2 wave equation

7/30/2019 2 Wave Equation

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Classification of partial differential equations of second order:

In the fields of wave propagation, heat conduction, vibrations, elasticity, boundary layer

theory, etc., second order partial differential equations are of particular interest.

The general form of a second order P.D.E. in the function u of the two independent

variables x, y is given by

( ) ( ) ( ) 0y

u,

x

u,y,xf

y

uy,xC

yx

uy,xB

x

uy,xA

2

22

2

2

=

+

+

+

. (i)

This equation is linear in second order terms. PDE (i) is said to be linear or quasi-linear

according as f is linear or non-linear.

PDE (i) is classified as Elliptic, Parabolic or Hyperbolic.

If 0AC4B2

, then PDE (i) is classified as hyperbolic.

2222ndndndnd TopicTopicTopicTopic

Applications ofApplications ofApplications ofApplications of

Partial Differential EquationsPartial Differential EquationsPartial Differential EquationsPartial Differential Equations

Wave equation

Vibration of a stretching string, solution of wave equation,

DAlemberts solution of wave equationPrepared by:

Dr. SunilNIT Hamirpur (HP)

(Last updated on 25-09-2007)


2/34

Applications of Partial Differential Equations: Wave Equation

Prepared by: Dr. Sunil, NIT Hamirpur (HP)

2

Examples:

Elliptic: 0AC4B2 :

One-dimensional wave equation:2

22

2

2

x

ua

t

u

=

.

Note:Laplaces equation, one dimensional heat-flow equation and one-dimensional

wave equation are homogeneous, whereas Poissons equation is non-

homogeneous.

Vibrations of a stretched string, One-dimensional wave equation:

The classical one-dimensional wave equation, which is hyperbolic, arises

in the study of transverse vibrations of an elastic (flexible) string or torsional

oscillations or longitudinal vibrations of a rod.

Vibrating string:

Consider a uniform elastic string of length l stretched tightly between two points O and

A, and displayed slightly its equilibrium position OA. Taking the end O as origin, OA as x-axis

and perpendicular line through O as the y-axis, we shall find the displacement y as a function

of the distance x and time t.

2

22

2

2

x

yc

t

y

=


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3

We shall obtain the equation of motion for the string under the following assumptions:

1. The motion takes place entirely in the xy-plane and each particle of the string moves

perpendicular to the equilibrium position OA of the string

2. The string is perfectly flexible and offers no resistance to bending.

3. The tension in the string is so large that the forces due to weight of the string can be

neglected.

4. The displacement y and the slopex

y

are small, so that their higher powers can be

neglected.

Let m be the mass per unit length of the string. Consider the motion of an element PQ

of length s . Since the string does not offer resistance to bending (by assumption), the tensions

T1 and T2 at P and Q, respectively are tangential to the curve.

Since there is no motion in the horizontal direction, we have

TcosTcosT 21 == (constant). (i)

Mass of element PQ is sm . Then, by Newtons second law of motion, the equation of

motion in the vertical direction is

=

sinTsinT

t

ysm 122

2

=

cosT

sinT

cosT

sinT

t

y

T

sm

1

1

2

2

2

2

[by using (i)]

( )

=

tantan

sm

T

t

y

2

2

=

+ xxx2

2

x

y

x

y

sm

T

t

y.

[Since xs = to a first approximation, and tan and tan are the slopes of the curve of the

string at x and xx + ]

P

x-axis

y-axis

O

Q

A

T1

T2

x

y

x

x+x

s


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4

2

2xxx

2

2

x

y

m

T

x

x

y

x

y

m

T

t

y

=

=

+ , as 0x

2

22

2

2

x

yc

t

y

=

, where

m

Tc

2 = .

Thus,

.

This is the partial differential equation giving the transverse vibrations of the string.

It is also called the one-dimensional wave equation or vibrations of a stretched string and y

y (x, y) is called displacement function.

Boundary conditions:The boundary conditions, satisfied by the equation

2

22

2

2

x

yc

t

y

=

are:

==

==

lwhen x0y(ii).

0when x0y).i(. This means that

( )

( )

=

=

0t,y

0t,0y

l.

These should be satisfied for every value of t.

Initial conditions:

If the string is made to vibrate by pulling it into a curve y =f(x) and then releasing it, the initial

conditions are:

(i). ( )xfy = , when t = 0,

(ii). 0t

y=

when t = 0.

Solution of the one-dimensional wave equation by separation of variables:The wave equation is

2

22

2

2

x

yc

t

y

=

. (i)

Assume that y is separable, therefore let ( ) ( )tTxXy = , (ii)

be a solution of (i).

Then TXt

y

2

2

=

and TX

x

y

2

2

=

.

2

22

2

2

x

yc

t

y

=


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5

Substituting in (i), we have TXcTX 2 = .

Separating the variables, we getT

T.

c

1

X

X

2

=

. (iii)

Now the LHS of (iii) is a function of x only and RHS is a function of t only.

Since x and t are independent variables, this equation can hold only when both sides reduce to

a constant say, k. Then, equation (iii) leads to the ordinary linear differential equations

0kXX = and 0TkcT 2 = . (iv)

Solving (iv), we get

(i).When k is positive and = p2, (say)

px2

px1 ececX

+= and cpt4cpt

3 ececT+= .

(ii).When k is negative and 2p= , (say)

pxsincpxcoscX 21 += and cptsinccptcoscT 43 += .

(iii).When k = 0,

21 cxcX += and 43 ctcT += .

Thus, the various solutions of the wave equation (i) are:

cpt4cpt

3px

2px

1 ececececy ++= ,

( )( )cptsinccptcoscpxsincpxcoscy 4321 ++= , ( )( )4321 ctccxcy ++= .

Of these three solutions, we have to choose that solution which is consistent with the physical

nature of the problem.

Since, we are dealing with a problem on vibrations, y most be a periodic function of x and t.

Therefore, the solution must involve trigonometric terms.

Accordingly, ( )( )cptsinccptcoscpxsincpxcoscy 4321 ++= , (v)

is only suitable solution of the wave equation and it corresponds to 2pk = .

Case 1: If boundary conditions are given:

Now applying boundary conditions that

y = 0 when x = 0 and y = 0 when l=x

Therefore ( )cptsinccptcoscc0 431 += (vi)


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6

and ( )( )cptsinccptcoscpsincpcosc0 4321 ++= ll . (vii)

From (vi), we have 0c1 = .

Then, equation (vii) reduce to ( ) 0cptsinccptcoscpsinc 432 =+l ,

which is satisfied whenl

ll === npnp0psin , where n = 1, 2, 3,

A solution of the wave equation satisfying the boundary condition is

lll

xnsin

ctnsinc

ctncosccy 432

+

= .

On replacing c2c3 by an and c2c4 by bn, we getlll

xnsin

ctnsinb

ctncosay nn

+

= .

Adding up the solutions for different values of n, we get

lll

xnsin

ctnsinb

ctncosay nn

1n

+

=

=

, (viii)

is also a solution.

Case 2: If boundary conditions and initial conditions are given:

Now applying the initial conditions.

y = f(x) and 0t

y=

, when t = 0, then from (viii), we have

( )l

xnsinaxf n

1n

=

=

(ix)

andll

xnsinb

cn0 n

1n

=

=

. (x)

Since equation (ix) represents Fourier series for f(x), we have

( ) dxxn

sinxf2

a

0

nll

l

= . (xi)

From (x), we get bn = 0, for all n.

Hence, (viii) reduces to

=

=ll

xnsin

ctncosay n

1n

, (xii)


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7

where ( ) dxxn

sinxf2

a

0

nll

l

= and ( ) ( )xf0,xy = .

DAlemberts solution of the one-dimension wave equation:

The wave equation is2

22

2

2

x

yc

t

y

=

. (i)

Let us introduce new independent variables ctxv,ctxu =+= , so that y becomes a

function of u and v. Thenv

y

u

y

x

y

+

=

and2

22

2

2

2

2

v

y

vu

y2

u

y

v

y

u

y

vv

y

u

y

uv

y

u

y

xx

y

+

+

=

+

+

+

=

+

=

Similarly,

+

=

2

22

2

22

2

2

v

y

vu

y2

u

yc

t

y.

Substituting in (i), we get 0vu

y2=

. (ii)

Integrating (ii) w.r.t. v, we get )u(fu

y=

, (iii)

where f(u) is an arbitrary function of u. Since the integral is a function of u alone, we may

denote it by ( )u . Thus

( ) ( )vuy += ( ) ( )ctxctx)t,x(y ++= . (iv)

This is the general solution of (i).

Now to determine and , suppose u(x, 0) = f(x) and( )

0t

0,xy=

.

Differentiating (iv) w.r.t. t, we get ( ) ( )ctxcctxct

y+=

At t = 0, ( ) ( )xx = (v)

and ( ) ( ) ( ) ( )xfxx0,xy =+= (vi)

(v) gives, ( ) ( ) kxx +=

(vi) becomes ( ) ( ) ( )xfkx20,xy =+=


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8

( ) ( )[ ]kxf2

1x = and ( ) ( )[ ]kxf

2

1x += .

Hence the solution of (iv) takes the form

( ) ( )[ ] ( )[ ] ( ) ( )ctxfctxfkctxf2

1kctxf

2

1t,xy ++=+++= , (vii)

which is DAlemberts solution of the wave equation (i).

**********************

FINAL CONCLUSIONS

1. The solution of one-dimensional wave equation2

22

2

2

x

yc

t

y

=

satisfying the boundary conditions

( )

( )

=

=

0t,y

0t,0y

l is given by

( )lll

xnsin

ctnsinb

ctncosat,xyy nn

1n

+

==

=

.

2. The solution of one-dimensional wave equation2

22

2

2

x

yc

t

y

=

satisfying the boundary conditions( )

( )

=

=

0t,y

0t,0y

l

and the initial conditions

( ) ( )

=

=

=0t

y

xf0,xy

0t

is

given by

( )

==

=ll

xnsin

ctncosat,xyy n

1n

,

where ( ) dxxn

sinxf2

a

0

nll

l

= and ( ) ( )xf0,xy = .

*****************************************

Now let us solve some problems related to one-dimensional wave equation:


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9

Q.No.1.: A string is stretched and fastened to two points l apart. Motion is started by

displacing the string in the forml

xsinay

= from which it is released at time t = 0.

Show that the displacement of any point at a distance x from one end at time t is

given by ( )ll

ctcos

xsinat,xy

= .

Sol.: The given is the problem of vibration of a stretched string, therefore we find the solution

of2

22

2

2

x

yc

t

y

=

. (i)

Boundary conditions: As the end points of the string are fixed, for all time, therefore

the boundary conditions are ( ) ( ) 0t,yt,0y == l . (ii)

Initial conditions: Since the initial transverse velocity of any point of the string is zero,

therefore, the initial conditions are

( )l

xsina0,xy

= and 0

t

y=

, when t = 0. (iii)

Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii).

Then ( )ll

xnsin

ctncosat,xy n

1n

=

=

, where ( ) dxxn

sin0,xy2

a

0

nll

l

=

dxxnsinxsina20 lll

l

= dxxnsinxsina2 0 lll

l = ,

which vanishes for all values of n except n = 1.

dxx

sina2

a2

01 ll

l = dx

x2cos1

a

0

=

ll

la

x2sin

2x

a

0

=

=

l

l

l

l.

Hence, the required solution is ( )ll

ctcos

xsinat,xy

= .

Q.No.2.: The points of trisection of a string are pulled aside through the same distance on

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

an expression for the displacement of the string at subsequent time and show that the

mid-point of the string always remains at rest.

Sol.: Let B and C be the points of trisection of the string OA of length l , say. Initially the

string is held in the form ACBO , where aCCBB == (say)


10/34



10

The equation of BO is xa3

yl

= .

The equation of CB is

=

3x

3

a2ay

l

li.e. ( )x2

a3y = l

l.

The equation of AC is ( )ll = xa3

y .

Here the boundary conditions are ( ) ( ) 0t,yt,0y == l .

And the initial conditions are

( ) ( )

( )

x3

2

,x

a3

3

2x

3

1,x2

a3

3

1x0,x

a3

t,0y

=

ll

ll

ll

l

l

and 0t

y=

when t = 0.

We have ( )ll

xnsin

ctncosat,xy n

1n

=

=

,

where ( ) dxxn

sin0,xy2

a0n ll

l =

( ) ( )

+

+

= dx

xnsinx

a3dx

xnsinx2

a3dx

xnsin

ax32

3/2

3/2

3/

3/

0 ll

lll

llll

l

l

l

l

l

=

3/

0

22

2

2

xnsin

n.1

xncos

nx

a6l

l

l

l

l

l

( ) ( )

3/2

3/22

2 xnsin

n.2

xncos

nx2

l

ll

l

l

ll

+

x-axis

y-axis

OB

C A

a

a

a,

3Bl

a,

3

2Cl

( )0,l


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11

( )

+

l

ll

l

l

ll

3/2

22

2 xnsin

n.1

xncos

nx

+

+

+= 3

ncosn33

n2sinn2

3n2cos

n33nsin

n3ncos

n3a6

2

22

22

22

22

2

lllll

l

+

+

3

n2sin

n3

n2cos

n33

nsin

n

2

22

22

22

2lll

( )[ ]n2222

2

211

3

nsin

n

a18

3

n2sin

3

nsin

n

3.

a6+

=

=

l

l

( ) ( )

=

=

=

=

3

nsin1

3

nsin10

3

nsinncos

3

ncosnsin

3

nnsin

3

2nsinSince

nn

0a n = , when n is odd.

3

nsin

n

a36

22

= , when n is even. i.e.

3

m2sin

m

a9

22

, taking n = 2m

Hence, ( )ll

xm2sin

ctm2cos

3

m2sin

m

1a9t,xy

21m

2

=

=

.

Also 0msinctm2

cos

3

m2sin

m

1a9t,

2

y

21m2

=

=

= l

l, since 0msin = .

The displacement of the mid-point of the string is zero for all values of t.

Thus, the mid-point of the string is always at rest.

Q.No.3.: A tightly stretched string with fixed end points x = 0 and l=x is initially at rest in

its equilibrium position. If it is set vibrating by giving to each of its points a velocity

( )xx l , find the displacement of the string at any distance x from one end at any

time t.

Sol.: Here the boundary conditions are ( ) ( ) 0t,yt,0y == l .

( )lll

xnsin

ctnsinb

ctncosat,xy nn

1n

+

=

=

. (i)

Since the string was at rest initially, y(x, 0) = 0. (Initial condition)


12/34



12

From (i), we have 0axn

sina0 nn1n

=

=

=l

.

Thus, ( )ll

xnsin

ctnsinbt,xy n

0n

=

=

(ii)

andllllll

xnsin

ctncosnb

cxnsin

ctncosb

cn

t

yn

1n

n

1n

=

=

=

=

But ( )xxt

y=

l when t = 0. (Initial condition)

( )llll

lxn

sinnbcxn

sinnbc

xx n1n

n

1n

=

=

=

=

,

where ( ) dxxnsinxx2nbc0n l

lll

l =

( ) ( ) ( )

l

l

l

l

ll

l

ll

l0

33

3

22

2 xncos

n2

xnsin

nx2

xncos

nxx

2

+

=

( ) ( )[ ]n33

2

33

2

11n

4ncos1

n

4

=

=

ll

=oddisnwhen,

n

8

evenisnwhen,0

33

2l

=( )33

2

1m2

8

l, taking 1m2n = .

( )44

3

n1m2c

8b

=

l.

From ( )ll

xnsin

ctnsinbt,xy n

0n

=

=

, the required general solution is

( ) ( ) ( ) ( )ll

l

x1m2sinct1m2sin1m2

1c8t,xy 4

1m4

3

=

=. Ans.

Q.No.4.: A tightly stretched string of length l with fixed ends is initially in equilibrium

position. It is set vibrating by giving each point a velocityl

xsinv 30

. Find the

displacement y(x, t).


13/34



13

or

A string of length l is initially at rest in equilibrium position and each of its points is

given the velocityl

xsinb

t

y 3

0t

=

=

. Find the displacement y(x, t).

Sol.: The given is the problem of a vibrating stretched string. So consider one-dimensional

wave equation

2

22

2

2

x

yc

t

y

=

, (i)

where( )

( )

=

=

0t,y

0t,0y

l. Boundary conditions (ii)

and

( )

=

=

= l

xsinb

t

y

00,xy

3

0t

. Initial condition (iii)

Now, we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii).

Now the solution of (i) satisfying (ii) is given by

( )lll

xnsin

ctnsinb

ctncosat,xy nn

1n

+

=

=

. (iv)

Now ( ) ll

xn

sin

ctn

cosat,xy n1n

=

=[from (iii)]

n0a n = .

Hence, (iv) reduces to ( )ll

xnsin

ctnsinbt,xy n

1n

=

=

. (v)

To find bn: Differentiating (v) w.r.t. t, partially, we have

lll

xnsin

ctncos.

cn.b

t

yn

1n

=

=lll

xsinb

xnsin.

cn.b

t

y 3n

1n0t

=

=

==

[from (iii)]

( )ll

l xnsinnb

xsin

c

bn

1n

3 =

=

,

which is a half-range sine series in ( )l,0 , hence


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14

dxxn

sinx

sin.c

b2nb 3

0

nll

l

l

l

= dx

xnsin

4

x3sin

xsin3

cn

b2b

0

nl

lll

=

[ Asin4Asin3A3sin

3

=Q

where we take l

x

A

= ]

dxx3

sinxn

sinx

sinxn

sin3cn2

bb

0

n

=

llll

l

(vi)

dxx3

sinxn

sin2cn4

bdx

xsin

xnsin2

cn4

b3

00llll

ll

=

( ) ( ) dxx

1ncosx

1ncos

cn4

b3

0

+

= ll

l

( ) ( ) dxx

3ncosx

3ncos

cn4

b

0

+

ll

l

( ) ( )[ ]BAcosBAcosBsinAsin2 +=Q

( )

( )

( )

( )

l

l

l

l

l

0

n

1n

x1nsin

1n

x1nsin

cn4

b3b

+

+

=

( )

( )

( )

( )

l

l

l

l

l

0

3n

x3nsin

3n

x3nsin

cn4

b

+

+

n0= except n = 1, n = 3. ( Zn0nsin =

Also n varies fro 1 to 3,0n .

Now for n = 1,

dxx3

sinx

sinx

sin3c2

bb 2

0

1

=

lll

l

dxsinx3

sin22

1

2

x2cos1

3c2

b

0

=

ll

ll

+

= dx

x4cos

x2cos

x2cos33

c4

b

0lll

l


+

=

l

l

l

l

l

l

4

x4sin

2

x2sin

2

x2sin

3x3c4

b[ ]03

c4

b

= l [ ]Zn0nsin ==


15/34



15

c4

b3

=

l.

Again from (vi), we have

dxx3

sinx

sinx3

sin3c6

bb

2

0

3

= lll

l

dxx3

sin2x

sinx3

sin23c12

b 2

0

= lll

l

= dx

x6cos1

x4cos

x2cos3

c12

b

0lll

l


l

l

l

l

l

l

l

06

x6sin

x

4

x4sin3

2

x2sin3

c12

b

+

=

c12

b

=

l. [ ]Zn0nsin ==

Hence, from (v), the required solution is given by

( )ll

xnsin

ctnsinbt,xy n

1n

=

=

........00x3

sinct3

sinbx

sinct

sinb 31 +++

+

=llll

ll

l

ll

l x3sin

ct3sin

c12

bxsin

ctsin

c4

b3

+

= [ n0bn = except n = 1, 3]

= llll

l x3sin

ct3sin

xsin

ctsin9c12

b. Ans.

Q.No.5.:Find the deflection y(x, t) of the vibrating string of length and ends fixed,

corresponding to zero initial velocity and initial deflection ( ) ( )x2sinxsinkxf = ,

given 1c2 =

Sol.: We know that the partial differential equation of the vibrating string is giving by

2

2

2

22

2

2

x

y

x

yc

t

y

=

=

,


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16

Now, we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii).

The required solution of (i) satisfying (ii) and 0t

y

0t

=

=

is given by

( ) ==

=xnsintncosat,xyy n

1n

nxsinntcosa n1n

== . (iv)

Now using ( ) ( )x2sinxsink0,xy = , (iv) gives

( )x2sinxsinknxsina n1n

=

=

,

which is half-range Fourier sine series in ( ),0 and hence,

( ) nxdxsinx2sinxsink2

a

0n

=

(v)

( ) ( )[ ]dxnxsinx2sin2nxsinxsin2k

0

=

( ) ( )( ) ( ) ( )[ ]x2ncosx2ncosx1ncosx1ncosk

0

+++

=

[ ])BAcos()BAcos(BsinAsin2 =Q

( ) ( ) ( ) ( )

+

++

+

+

=

02n

x2nsin

2n

x2nsin

1n

x1nsin

1n

x1nsink [ ]Zn0nsin =Q

n0= except n = 1, 2 (As n varies from 1 to 2n )

Also n = 1, (v) gives

( ) xdxsinx2sinxsink2

a

0

1

=

( )dxxsinx2sin2xsin2k 20

=

( )dxx3cosxcosx2cos1k0

+

=

+=

03x3sinxsin

2x2sinxk

[ ] k000k

=+

= . [ ]Zn0nsin =Q

For n = 2, (v) gives


17/34



17

( ) xdx2sinx2sinxsink2

a

0

2

=

( )dxx2sin2x2sinxsin2k 20

=

( )dxx4cos1x3cosxcosk

0

+

=

Asin21A2cos 2=Q

+

=

04

x4sinx

3

x3sinxsin

k[ ] k

k=

= [ ]Zn0nsin =Q

Therefore, from (iv), we have

( ) nxsinntcosat,xy n1n

=

== .........x2sint2cosaxsintcosa 21 ++=

x2sint2coskxsintcosk = [ ]21,nexceptn0a n ==

( )x2sint2cosxsintcosky = , is the required solution of (i).

Q.No.6.: A tightly stretched flexible string has its ends fixed at x = 0 and l=x . At time t = 0,

the string is given a shape defined by ( )xx)x(f = l , where is a constant, and

then released. Find the displacement of any point x of the string at any time t > 0.

Sol.: The given is the problem of vibration of a stretched string, therefore we find the solution

of2

22

2

2

x

yc

t

y

=

. (i)

Also the boundary conditions are( )

( )

=

=

0t,y

0t,0y

l(ii)

and the initial conditions are

( )

=

=

=

0t

y

xx)0,x(y

0t

l

. (iii)


Here the solution of (i) satisfying (ii) and (iii) is given by

( )ll

tnsin

ctncosat,xyy n

1n

==

=

, l


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18

Now ( ) dxxn

sinxx2

aIII0

nl

ll

l

= ,

Integrating by parts, we get

( ) ( ) ( )

l

l

l

l

ll

l

ll

l

03

33

2

22n n

xncos

2n

xnsin

x2n

xncos

.xx2

a

+

=

=

33

3

33

3

n

20ncos

n

200

2 ll

l(As n varies from 1 to 0n )

( )

= ncos1n

2.233

3l

l( )

3n

32

n

114 = l

=odd.isn,

n

8

evenisn,0

33

2l

Hence, from (iv), the required solution is given by

( )ll

l xnsin

ctncos

n

1.

8t,xyy

33

2

oddn

==

=

( )

( ) ( )

ll

l x1m2sin

ct1m2cos

1m2

18

31m

3

2

=

=

, Ans.

where 1m2n = , odd.

Q.No.7.: A string is stretched between the fixed points (0, 0) and ( )0,l and released at rest

from the initial deflection given by

( ) ll

ll

l

l


19/34



19

Also given the boundary conditions are( )

( )

=

=

0t,,y

0t,0y(ii)


( )


20/34



20

string is at rest with the point x = b drawn aside through a small distance d and

released at time t = 0. Show that

( )( ) llll

l ctncos

bnsin

bnsin

n

1

bb

d2t,xy

2

1n

2

2

=

=

.

Sol.: The given is the problem of a vibrating stretched string of length l . So consider

2

22

2

2

x

yc

t

y

=

. (i)


( )

=

=

0t,y

0t,0y

l(ii)


=

=

0t

y

0t

. (iii)


To find y(x, 0), equation of OA is given by

( ) xb

d0x

0b

0d0y =

= .

Also, equation of AB is given by

( )ll

= x

b

0d0y ( )l

l

= x

b

d.

Hence,

( )

=

lll

xb,x-b

d

bx0,xbd

)0,x(y (iv)

Therefore, the solution of (i), satisfying (ii), (iii) and (iv) is given by

y-axis

x-axis

A b d

O 0 0

d

b

BL


21/34



21

ll

xnsin

ctncosa)t,x(y n

1n

=

=

, l


22/34



22

2

22

2

2

x

yc

t

y

=

. (i)


( )

=

=

0t,y

0t,0y

l(ii)


=

=

0t

y

0t

. (iii)


We find y(x, 0). Now equation of OB is given by

( )0x0

3

0h0y

=l l

hx3y = ,

3x0

l

Also equation of AB is given by

( )ll

l

= x

3

0h0y ( )l

l

= x

2

h3y , l

l x

3

Hence,

( )

=

ll

ll

l

l

xb3

,x2

3h-

3x0,x

h3

)0,x(y (iv)

Now, the solution of (i) satisfying (ii) and (iii) is given by

( )ll

xnsin

ctncosat,xyy n

1n

==

=

, l


23/34



23

where ( ) dxxn

sinxf2

a

0

nll

l

= , y(x, 0) = f(x)

Now ( )

+

= dxxn

sinx2

h3

dx

xn

sin

hx32

aIII3/III

3/

0n l

lllll

l

l

l

Integrating by parts, we get

( )

3/

02

222n n

xnsin1

n

xncosx

h6a

l

l

l

l

l

l

= ( )( )

l

ll

l

l

ll

l

3/2

222 n

xnsin1

n

xncos

xh3

+

=

3nsin.

n3ncos.

n.

3h6

22

2

2lll

l

3nsin

n3ncos

n320h3

22

22

2ll

l

3

nsin

n

h3

3

ncos

n

h2

3

nsin

n

h6

3

ncos

n

h2

2222

+

+

+

=

3

nsin

n

h9

22

= , 0n . (vi)

Hence, from (v) and (vi), the required solution is given by

llxnsinctncos

3nsin

n1h9)t,x(yy 2

1n2

==

=. Ans.

Q.No.10.: A tightly stretched string with fixed end points x = 0 and l=x is initially in a

position given by

=

l

xsinyy 30 . If it is released from rest from this position,

find the displacement y(x, t).

Sol.: The given is the problem of a one dimensional wave equation

2

22

2

2

x

yc

t

y

=

. (i)

Also, the boundary conditions are( )

( )

=

=

0t,y

0t,0y

l(ii)


=

=

0t

y

0t

. (iii)


24/34



24


We know that the solution of (i), satisfying (ii) and (iii) is given by

( )ll

xnsin

ctncosat,xy n

1n

=

=

, l


25/34



25

[ ])BAcos()BAcos(Bsinasin2 +=Q

ll

l

l

l

l

l

l

l

l

0

0

0

0

4

x4sin

2

x2sin

4

y

2

x2sin

x4

y3

= ( )4

y3

4

y3 00 == ll

[ ]Zn0nsin =Q

Again from (iv), we have

dxx3

sinx3

sinx3

sinx

sin32

ya

0

03

=

lllll

l

= dxx3

sin2

ydx

xsin

x3sin2

4

y3 2

0

0

0

0

lllll

ll

dxx6

cos14

ydx

x4cos

x2cos

4

y3

0

0

0

0

=

lllll

ll

== lQ

x3AtakewewhereAsin21A2cos

2

ll

l

l

l

l

l

l

l

l

0

0

0

0

6

x6sin

x4

y

4

x4sin

2

x2sin

4

y3

=

Hence from (*), we get

( )ll

xnsinctncosat,xyy n1n

== =

..........00x3sinct3cosaxsinctcosa 31 ++++=llll

llll

x3sin

ct3cos

4

yxsin

ctcos

4

y3 00

= [Since n0an = except n = 1, 3]

=

llll

x3sin

ctcos

xsin

ctcos3

4

y0 .

Thus,

=

llll

ct3cos

x3sin

ctcos

xsin3

4

y)t,x(y 0 . Ans.

Q.No.11.: Solve the boundary value problem2

2

2

2

x

y4

t

y

=

,

y(0, t) = y(5, t) = 0,

y(x, 0) = 0, ( )xft

y

0t

=

=

,


26/34



26

where (i) x5sin2x2sin3)x(f = , (ii) xsin5)x(f = .

Sol.: (i). We know that the solution of one dimensional wave equation

2

2

2

2

x

y4

t

y

=

, (i)

satisfying the boundary conditions

y(0, t) = y(5, t) = 0 (ii)

is given by

5

xnsin

ctnsinb

ctncosa)t,x(yy nn

1n

+

==

=ll

, 0 < x < 5,

where c2

= 4 and 5=l . Hence, we have

5xnsin

5tnsinb

5tncosa)t,x(yy nn

1n

+==

=

(iii)

To find an: Use y(x, 0) = 0, (iii) gives

( ) 05

xnsin00cosa)0,x(y n

1n

=

+=

=

(given)

n0an = .

Hence (iii) reduces to

5

xnsin

5

tn2sin.b)t,x(yy n

1n

==

=

(iv)

To find bn: Differentiate (iv) w.r.t. t partially, we have

5

xnsin.

5

tn2cos.

5

n2.b

t

yn

1n

=

=

x5sin2x2sin35

xnsinb

5

n2

t

yn

1n0t

=

=

==

, (given)

which is a Fourier half-range sine series in (0, 5), and hence, for 0 < x < 5, we have

( ) dx5

xnsinx5sin2x2sin3

5

2b

5

n25

0

n

=

dx5

xnsinx5sin2

5

xnsinx2sin3

n

1b

5

0

n

= (v)


27/34



27

xdx5sin5

xnsin2

n

1xdx2sin

5

xnsin2

n2

35

0

5

0

=

( )[ ])BAcos(BAcosBsinAsin2 +=Q

5

0

5

0

dxx25

ncosx2

5

ncos

n2

3

+

=

5

0

5

0

dxx55

ncosx5

5

ncos

n

1

+

n0= except 025

n=

or n = 10 [As n varies from 1 to , 25n ]

and 25n055

n==

. [ ]Zn0nsin =

Now ( )dxx2sinx5sin2x2sinx2sin310

1b

5

0

10

=

( )dxx7cosx3cosdx10

1xdx2sin2

2

1.

10

35

0

25

0

=

( ) ( )dxx7cosx3cos10

1dxx4cos1

20

35

0

5

0

=

x2ATake,Asin21A2cos 2 ==Q

5

0

5

0 7

x7sin

3

x3sin

10

1

4

x4sinx

20

3

=

( )

=

=4

35

20

3 [ ]Zn0nsin =Q .

Also

( )dxx5sinx5sin2x5sinx2sin325

1b

5

0

25

= [Using (v)]

xdx5sindx25

1xdx5sinx2sin2

2

1.

25

3 25

0

5

0

=

( ) ( )dxx10cos125

1dxx7cosx3cos

50

35

0

5

0

=


28/34



28

( )[ ]BAcos)BAcos(BsinAsin2 +=Q

5

0

5

0 10

x10sinx

25

1

7

x7sin

3

x3sin

50

3

=

=

5

1 [ ]Zn0nsin =

Therefore, from (iv), we have

( )5

xnsin

5

tn2sinbt,xyy n

1n

==

=

.......00x5sint10sinbx2sint4sinb 2510 +++=

[ ]2510,nexceptn0bn ==Q

x5sint10sin5

1x2sint4sin

4

3y

= . Ans.

(ii). Proceed similarly up to equation (iv), we have

5xnsin

5tn2sin.b)t,x(yy n

1n

==

=

(iv)

To find bn: Differentiate (iv) w.r.t. t, partially, we have

x5

nsin.t

5

n2cos.

5

n2.b

t

yn

1n

=

=

xsin3x5

nsin.

5

n2b

t

yn

1n0t

=

=

==

, (given)

which is a Fourier half-range sine series for 0 < x < 5, we have

( ) xdx5

nsin.xsin5

5

2b

5

n25

0

n

=

dxxdx5

nsin.xsin5

n

1b

5

0

n

= (v)

xdxsinx

5

nsin2

n2

55

0

= dxnx15

ncosnx1

5

ncos

n2

55

0

+

=

( )[ ])BAcos(BAcosBsinAsin2 +=Q


29/34



29

5

0

15

n

x15

nsin

15

n

x15

nsin

n2

5

+

+

= n0= except 5n01

5

n==

[ ]Zn0nsin =Q

015

n

+ as n varies from 1 to ]

Put n = 5, using (v), we have

dxnsinxsin55

1b

5

0

5

= dx2x2cos11

xdxsinn

15

0

25

0

==

xsin21x2cos 2=Q

=

=

=

2

55

2

1

2

x2sinx

2

15

0

[ ]Zn0nsin =Q

Hence, from (iv) the required solution is given by

x5

nsint

5

n2sin.b)t,x(yy n

1n

==

=

..........0xsint2sinb......00 n +++++= [ n0bn =Q except n = 5]

xsint2sin2

5

= . Ans.

Q.No.12.: 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; ( ) 0t,y =l ; (Boundary conditions)

( ) )x(f0,xy = , ( ) l


30/34



30

( )

l


31/34



31

0pm 22 =+ pim = .

CF pxsincpxcosc 21 += .

Also PI = 0.

PICFX += pxsincpxcoscX 21 += .

0cpm 222 =+ pcim = .

CF pctsincpctcosc 43 += .

Also PI = 0.

PICFT += pctsincpctcoscT 43 += .

Case 3: When 0k= , then

(vi) gives 0XD2 = .

Its auxiliary equation is 0m2 = 0,0m = .

CF ( ) xccexcc 21x0

21 +=+= .

Also PI = 0.

PICFX += xccX 21 += .

(vii) gives 0TD2 = .

Its auxiliary equation is 0m2 = 0,0m = .

CF ( ) tccetcc 43t0

43 +=+= .

Also PI = 0.

PICFT += tccT 43 += .

Hence, from (iv), the various possible solutions of (i), are given by

Y = XT

( )( )pct4pct3px2px1 ececececy ++= (viii)

( )( )pctsincpctcoscpxsincpxcoscy 4321 ++= (ix)

( )( )tccxccy 4321 ++= (x)

Now,, out of these solutions, we must choose that solution which is consistent with the

physical nature of the given problem. As we are dealing with the problem of a wave equation

and since waves are periodic in nature therefore y must be a periodic function i.e., y must

involve trigonometric identities. Hence from (ix),

)t,x(yy = ( )( )pctsincpctcoscpxsincpxcosc 4321 ++= (xi)

is the required solution where 2pk =

Using the given condition (2), we have from (11)

( )pctsincpctcoscc0)t,0(y 431 +== 0c1 = .

Also ( )pctsincpctcoscpsinc0)t,(y 432 +== ll

0psin =l [ ,0c2 otherwise (xi) reduces to y = 0, which is not possible]

= npl l

=

np , n = 0, ...,.........2,1

Hence (xi) reduces to


32/34



32

( )

+

= t

cnsinct

cncoscx

nsinct,xy 432

lll

xn

sintcn

sincctcn

coscc 4232lll

+

Replacing c2c3, by an, c2c4 by bn, and adding all solutions, we have

( ) xn

sintcn

sinbtcn

cosat,xy nn1n lll

+

=

=

. (xii)

Further, we find an and bn, by using (iii)

Now, from (xii) xn

sin.cn

.tcn

cosbtcn

sinat

ynn

1n llll

+

=

=

0x

n

sin.

cn

.bt

yn

1n0t=

=

== lln0bn =

Hence (xii) reduces to

( ) xn

sintcn

cosat,xy n1n ll

=

=

(xiii)

Finally, ( ) xn

sina)x(f0,xy n1n l

=

=

,

Which is a Fourier half-range sine series for l


33/34



33

Home Assignments

Q.No.1.: Find the solution of the wave equation2

22

2

2

x

yc

t

y

=

, corresponding to the

triangular initial deflection ( ) xk2

xfl

= when2

x0l


34/34



34

( )[ ] ( )

( )

+

21n4

4at1n4cosx1n4sin

.

Ans.:

Q.No.4.: A taut string of length 20 cms. fastened at both ends is displaced from its position

of equilibrium, by imparting to each of its points an initial velocity given by

v = x in 10x0

x20v = in 20x10 , x being the distance from one end . Determine the

displacement at any subsequent time.

Ans.:

+

= .....

20

at3sin

20

x3sin

3

1

20

atsin

20

xsin

a

1600y

33

Q.No.5.: Show that the wave equation2

22

2

2

x

uc

t

u

=

, under the conditions u(0, t),

0)t,(u =l for all t, u(x, t) = f(x), ( )xgt

u

0t

=

=

has the solution of the form

( ) ( ) xn

sintsinCtcosBt,xu nnnn1n

l

+=

=

, where dxxn

sin)x(f2

B1

0n ll

= ,

dxxn

sin)x(gcn

2C

1

0n l

= .

Ans.:

Q.No.6.: Using DAlemberts method, find the deflection of a vibrating string of unit length

having fixed ends, with initial velocity zero and initial deflection:

(i) ( ) ( )3xxaxf = , (ii) ( ) xsinaxf 2 = .

Ans.: (i). 222 tc3x1ax)t,x(y = , (ii). ( )ct2cosx2cos1

2

a)t,x(y = .

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2 wave equation

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