x-l) in (2l/3, l)

l

23


(i) y(0,t) =0

(ii) y(l,t)=0

(iii) ∂y(x,0) =0

∂t

(iv) y(x-0) = 3a x, 0≤x<l/3

l

=3a(l-2x) ,l/3 ≤x<2l/3

l

=3a(x-l), 2l/3≤x<l/3

L

then proceed this problem same as previous problems,

Problems on vibrating string with non-zero velocity

1) A tightly stretched string with fixed end points x=0 and x=l is initially at rest in its equilibrium

position and each of its points is given the velocity.

∂ y

∂ t t=0 = V0 sin3 πx , 0<x<l

l

Determine the displacement function y(x,t)

Soln : -

Here we have to solve the wave equation

∂ 2y = a

2 ∂ 2y

∂ t2 ∂ x

2


i) y(0,t) =0

ii) y(l,t) =0

iii) y(x,0) =0

iv) ∂ y(x,0) = V 0 sin

3πx = V0

( 3/4 sin πx -1/4 sin 3πx )

∂ t l l l

The correct soln of (1) is

y(x,t) = [C1 cospx + C2 sinpx ] [C3 cospat + C4 sinpat]

1

2

24


y(0,t) = c1 [c3 cospat _ c4 sinpat] =0

=> c3 cospat + C4 sinpat ≠ 0

∴

Sub c1 =0 in (2) , we get


y(l,t) = c2 sinpx [c3 cospat + c4 sinpat]=0

∴ sinpl =0

ie., pl=sin -1

0

pl=nπ

p =nπ l

sub p =nπ in 3, we get

l

y(x,t)=c2sin nπx c3 cos nπat + c4sin nπat

l l l

Apply condition (3) in 4 we get

y(x,0)= c2sin nπx c3=0

l

c2c3 sin nπx c3=0

l

c2≠0 & sin nπx =0

l

∴c3=0

c1 =0

y(x,t) = [c2 sinpx + [c3 Cospat + c4 sinpat]

4

4

25

Sub c3=0 in 4 , we get

y(x,t)=c2sin nπx c4sin nπat

l l

=c2c4 sin nπx sin nπat

l l

= cn sin nπx sin nπat

l l


Partially w.r.to ;’t’ we get

Apply condition (iv) we get

Equate the coefficients, we get

C2=0

∝

y(x,t)= ∑ cn sin nπx sin nπat

n=1 l l

5

=

∂

∂∑

∞

= l

atn

l

xn

l

anc

t

txy

n

n

πππcossin

),(

1

−

=

=

∂

∂∑

∞

= l

x

l

xV

l

xn

l

anc

t

xy

n

n

ππππ 3sin

4

1sin

4

3sin

)0,(0

1

−

=

+

+

+

=>

l

xV

l

xV

l

x

l

ac

l

x

l

ac

l

x

l

ac

ππ

ππ

ππππ

3sin

4sin

43

.........3

sin3

2sin

2sin

00

3

21

01 43 V

l

ac =

π

a

lVc o

π3

31 =

4

3 03

V

l

ac

−=

π

26

Sub these values of c’s in (5), we get

−

=

l

at

l

x

a

Vol

l

at

l

x

a

lVtxy

ππ

π

ππ

π

3sin

3sin

12sinsin

3

3),( 0

2) A string is stretched between two fixed points at a distance 2l apart & the points

of the string are given initial velocities V where

V = cx is 0<x<l

l

= c/l (2l-x) in l < x <2l

x being the distance from an end point.. find the displacement of the string as any time

Solution:-

Here we have to solve the wave – equation

∂2y = a

2 ∂2y

∂t2

∂x2

Where 2l =L for convenience

With boundary conditions

i) y(0,t) =0

ii) y(Ll,t)=0

iii) y(x,0)=0

iv) ∂y(x,0) = 2cx , 0<x<L/2

L

&

c4=c5=……..=0

a

lVc o

π123

−=

1

27

= 2c(L-x), L/2 < x<L

L

The correct Soln of (1) is

y(x,t)= c1cospx+c2sinpx c3 cospat + c4 sinpat

Apply condition (i) is(2), we get

y(0,t)=c1 c3cospat+c4sinpat = 0

=> c3 cospat + c4sinpat ≠ 0

∴c1=0

sub c1 = 0 in (2) , we get

y(x,t) = c2 sinpx c3cospat + c4 sinpat


y(L,t) = c2 sinpL c3cospat + c4 sinpat

=> c3cospat + c4 sinpat≠0 & c2≠0

∴ sinpL=0

pL=sin-1

0

pL=nπ

p=nπ

L

Sub p= nπ in (3), we get

L

2

3

28

y(x,t)=c2sin nπx c3 cos nπat + c4 sin nπat

L L L


y(x,0)=c2 sin nπx c3=0

L

=>c2 c3 sin nπx =0

L

=> c2≠ 0 & sin nπx ≠0

L

∴

sub c3=0 in 4 , we get

y(x,t)=c2sin nπx c4sin nπat

L L

=c2 c4 sin nπx sin nπat

L L


∝

y(x,t)= ∑ cn sin nπx sin nπat

n=1 L L

Partially w. r. to ‘t’, we get

∝

∂y(x,t) = ∑ cn nπa sin nπx sin nπat

∂t n=1 L L L

Apply condition (iv) , we get

c3=0

4

5

29

∝ 2cx , 0<x<L/2 ∝

∂y(x,0)= ∑ cn nπa sin nπx = L = ∑ bn sin nπx

∂t n=1 L L 2c (L-x), L/2<x<Ll n=1 L

L

L/2 L

where cn nπa = bn = 2 ∫ 2cx sin nπx dx + ∫ 2c(L-x) sin nπx dx

L L 0 L L/2 L L

u=x v= sin nπx u=L-x

L

u1=1 v1 = -cos nπx nπ u1=

-1

L L

v2= -sin nπx n2π2

L L2

= 4c

L2

−+

= ∫ ∫

2

0 2/

2sin)(sin

4L

L

L

dxL

xnxLdx

L

xnx

L

c ππ

( )

−

−−−

+

+

−

L

L

LL

n

L

xn

L

nL

xnxL

L

n

L

xn

L

n

L

xnx

2/

0

2

22

2

2

22

sincos)(

sincos

π

π

π

π

π

π

π

π

30

=>


Replace L by l, we get

+++−

=

2

22

2

22

2

2

2sin

2cos

2.

2sin

2cos44

L

n

n

L

n

nL

L

n

n

Ln

n

L

c

π

π

π

π

π

π

π

π

×= 2

222

2sin24

Ln

n

L

c

π

π

22

2sin8

π

π

n

nc

=

22

22

2sin8

2sin8

π

π

π

π

ππ

n

nc

an

Lc

n

nc

L

anc

n

n

=

=

Cn = 8cL sin( nπ/2)

n3π3

a

( )∑

∞

=

=

133

sinsin2sin8

),(n L

atn

L

xn

an

ncLtxy

ππ

π

π

( )∑

∞

=

=

133 2

sin2

sin2sin16

),(n l

atn

l

xn

an

ncltxy

ππ

π

π

31


3) Find the displacement of a tightly stretched string of length 7 cms vibrating

between fixed end points if initial displacement is 10 sin

and initial velocity is 15 sin

Soln

Here we have to solve the wave equation

Here

with boundary conditions, we get

i) y(0,t) =0

ii) y(l,t) = 0

here both initial velocity

iii) y(x,0)=10 sin 3πx & initial displacement

l are given.

iv) ∂ y(x,0) = 15 sin 3πx

∂ t l


Apply condition (i) in (2) we get

y(0,t) =c1[c3cospat + c4 sinpat] =0

7

3 xπ

7

9 xπ

2

2

2

2

x

ux

t

u

∂

∂=

∂

∂

l=7 cm

y(x,t)= [ c1cospx+c2sinpx] [c3 cospat + c4 sinpat]

1

2

32

=> c3 cospat + c4 sinpat ≠0

Sub c1=0 in (2) we get


y( l,t) = c2 sinpl [c3 cospat + c4 sinpat] =0

=> c3 cospat + c4 sinpat ≠ 0 & c2 ≠ 0

∴ sinpl =0

pl =sin-1

0

pl = nπ

Sub p= nπ in (3) , we get

l

∴c1=0

y(x,t)= c2sinpx (c3 cospat + c4 sinpat)

p= nπ

l

+

=

l

atnc

l

atnc

l

xnctxy

πππsincossin),( 432

+

==>

+

==>

l

atn

l

xnd

l

atn

l

xnctxy

l

atn

l

xncc

l

atn

l

xncctxy

nn

ππππ

ππππ

sinsincossin),(

sinsincossin),( 4232

3

33

By super position principle, we get

∞ ∞

y(x,t) = ∑ cn sin(nπx) cos(nπat) + ∑ dn sin (nπx)sin (nπat)

n=1 l l n=1 l l

∞

y(x,t) = ∑ cn sin(nπat) + dn sin (nπat) sin (nπx) n=1 l l l

Apply condition (iii) in (4) . we get

∞

y(x,0) = ∑ cn sin(nπx) = 10sin(3πx) N=1 l l

c1 sinπx + c2 sin 2πx + c3sin3πx + ……………. = 10 sin (3πx)

l l l l

Equate like coefficients, we get

c3 = 10 & c1 = c2 = c4 =…………………………………….0 (A)

From (4) , we get

∞

∂y(x,t) = ∑ cn (nπa) -sin(nπat) + dn (nπa)cos (nπat) sin nπx

∂t n=1 l l l l l

Apply condition (iv) , we get

∞

∂y(x,0) = ∑ dn (nπa ) sin (nπx) = 15 sin (9πx)

∂t n=1 l l l

d9 ( 9πa) = 15 [Equate like coefficients]

l

d9 = 15l & d1= d2 …………………………………….d8 = d10 =0_______________(B)

9πa

sub (A) & (B)in ( 4), we get

34

where l =7

ONE – DIMENSIONAL HEAT FLOW EQUATION

One dimensional heat equation is

∂u =α2 ∂2

u

∂t ∂x2

The various solutions of the heat equations is

u(x,t) = c1 [ c2 x+c3 ]

Note

The correct soln of one dimensional heat flow eqn as

Problems

1. Solve the equation ∂u =α2 ∂2u subject to the boundary conditions u(0,t)=0 u( l,t)=0, u(x,0) = x.

∂t ∂x2

Soln:-

Here, we have to solve the heat flow equation ∂u =α2 ∂2u

∂t ∂x2

+

=

l

x

l

at

a

l

l

x

l

attxy

ππ

π

ππ 9sin

9sin

9

153sin

3sin10),(

[ ]pxececectxu pxtp −+= 543

22

),( α

[ ]pxcpxcectxu tp sincos),( 876

22

+= −α

[ ]pxBpxAetxu tp sincos),(22

+= −α

35

with given boundary conditions

i) u(0,t)=0

ii) u(l,t)=0

iii) u(x,0)=x


u(x,t) = ( A cospx + B sinpx) tpe

22α−


u(0,t) = A tpe

22α− = 0

=> tpe

22α− =0

∴ A=0

Sub A=0 in (2) , we get

u(x,t) = B sinpx tpe

22α−

Apply condition (ii) is (3) , we get

u(l,t) = Bsinpl tpe

22α− =0

=> B≠ 0 tpe

22α− ≠0

=> sinpl=0

∴pl = nπ

p= nπ

l

2

36

Sub p = nπ in (3), we get

l

u(x,t) = Bsin nπx

− 2

222

l

tn

eπα

l

In general,


Apply condition (iii) is (4) we, get

∞ ∞

u(x,0) = ∑ Bn sin(nπx) = x= ∑ bn sin nπx

n=1 l n-1 l

where Bn = bn =2/l ∫l

0

x sin l

xnπ

u=x v= sin nπx

l

u1=1 v1=-cos nπx nπ

l l

v2 = sin nπx (nπ)2

l l2

l

∴Bn= 2 -x cosnπx + sin nπx

L nπ/l (nπ/2)2 0

=2/l -l(-1)n

nπ/l

−

= 2

222

sin),( l

tn

n el

xnBtxu

παπ

−

∞

=

=∑ 2

222

sin),(1

l

tn

n

n el

xnBtxu

παπ

37

Bn = 2 l( -1)n+1

nπ

Sub Bn in (4) we get

∝

u(x1t) = ∑ 2l (-1)n+1

sin nπx

− 2

222

l

tn

eπα

n-1 nπ l

STEADY STATE CONDITIONS AND ZERO BOUNDARY CONDITIONS

Problems:

1) A rod of length l has its ends A and B kept at 00 c of 100

0 c until steady state condition prevail.

if the temperature at B is reduced suddenly to 00 c & kept so while that of A is maintained, find

the temperature u(x,t) at a distance x from A and time t.

SOLUTION:

Steady state

l A B

00 c 100

0c

In given problem,

When steady state prevails the end A is at 00 c and the end B is at 100

0 c

Now, we have to find the temperature when steady state condition prevails

ie., to find u(x, t) before the end B is reduced to 00 c

Hence when steady state conditions prevails, the heat flow equation becomes

∂ 2u =0 => d

2 u =0

∂x2 dx

2

38

=>


(i) u(0) =0

(ii) u(l) =100

Apply (i) in (1) , we get u(0) = b=0

=> b=0

Sub b=0 is (1) we get

Apply (ii) in we get u(l) = al = 100

=>a=100/l

Sub a is , we get

∴ In steady state, the temperature distribution

is

Now, the temperature distribution reached at the steady state becomes the initial distribution for the unsteady state

ie ,

u(x)=ax+b

u(x) =ax

u(x) =100x

l

u(x) = 100x

l

u(x,0) = 100x is 0<x<l

l

2

2

2

3

4

1

39

Unsteady state

l

A B

00 c 0

0c

(the end A is at 00 C & B is at 0

0C)

Hence , when unsteady state condition prevails,

∂u = ∝2 ∂ 2u

∂t ∂x2


(i) u(0,t) = 0

(ii) u(l,t)=0

(iii) u(x,0) = 100x in 0<x<l

l


u(x,t) = ( A cospx + Bsinpx) tpe

22α−

Apply condition (i) is ,we get

u(0,t) = A tpe

22α−

=> tpe

22α− ≠ 0

∴

Sub A=0 is , we get

u(x,t) = Bsinpx tpe

22α−

6

6

A=0

6

7

40

Apply condition (ii) is , we get

u(l,t) = B sinpl tpe

22α− = 0

=> B ≠ 0 & tpe

22α− ≠ 0

∴ sinpl=0

� pl = nπ

p=nπ

l

Sub p=nπ in , we get

l

u(x,t) = Bsin nπx

− 2

222

l

tn

eπα

l

In general,

u(x,t) = Bn sin nπx

− 2

222

l

tn

eπα

l

By superposition principle , we get


∞ ∞

u( x, 0) = ∑ Bn sin nπx = 100x=∑ bnsin nπx

n=1 l l n=1 l

Where Bn=bn = 2/l ∫

l

l

xn

l

x

0

sin100 π

dx

u=x v = sin nπx

l

u1=1 v1= -cos nπx / nπ

l l

7

7

−

∞

=

∑

= 2

222

1

sin),( l

tn

n

ne

l

xnBtxu

παπ8

41

v2 = sin nπx nπ 2

l l

Bn = 200 (-1)n+1

nπ

Sub Bn in (8) , we get

2

222

1

1

sin)1(200

),(l

tne

l

xn

ntxu

n

n παπ

π

−−=∑

∞

=

+

STEDY STATE CONDITIONS AND NON-ZERO BOUNDARY CONDITIONS

Problems : A bar 10 cm long with insulated sides, has its ends A&B kept at 20

0C and 40

0C respectively

until steady state conditions prevail. The temperature at A is then suddenly raised to 500C & at the

sometime that at B is reduced at 100C .Find the temperature at any point of the bar at any time

Soln: -

In steady state condition,

the heat flow equation becomes

Steady state

d2u

dx2

10cm u(x)=ax+b A B

200C 40

0C


(i) u(0)=20

0

(ii) u(10)=400

Apply (i) in , we get

( )

l

ln

l

xn

ln

l

xnx

lBn

0

22

sincos200

+−

−=π

π

π

π

( )

−−=

ln

l

l π

2

2

1200

1

1

42

u(0)=b=20 =>

Sub b=20 in (1) , we get

u(x)=ax+20

Apply (ii) in (2) , we get

u(10)=10a+20=40

=>a+2=4

=>a =2

sub a in (2) , we get

∴ In steady state , the temperature distribution is

Now , the temperature at A & B are changed

At this stage , the steady state is changed into unsteady state.

From this stage, the initial temperature distribution is

in 0<x<20

Hence , when unsteady state condition prevails the heat flow equation is

∂u = α2 ∂2

u

∂t ∂x2


(i) u(0,t)=50 10cm

(ii) u(10,t)=50 A B

(iii)u(x,0)=2x+20 500 c 10

0c

The correct solutions of is

u(x)=2x+20

u(x)=2x+20

2

b=20

3

u(x,0)=2x+20 4

5

5

u(x,t)=(Acospx + Bsinpx) tpe

22α−

6

43

Apply (i) in , we get

u(0,t)=A tpe

22α− = 50

Apply (ii) in , we get

u(10,t)=(Acos10p +Bsin10p) tpe

22α− = 10

From & , it is not possible to find the constants A & B.

In this case , we spilt the solution u(x,t) into two parts.

i.e

where us(x) is a solution of the ∂u = α2 ∂2

u

∂t ∂x2

and is a function of x alone

and satisfying the conditions

us(0) = 50

& us(10) = 10

and ut(x,t) is a transient solution satisfying which decreases as t increases.

To find us(x)


(i) us(0)=50

(ii) us(10)=10

Apply (i) is , we get

us(0)= b1=50

=>

Sub b1 in , we get

6

7

6

8

7 8

u(x,t)=u s(x) + ut (x,t) 9

9

us(x)=a1x+b1 10

10

us(x) = a1x+50

b1=50

11

44

Apply (ii) in(11) , we get

us(10)=a1(10)+50=10 a1+5=1

a1=1-5

sub a1 in (11), we get

To find ut(x,t)

� => ut(x,t) = u(x,t)-us(x)

is a transient solution of ∂u = α2 ∂2

u

∂t ∂x2

Now, we have to find the boundary conditions for ut(x,t)

(i) ut(0,t) = u(0,t)-us(0) = 50-50 = 0

(ii) ut(10,t) = u(10,t)-us(10) = 10-10 = 0 (iii) ut(x,0) = u(x,t)-us(x) = 2x+20+4x-50 = 6x-30


ut(x,t)=(Acospx+ Bsinpx ) tpe

22α−

Apply condition (i), we get


∴A=0

0

0),0(22

22

≠=>

==

−

−

tP

tp

t

e

Aetu

α

α

a1= - 4

us(x)= - 4x+50 12

9

13

15

tp

t pxeBtxu22

sin),( α−= 16

45

Apply (ii), in (16) we get

=> B ≠0 & etP22α−=0

∴ sin 10 p =0

10p =nπ

Sub

In general,

By superposition principles, we get


u=2x-10 v=

u1=2

v1=

010sin),10(22

== − tP

t peBtuα

10πnp =

wegetinnp ),16(10

π=

2

222

10

10sin),(

tn

te

xnBtxu

παπ

−

=

100

222

10sin),(

tn

t exn

Btxu

παπ

−

=

0

222

10

1 10sin),(

tn

n

n

t exn

Btxu

παπ

−∞

=

∑=

10sin306

10sin),(

11

xnbx

xnBtxu

n

n

n

nt

ππ∑∑

∞

=

∞

=

=−==

( )∫ −==10

010

sin30610

2dx

xnxbWhereB nn

π

( ) dxxn

x10

sin1025

310

0

π∫ −=

10

10cosπ

π

n

xn

−

10sin

xn π

17

46

v2 =

Sub Bn in (17), we get

But u(x,t) =us(x) + ut(x,t)

which is the required soln :-

Thermally insulated ends: -

If there will be no heat flow passes through the ends of the bar then that two ends are said

to be thermally insulated.

Note :

If the end, say x=0 is thermally insulated then we have ∂y =0

Bn = -60 (1+(-1)

n

nπ

2

10

10sin

−

π

π

n

xn

( )10

0

22

100

10sin2

10

10cos102

5

3

×+×−−=∴

π

π

π

π

n

xn

n

xnxBn

+−−=

−+

−=

−

π

ππ

n

nX

n

n

n

1)1(

5

300

10)10(

100

53

)1(

100

1

222

10sin)11(

60),(

tn

n

n exn

ntxu

παπ

π

−∞

=

∑ −+−

=

100

1

222

10sin)11(

60504),(

tn

n

n

t exn

nxtxu

παπ

π

−∞

=

∑ −+−

++−==>

47

∂x x=0

∂y =0

and if the end say x=l is thermally insulated then we have ∂x x=0

Problems :

The temperature at one end of a bar 50 cm long with insulated sides is kept at 00C and that at

the other end is kept at 1000C until steady –state conditions prevail. The two ends are then suddenly

insulated, so that the temperature gradient is zero at each end thereafter . Find the temperature

distribution.

Soln :- A 50cm B

When steady state condition, prevails, 00C 100

0C

The heat flow equation becomes

d2u =0 _____________________(1)

dx2

=> u(x) =ax+b


i) u(0) =0

ii) u(50) =100

Apply (i) in (1) , we get

u(0) =b=0 => b=0

sub b=0 in (i) ,we get

u(x) = ax


u(50) = a[50] =100

a=2

2

48

=>

Sub a=2 in (2) , we get

u(x)=2x

∴ In steady state, the temperature distribution

Now , the temperature at A & B are changed.

At this stage, the steady state is changed into unsteady state. From this stage, the initial

temperature distribution is

u(x,0) = 2x

Where unsteady state conditions prevails, the heat flow equation is

∂ u = α2 ∂ 2u

∂ t ∂x2


i) ∂u(0,t) = 0

∂ x ∴the two ends are thermally insulated

ii) ∂u(50,t) = 0

∂ x

iii) u(x,0) =2x, 0<x<50


Diff . Paratially w. r.to. ‘x’ we get

Apply condition (i) , we get

u(x) =2x

∴ B=0

tpepxBpxAtxu22

)sincos(),( α−+=

[ ] tpepxBppxAp

x

txu 22

cossin),( α−+=

∂

∂

3

4

5

6

49

Sub B in (6) , we get

Partially diff w.r.to ‘x’ we get

Apply condition (ii) , we get

∴ sin 50P =0

=> 50 p =nπ

Sub p= nπ in (7) , we get

50


p= nπ

50

0&0),0( 22

≠≠=∂

∂ −pBpe

x

tu tpα

0&022

≠≠=> − pe tpα

tp

pxeAtxu22

cos),(α−=

tppxeA

x

txu 22

sin),( α−−=

∂

∂

0&0

050sin),50(

22

22

≠≠=>

≠−−=∂

∂

−

−

tp

tp

eA

epAx

tu

α

α

2500

222

50cos),(

tn

exn

Atxu

παπ −

=

7

50

Apply condition (iii) , we get

Where A0 = a0 & An=an

2

Now ,

a0 = 100 A0 = 50

u=x v =cos nπx

50

2500

0

222

50cos),(

tn

n

n

exn

Atxu

παπ −∞

=

∑=

xxn

Axu n

n

250

cos)0,(0

== ∑∞

=

π

50cos

22

50cos

1

0

1

0

xna

ax

xnAA

n

n

n

n

ππ∑∑

∞

=

∞

=

+==+

×=

=

= ∫

2

25002

502

2

2

50

2

250

2

502

50

0

0

0

x

xdxa

∫

∫

=

=

50

0

50

0

50cos

504

50

2cos2

502

dxxn

x

dxx

xa n

π

π

50

50sin π

π

n

xn

8

51

u1=1 v1=

v2 =

Sub A0 & An in (8) we, get


TRY YOURSELF:

(9)The temperature at one end of a bar 20cm long & with insulated sides is kept at 00C & that at the

other end is kept at 600C until steady state conditions prevail. The two ends are then suddenly

insulated, so that the temperature is 00C at each end there after. Find the temperature distribution of

the bar.

( )

( ) [ ]

( )1)1(200

1)1(50

504

50)1(

50

504

50

50cos

50

50sin

50

4

22

22

2

22

2

22

2

50

22

2

0

−−=

−−=

−+=

×

+×

=

n

n

n

n

na

nX

nn

n

xn

n

xnx

π

π

ππ

π

π

π

π

2

50

50cos

−π

π

n

xn

( )1)1(200

22−−=∴ n

nn

Aπ

( ) 25001

221

222

50cos)1(

20050),(

tn

n

n

exn

ntxu

παπ

π

−

−+= −

∞

=

∑

52

10) An insulated metal rod of length 100cm has one end ‘A’ kept at 00C & the other end ‘B’ at

1000C until steady state conditions prevail. At time t=0, the end B is suddenly insulated

while the temperature at A is maintained at 00C . Find the temperature any point of the rod

at any subsequent time.

Soln

When steady state conditions prevail in the rod, the temp distribution is given by

d2y = 0

dx2

the corresponding boundary conditions are

u (0) =0

u (100) =0

soling (1), we get u(x) = c1x+c2

using (2) & (3) in (4) ,we get c1=1 & c2=0

∴

One end ‘B’ is insulated, though the temperature at ‘A’ is not altered, the heat flow

under unsteady state conditions & the subsequent temperature distribution is given by


i) u(0,t) =0 for all t ≥ o

ii) ∂u(l,t) =0 for all t ≥ 0

∂x

iii) u(x,0) =x for o<x<l

The correct solution is

u(x) =x

1

2

3

4

22

2

u u

t xα

∂ ∂=

∂ ∂6

2 2

( , ) ( cos sin ) p tu x t A px B px e

α−= +

5

53


=>

∴

sub A=0 in (7) , we get

Partially diff (8) w.r.to ‘x’, we get

Apply condition (ii) (8), we get

∴ cospl =0

=> pl cos-1

0

=an odd multiple of π /2

(or)

(2n-1) π /2

sub l

np 2

)12( π−= in (8), we get

A=0

2 2

(0, ) 0p tu t Ae α−= =2 2

0p te

α− ≠

2 2

( , ) sin p tu x t B pxe α−=8

2 2( , )cos p tu x t

pB pxex

α∂= −

∂2 2( , )

cosp tu x t

pB pxex

α∂= −

∂

2 2

0; 0; 0p tp e B

α=> ≠ − ≠ ≠

,....2,12)12(

=−

= wherenl

np

π

2 2 2(2 1)

242 1)( , ) sin

2

n t

ln xu x t B e

l

α π

π− −

− =

54

the most general solution is


This is of the Fourier fine series form

∫

−==−

l

nn dxl

xnx

lbwhereB

0

122

)12(sin2 π

=

=

= 1)1()12(

8 +−−

n

n

l

π

Sub (10) in (9), we get

2 2 2

2

(2 1)

42 1

1

(2 1)( , ) sin

2

n t

ln

n

n xu x t B e

l

π απ

− −∞

−=

− =

∑ 9

2 1

1

(2 1)( ,0) sin

2n

n

n xu x B x

l

π∞

−=

− = =

∑

l

n

l

l

xn

n

l

l

xnx

l0

2

)12(

2

2

)12sin(

)12(

2

2

)12(cos2

−

−+

−

−−

π

ππ

−+

− 2)12sin(

)12(

222

ππ

nn

l

l

π)12(

)1(8 1

12−

−=

+

−n

lB

n

n10

55

Which is the required solution

PROBLEMS WITH NON ZERO BOUNDARY VALUES & STEADY STATE

CONDITIONS

1. A bar 10 cm long with insulated sides, has its ends A & B kept at 200C & 40

0 C

respy., until steady state. Conditions prevail. The temperature at A is then suddenly

raised to 500C & at the same time that at B is reduced at 10

0C .Find the temperature

at any point of the bar at any time.

200C 40

0C

A B

500C 10

0C

Soln

In steady state conditions,

the heat flow equation

becomes d2y = 0

dx2

=> u(x) =ax +b


i) u(o) =20

ii) u(10) =40

Apply (i) in (1) , we get

u(0)= b= 20 =>

sub b=20 in (1), we get

2 2 2

2

(2 1)1

4

1

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

(2 1) 2

n tn

l

n

l n xu x t e

n l

π απ

π

− −+∞

=

− − =

− ∑

1

b=20

56

Apply (ii) in (2), we get

u( 10) = 10 a + 20 =40

=>a+2 =4

=>

Sub a in (2), we get

In steady state, the temperature distribution is

Now, the temperature at A & B are changed.

At this stage, the steady state is changed into unsteady state.

From this stage, the initial temperature distribution is

in 0< x <10

Hence, when unsteady state condition prevails, the heat flow equation is

10cm


A B

i) u(0,t) =50 500C 10

0C

ii) u(10,t) =10

iii) u(x,0) =2x +20


u(x) = ax +20

a=2

u(x) = 2x + 20

2

3

u(x) =2x +20

u(x,0) =2x +20 4

2

22

t

u

t

u

∂

∂=

∂

∂α 5

tpepxBpxAtxu22

)sincos(),( α−+= 6

57

Apply (i) in (6), we get


From (7) & (8) , it is not possible to find the constants A & B.

In this case, we split the soln u(x,t) into two parts.

ie.,

Where us (x) is a soln of the eqn

& is a function of x alone

[

]0,

0,

2

2

2

2

=

=∂

∂

dx

udie

x

uie

& satisfying the conditions

us(0) =50

& us(10)=10

and ut(x,t) is a transient solution satisfying (9) which decreases as t increases

To find us(x)


(i) us(0) =50

(ii) us(10) =10

Apply (i) in (10), we get

50),(22

== − tpAetou α 7

10)10sin10cos(),10(22

=+= − tpepBpAtu α 8

u(x,t) = us(x) + ut(x,t) 9

us(x)=a1x +b1 10

2

22

x

u

t

u

∂

∂=

∂

∂α

58

us(0) = b1 =50

=>

sub b, in (10), we get

Apply (ii) in (11), we get

us (10) =a1(10) +50 = 10

a1+5 =1

a1 = 1-5

sub a1 in (11) , we get

To find ut (x,t)

9) => ut (x,t) = u(x,t) –us(x)

is a transient solution of 2

22

t

u

t

u

∂

∂=

∂

∂α

Now, we have to find the boundary conditions for ut(x,t)

i) ut (0,t) = u(0,t) – us(0) =50 -50 =0

ii) ut(10,t) = u(10,t) –us(10) =10-10 =0

iii) ut(x,0) = u(x,0) –us(x) =2x +20 +4x -50 = 6x -30


Apply conditions (i), we get

b1=50

us(x) =a1x + 50 11

a1= -4

us (x) = -4x +50 12

13

tp

t epxBpxAtxu22

)sincos(),( α−+= 15

59



Sub

In general,


0

0),(

22

22

≠=>

==

−

−

tp

tp

t

e

Aetou

α

α

∴ A=0

tp

t pxeBtxu22

sin),(α−=

16

π

α

α

np

p

eB

peBtu

tp

tp

t

=

=∴

≠≠=>

==

−

−

10

010sin

0&0

010sin),10(

22

22

10πnp =

wegetinnp ),16(10

π=

2

222

10

10sin),(

tn

t exn

Btxu

παπ

−

=

100

1

222

10sin),(

tn

n

nt exn

Btxu

παπ −∞

=

∑=

16

100

222

10sin),(

tn

nt exn

Btxu

παπ

−

=

60

Apply conditions (iii) in (16), we get

u=2x -10 v= sin

u1 =2 v1 =

v2 =

Sub Bn in (16) , we get

dxxn

xbwhereB

xnbx

xnBxu

nn

n n

nnt

10sin)306(

10

2

10sin306

10sin)0,(

10

0

1 1

π

ππ

∫

∑ ∑

−==

=−==∞

=

∞

=

dxxn

x10

sin)102(10

310

0

π∫ −=

10

xnπ

−

10

10cos

π

π

n

xn

2

10

10sin

−

π

π

n

xn

10

0

22

100

10sin2

10

10cos)102(

5

3

×+×−−=∴

π

π

π

π

n

xn

n

xnxB

n

+−−−=

×−+−−=

π

ππ

n

nn

n

n

1)1(

5

300

10)10()1(

100

53

))1(1(60 n

nn

B −+−

=π

100

1

222

10sin))1(1(

60),(

tn

n

n

te

xn

ntxu

παπ

π

−∞

=

∑ −+−

=∴

61

But u(x,t) =us(x) + ut (x,t)

=>


TWO DIMENSIONAL HEAT FLOW EQUATION

The two dimensional heat flow equation (or) Laplace equation is

=0

The various solutions of equation (1) is

Problems

(1)A square plate is bounded by the lines x=0, y=o, x=20 & y=20. Its faces insulated. The

temperature along the upper horizontal edge is given by u(x,20) =x(20-x)when 0<x<20 while the

other three edges are kept at 00C. Find the steady state temperature in the plate.

Soln

100

1

222

10sin))1(1(

60504),(

tn

n

ne

xn

nxtxu

παπ

π

−∞

=

∑ −+−

++−=∴

1 2

22

t

u

t

u

∂

∂=

∂

∂α

))((),()

))(sincos(),()

)sincos)((),()

1211109

8765

4321

cyccxcyxuiii

ececpxcpxcyxuii

pycpycececyxui

pypy

pxpx

++=

++=

++=

−

−

62

let us take the sides of the plate be l=20 (for convenience)

The temperature distribution is given by

Y


y=l

(i) u(0,y) =0

(ii) u(l,y) =0 x=0 0 x=l

(iii) u(x,0) =0

(iv) u(x,l) =x(l-x), 0<x<l

X

y=0

The correct solution of eqn (1) is

Apply conditions (i) in (2), we get

u(0,y) =c1 ( c3epy

+ c4 e-py

) = 0

=> c3epy

+ c4 e-py ≠ 0

Sub c1=0 in (2) ,we get

Apply condition (ii) in (3), , we get

u(l,y) =(c2 sinpl ) ( c3epy

+ c4 e-py

)=0

=> c2≠0 & c3epy

+ c4 e-py

≠ 0

∴ sinpl =0

pl = sin -1

0

pl =nπ

p =nπ

02

2

2

2

=∂

∂+

∂

∂

y

u

x

u1

u(x,y) =(c1cospx + c2 sinpx ) ( c3epy

+ c4 e-py

) 2

∴ c1=0

u(x,y) =(c2 sinpx ) ( c3epy

+ c4 e-py

)

3

u(x,l)=x(l-x)

0

0 C 0

0 C

63

l

sub p = nπ in (3), we get

l


∴ c3 +c4 =0

=> c3 = -c4

Sub c3 = -c4 in (4), we get

The most general solution is

Apply condition (iv) in (6) ,we get

l

yn

l

yn

ececl

xncyxu

ρππ

−

+

= 432 sin),( 4

0)(sin),( 432 =+

= cc

l

xncyxu

π

0sin&02 ≠

≠=>

l

xnc

π

+−

=

−

l

yn

l

yn

ececl

xncyxu

πππ

442 sin),(

−

−=

−

l

yn

l

yn

eel

xncc

πππ

sin42

−=

l

yn

l

xncc

ππsinh2sin42

−=

l

yn

l

xncc

ππsinhsin2 42

=∴

l

yn

l

xncyxu n

ππsinhsin),(

∑∞

=

=

1

sinhsin),(n

nl

yn

l

xncyxu

ππ6

64

which is of the form of F.S. series

where

ie.,

u= xl-x2

u

1 =l-2x

u11

= -2

)()(sinhsin),(1

xfxlxl

ln

l

xnclxu

n

n =−=

=∑

∞

=

ππ

∑∞

=

=

1

sin)(n

nl

xnbxf

π

∫

−=

=

l

nndx

l

xnxlx

ll

lncb

0

sin)(2sinhππ

( ) ∫

−=

l

ndx

l

xnxxl

lnc

0

2 sin)(2sinhπ

π

33

22

1

cos

sin

cos

sin

=

−=

−

=

=

l

nl

xnv

l

nl

xnv

l

nl

xn

v

l

xnv

π

π

π

π

π

π

π

( ) ( )[ ]l

nn

l

l

xn

n

l

l

xnxl

n

l

l

xnxxl

lnc

033

3

22

22 cos2sin)2(cos

2sinh

π

π

π

π

π

ππ ×

−×

−+×

−−=∴

( )

( )

( )ππ

π

π

ππ

nn

lc

n

l

l

n

l

n

l

l

n

n

n

n

n

sinh

1)1(1

4

)1(14

)1(1ln

4

2)1(22

33

2

33

2

33

3

33

3

33

3

−−=

−−=

−−=

+−−=

65

Sub cn in (6), we get

Replace l by 20, we get

(2) Find the steady state temperature at any point of a square plate whose two adjacent

edges are kept at 00C & the other two edges are kept at the constant temp 100

0C.

Soln

Let the side of the square plate be ‘l’

The temperature u(x,y) is given by

Y=l

x=0 x=l


i) u(x,0) =0 for 0<x<l

ii) u(l,y) =100 for 0<y<l y=0

iii) u(x,l) =100 for 0<x<l

iv) u(0,y) =0 for 0<y<l

let u(x,y) = u1(x,y) + u2(x,y)

where u1(x,y) & u2(x,y) are satisfying the following boundary conditions.

( ) ππ

echnn

lc n

n cos)1(14

33

2

−−=

( )

−−=∑

∞

= l

xn

l

ynechn

n

lyxu

n

n

πππ

πsinsinhcos)1(1

4),(

133

2

−−

=∑

∞

= 20sin

20sinhcos

)1(1)20(4),(

31

3

2 xnynechn

nyxu

n

n

πππ

π

( )

−−= ∑

∞

= l

xn

l

ynechn

nyxu

n

n

πππ

πsinsinhcos

)1(11600),(

31

3

02

2

2

2

=∂

∂+

∂

∂

y

u

x

u

1

1000 C

0

0 C 100

0 C

66

(a1) u1 (0,y) =0 (a2) u2(x,0) =0

(b1) u1(l,y) =0 (b2) u2 (x,l) =0

(c1) u1(x,0) =0 (c2) u2(0,y) =0

(d1) u1(x,l) =100 (d2) u2(l,y) =100

To find u1 (x,y)


u1(x,y) =(c1cospx + c2 sinpx ) ( c3epy

+ c4 e-py

)

Apply condition (a1) in (2), we get

u1(0,y) =c1( c3epy

+ c4 e-py

)=0

=> c3epy

+ c4 e-py

≠ 0


u1(x,y) =(c2 sinpx ) ( c3epy

+ c4 e-py

)

Apply condition (b1) in (3) , we get

u1(l,y) = c2 sinpl ( c3epy

+ c4 e-py

) = 0

=> c2≠0 & c3epy

+ c4 e-py ≠ 0

∴ sinpl =0

pl =sin-1

0=nπ

Sub p=nπ in (3) , we get

l

Apply condition (c1) in (4), we get

∴c1=0

p=nπ

l

2

3

+

=

−

l

yn

l

yn

ececl

xncyxu

πππ

4321 sin),( 4

67

=> c2 ≠ 0 &

l

xnπsin ≠0

∴ c3 +c4 =0

=> c3 = -c4

Sub c3 = -c4 in (4), we get


Apply condition (d1) in (6) , we get

which is of the form of Fourier sine series f(x)

=∑ bn

l

xnπsin

where

0)(sin),( 4321 =+

= cc

l

xncoxu

π

+−

=

−

l

yn

l

yn

ececl

xncyxu

πππ

4421 sin),(

−

=

−l

yn

l

yn

eel

xncc

πππ

sin42

−=

l

yn

l

xncc

ππsinh2sin42

−=

l

yn

l

xncc

ππsinh2sin2 42

=

l

yn

l

xncn

ππsinhsin

∑∞

=

=

1

1 sinhsin),(n

nl

yn

l

xncyxu

ππ

6

100)(sinhsin),(1

=

=∑

∞

=

ππ

nl

xnclxu

n

n

∫

==

l

nn dxl

xn

lncb

0

sin1002sinhπ

π

68

cn sinhnπ =


To find u2 (x,y)


u2(x,y) =(c1cospy + c2 sinpy ) ( c3epx

+ c4 e-px

)

Applying the boundary conditions (a2), (b2), (c2), & (d2) we get

∴ The solution of (1) is

u(x,y) =u1 (x,y) + u2(x,y)

( )

))1(1(200

)1(200

cos200

0

n

n

l

nl

l

nl

nl

l

n

l

l

xn

l

−−×=

+×−−=

×−=

π

ππ

π

π

( )( )n

n11

200−−

π

ππ

echnn

c n

n cos)1(1(200

−−==>

( )

−−= ∑

∞

= l

yn

l

ynechn

nyxu

n

n πππ

πsinhsinhcos

)1(1200),(

1

1

( )

−−= ∑

∞

= l

xn

l

ynechn

nyxu

n

n πππ

πsinhsinhcos

)1(1200),(

1

2

69

Try yourself:

1) A rectangular plate is bounded by the lines x=0, x=a, y=0 & y=b, the edge temperature are

u(0,y)=0, u(a,y) =0, u(x,b) =0 , u(x,0) =sin3

πx

a

find the temperature distribution.

2) A rectangular plate is bounded by the lines x=0, x=a, y=0 & y=b & the temperatures at the

edges are given by.

u(0,y) = y in 0<y<b/2

; u(a,y) =0 ; u(x,b) =0 &

b-y in b/2 <y<b

. Find the steady state temperature distribution inside the

plate.

Infinite plates

1) An infinitely long rectangular plate has its surfaces insulated & the two sides as well as one

of the short sides are maintained at 00C. find an expression for the steady state temp u(x,y) if the short

side y=0 is π cm long & is kept at u00C.

Soln :-



i) u(0,y) =0

ii) u(π,y) =0 x=0

iii) u(π,∞)=0

iv) u(x,o) =u0

The correct soln is

u(x,y) =(c1cospx + c2 sinpx ) ( c3epx

+ c4 e-px

)

00 C

00 C 0

0C

u00 C

( )

+

−−= ∑

∞

= l

xn

l

yn

l

yn

l

xnechn

nn

n πππππ

πsinhsinsinhsinhcos

)1(1200

1

+

=

a

x

a

xxu

ππ 3sin3

4sin5)0,(

02

2

2

2

=∂

∂+

∂

∂

y

u

x

u 1

2

y=0

x=π

y=∞

70

Apply condition (i) in (2), we get

u(0,y) =c1( c3epy

+ c4 e-py

)=0

Here c3epy

+ c4 e-py ≠ 0



u (π,y) = c2 sinpπ ( c3epy

+ c4 e-py

) =0

Here c3epy

+ c4 e-py≠ 0 & c2 ≠ 0

∴ sinpπ =0

pπ =sin-1

0

pπ =nπ

=>

sub p=n in (3), we get


u(x,∞) = c2 sinnx ( c3e∞)=0

=> e∞ ≠ 0 ; sinnx ≠ 0 & c2 ≠ 0

∴c3 =0


u(x,y) = c2 sinnx c4 e-ny

u(x,y) = c2 c4 e-ny

sinnx

u(x,y) = cn e-ny

sinnx

p=n

∴c1=0

u(x,y) = c2 sinpx ( c3epy

+ c4 e-py

)

3

u(x,y) = c2 sinnx ( c3eny

+ c4 e-ny

)

4

71


Apply condition (iv) in (5), we get

This is of the form of Fourier sine series

where

Sub cn in (5), we get

nxecyxuny

n

n sin),(1

−∞

=

∑= 5

0

1

sin),( unxecoxuny

n

n == −∞

=

∑

nxbxfn

n sin)(1

∑∞

=

=

nxdxxfcb nn sin)(2

0

∫==π

π

nxdxu sin2

0

0∫=π

π

−−=

+

−−=

−=

n

uc

nn

u

n

nxu

n

on

n

o

o

)1(12

1)1(2

cos2

0

π

π

π

π

nyn

n

nxen

uyxu −

∞

=

−−=∑ sin

)1(12),(

1

0

π

nyn

n

nxen

uyxu −

∞

=

−−= ∑ sin

)1(12),(

1

0

π

72

2) An infinitely long rectangular plate with insulated surface is 10 cm.The two long edges & one

short edge are kept at 00C. while the other edge x=0 is kept

at u= 20 y , for 0 < y< 5

20 (10-y) , for 5<y<10

Find the steady state temperature distribution in the plate.

Soln



i) u(x,0) =0 x=0

ii) u(x,l) =0

iii) u( ∞ , y) =0

iv) u(0,y) = 2ly 0< y< l/2

2l(l-y) l/2 <y<l

(let l=10 for the convenience)


Apply condition (i) in (2), we get

u(x,o) = c3( c1epx

+ c2e-px

) =0

=>

sub c3 =0 in (2), we get

00 C

f(y) 00C

O C

c3=0

02

2

2

2

=∂

∂+

∂

∂

y

u

x

u1

u(x,y) = ( c1epx

+ c2e-px

) (c3cospy + c4 sinpy )

u(x,y) = ( c1epx

+ c2e-px

) c4 sinpy

3

x=∞

y=0

y=l

73


u(x,l) = ( c1epx

+ c2e-px

) c4 sinpl =0

Here c4 ≠ 0 ∴ sinpl =0 ie, pl =nπ

=>

Sub p=nπ in (3), we get

l


u(∞,y) = ( c1e∞ + c2e

-∞) c4 sin

l

ynπ=0

=>


The most general soln is

∴c1=0

p=nπ

l

l

yncececyxu l

xn

l

xnπ

ππ

sin),( 421

+=

−

4

0sin41 =∞

l

yncec

π

0sin&00 4 ≠≠≠∞

l

ynce

π

=

−

l

yncecyxu l

xnπ

π

sin),(42

=

=

−

−

l

yncec

l

yncecc

l

xn

n

l

xn

π

π

π

π

sin

sin

4

442

74

Apply condition (iv) in (5), we get

where f(y) = 2ly , l < y<l/2

2l(l-2), l/2< y<l

This is form of the F.S.Series

=

−∞

=

∑l

ynecyxu

l

xn

n

n

ππ

sin),(1

5

∑∞

=

=

=

1

sin),0(n

n flyl

yncyu

π

∑∞

=

=

1

sin)(n

nl

ynbyf

π

∫

==

l

nndy

l

ynyf

lcb

0

sin)(2 π

( )

}2

sin2

cos22

sin2

cos2

4

sincos)

sincos

4

sin)(2sin22

22

22

22

22

2

2

2

0

2

2

0

`

2

π

π

π

π

π

π

π

π

ππ

ππ

π

π

π

π

ππ

n

n

ln

n

ln

n

ln

n

l

nl

l

yn

nl

l

ynyl

n

l

l

yn

n

l

l

yny

dyl

ynylldy

l

ynly

l

l

l

l

ll

l

+++ −

=

×

−×

−+

+×

−

=

−+

= ∫ ∫

=

2sin

24

22

2 π

π

n

n

l

75

Sub (6) in (5), we get

Replacing l by 10, we get

which is the required solution.

Objective Questions

1) The suitable solution of ODWE is ---------------------------

a) y(x,t) = (A epx

+ B e-px

) (C epax

+ D e-pax

)

b) y(x,t) = (A epx

+ B e-px

) (C cos pat + D sin pat)

c) y(x,t) = (A cos pax + B sin pax) (C cos pat + D sin pat)

d) y(x,t) = (A x+B) (Ct+D)

2) In one dimensional wave equation utt = a2

u xx , a2 stands for -------------

(a) (b) (c) 0 (d) None of these [T-Tension,m-Mass per unit length]

3) In ODWE mass of the string per unit length is a ------------------

(a)0 (b) 1 (c) Constant (d)Variable

4) Among the three possible solution in ODWE , the suitable solution is taken because

(a) Displacement is periodic in nature

(b) It contains trigonometric terms

(c) Both (a) & (b)

(d) None of these

2sin

822

2 π

π

n

n

lcn =

6

( )

=

−∞

=

∑l

ynen

n

lyxu l

xn

n

πππ

π

sin2

sin8

),(1

22

2

( )

=

−∞

=

∑10

sin2

sin1800

),( 10

122

ynen

nyxu

xn

n

πππ

π

76

5) Which boundary condition in ODWE should be taken last one and used after finding most

general solution

a. boundary condition with zero value

b. boundary condition with non zero value

c. both (a) and (b)

d. none of these

6) The steady state temperature of a rod of length ‘ℓ’ whose ends are kept at 30 and 40 is ----------

-------------------

+30 (b) +30 (c) +10 (d) +10

7) The general solution y(x,t) of vibratory motion of a string of length ‘ℓ’ with fixed end points

and zero initial velocity is

(a) y(x,t) = ∑ bn sin (nлx/ ℓ) sin (nлat/ ℓ) (b) y(x,t) = ∑ bn cos (nлx/ ℓ) cos (nлat/ ℓ)

(c) y(x,t) = ∑ bn cos (nлx/ ℓ) sin (nлat/ ℓ) (d) y(x,t) = ∑ bn sin (nлx/ ℓ) cos (nлat/ ℓ)

8) Two dimensional steady state heat conduction equation is --------------

(a) Utt+Uxx=0 (b)Utt=a2Uxx (c)Ut=a

2Uxx (d) None of these.

9) The quantity of heat required to produce a given temperature change in a body is proportional

to the mass of the body and to the ----- Change

(a)Temperature (b)Distance (c)Time (d)None of these

10) In one dimensional heat flow equation , if the temperature function u is independent of time ,

then the solution is ------------------

(a)Constant (b)Steady (c) Unsteady (d)None of These

11) One of the possible solution to two dimensional heat flow equation in Cartesian system of

coordinates is

(a)u(x,y) = (C1 x + C2) (C3 y + C4) (b) u(x,y) = (C1 + C2) (C3 y + C4)

(c)u(x,y) = (C1 x + C2) (C3 + C4) (d)None of these

12) In two dimensional heat flow the rate of heat flow across an area is proportional to -------------

-------------

(a) Area and the temperature gradient parallel to the area

(b) Area and the temperature normal to the area

(c) Area and the temperature gradient normal to the area

(d) Area and the temperature parallel to the area

13) In 2D heat flow, the temperature at any point is independent of

(a) Time (b) X-Coordinate (c) Y-Coordinate (d) Z-Coordinate

77

14) In one dimensional heat flow equation ut = α2 uxx , α

2 stands for

(a) Diffusivity of the material

(b) Thermal conductivity of the material

(c) Specific heat of the material

(d) Density of the material

(15) When the ends of a rod in non zero for one dimensional heat flow equation. The temperature

function u(x,t) is modified as the sum of the steady state and transient state temperature. The transient

part of the solution which

(a) Increase with increase of time

(b) Decrease with increase of time

(c) Decrease with decrease of time

(d) Increase with increase of time

16)The One Dimensional Wave Equation is

(a)Utt+Uxx=0 (b)Utt=a2Uxx (c)Ut=a

2Uxx (d)None of these.

17)The One Dimensional Heat Equation is

(a)Utt+Uxx=0 (b)Utt=a2Uxx (c)Ut=a

2Uxx (d)None of these.

18)A Second order Partial Differential equation is said to be elliptic if

(a)B2-4AC<0 (b)B

2-4AC=0 (c) B

2-4AC>0 (d)None of these

19) A Second order Partial Differential equation is said to be hyperbolic if

(a)B2-4AC<0 (b)B

2-4AC=0 (c)B


20) A Second order Partial Differential equation is said to be parabolic if

(a)B2-4AC<0 (b)B

2-4AC=0 (c)B


21)In deriving One Dimensional Wave Equation we assume that motion takes place entirely in ------

plane.

(a)One (b)two (c)three (d)None of these

22) In deriving One Dimensional Wave Equation we assume that the Tension T is ------------ at all

points & at all time of the deflected string.

(a)Constant (b)Varies (c)Zero (d)none of these.

23) In deriving One Dimensional Heat Equation we assume that heat flows from --------- to ------------

-- temperature.

(a)High to low (b)low to high (c)high to high (d)None of these.

24)We require ----------- number of boundary conditions to solve One Dimensional Heat Equation

(a)1 (b)2 (c)3 (d)4

25) We require ----------- number of boundary conditions to solve One Dimensional wave Equation

(a)1 (b)2 (c)3 (d)4

78

26)The suitable solution of the One Dimensional Heat Equation is

(a)U(x,t)=(Acos px + B sin px)

(b) U(x,t)=(Acos px + B sin px)

(c)U(x,t)=(Acos px + B sin px)

(d)None of these.

PART-A

1. Classify the partial differential equation

(x + 1)zxx + p2(x + y + 1)zxy + (y + 1)zyy + yzx ¡ xzy + 2 sin x = 0:

2. Write down all possible solutions of one dimensional wave equation.

3. A taut string of length 50 cm fastened at both ends, is disturbed from its position of equilibrium

by imparting to each of its points an initial velocity of magnitude kx for 0 < x < 50. Formulate

the problem mathematically.

4.Write down all possible solutions of the one dimensional heat °ow equation.

5. If the temperature at one end of a bar, 50 cm long and with insulated sides, is kept at 0±C and

that the other end is kept at 100±C until steady state conditions prevail, ¯ nd the steady state

temperature in the rod.

PART-B

1. A taut string of length ‘L’ is fastened at both ends. The midpoint of this string is taken to a

height of ‘b’ and then released from rest in this position. Find the displacement of the string at any

time.

2. A taut string of length 2l is fastened at both ends. The midpoint of the string is taken to a

height b and then released from rest in that position. Find the displacement of any point of

the string at any subsequent time.

3. A tightly stretched string with fixed end points x=0 and x= l is initially at rest in its equilibrium

position. If it is set vibrating by giving each point a velocity ( ).x l xλ − find the displacement of

the string at any distance x from one end at any time t.

4. The end A and B of a rod l cm long have the temperatures400c and 90

0c a until steady state

prevails. The temperature at A is suddenly raised to 900c and at the same time that at B is

lowered to 400c.Find the temperature distribution in the rod at time ‘t’.Also show that the

79

temperature at the mid point of the rod remains unaltered for all time, regardless of the

materials of the rod.

5. A metal bar 10 cm long with insulated sides has its ends A and B kept at 20 co and 40 co

respectively until steady state conditions prevail. The temperature at A is then suddenly raised

to 50 co and at the same instant that at B is lowered to 10 co . Find the subsequent temperature

at any point at the bar at any time.

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

that it may be considered infinite length. If the temperature along short edge y=0 is given by

u(x, 0) =8 sin (10

xΠ) where 0<x<10,while the two long edges x=0 and x=10 as well as the

other short edge are kept at 0 c◦.Find the steady state temperature function u(x,y).

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