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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)
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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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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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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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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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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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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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( ) ( )[ ]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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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)
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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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( )
+
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)
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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).
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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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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
( ) ( )[ ]BAcosBAcosBsinAsin2 +=Q
+
=
l
l
l
l
l
l
4
x4sin
2
x2sin
2
x2sin
3x3c4
b[ ]03
c4
b
= l [ ]Zn0nsin ==
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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
( ) ( )[ ]BAcosBAcosBsinAsin2 +=Q
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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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
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( ) 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)
Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii).
Here the solution of (i) satisfying (ii) and (iii) is given by
( )ll
tnsin
ctncosat,xyy n
1n
==
=
, l
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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
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Also given the boundary conditions are( )
( )
=
=
0t,,y
0t,0y(ii)
and the initial conditions are
( )
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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)
Also the boundary conditions are( )
( )
=
=
0t,y
0t,0y
l(ii)
and the initial conditions are
=
=
0t
y
0t
. (iii)
Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (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
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ll
xnsin
ctncosa)t,x(y n
1n
=
=
, l
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2
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
0t
. (iii)
Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (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
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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)
and the initial conditions are
=
=
0t
y
0t
. (iii)
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Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii).
We know that the solution of (i), satisfying (ii) and (iii) is given by
( )ll
xnsin
ctncosat,xy n
1n
=
=
, l
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[ ])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
=
=
,
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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)
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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
=
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( )[ ]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
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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
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( )
l
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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
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( )
+
= 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
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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
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( )[ ] ( )
( )
+
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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