One Dimensional Wave Equation

Let OA be a stretched string of length ‘l’ with fixed ends O and A. Let us take X-
axis along OA and Y-axis along OB perpendicular to OA, with O as origin. Let us
assume that the tension T in the string is constant and large when compared
with the weight of the string so that the effects of gravity are negligible. Let us
pluck the string in the BOA plane and allow it to vibrate.

Let ‘P’ be any point of the string at time‘t’. Let there be no external forces
acting on the string. Let each point of the string make small vibrations at right
angles to OA in the plane of BOA. Draw PP1 perpendicular to OA. Let OP1 =x
and PP1 =y. Then y is a function of x and t. Under the assumptions using
Newton’s second law of motion, it can be proved that y(x,t) is governed by the

equation

∂2 y 1 ∂ 2 y
= ................... (1)
∂ x 2 c 2 ∂ t2

∂2 y 2 ∂ 2 y
i.e., 2 =c 2
∂t ∂x

2 T
where c = m with T = tension in the string at any point and m is mass per unit
length of the string.

→ Since the points O and A are not disturbed from their original positions for
any time‘t’ we get

y(0,t) =0 .................. (2)

y(l,t) = 0 ................... (3)

These are referred to as the end conditions or boundary conditions.

Further it is possible that, we describe the initial position of the string as well
as the initial velocity at any point of the string at time t= 0 through the
conditions

y(x,0) =f(x), 0 ≤ x ≤ l ............... (4)

and ( ∂∂ yt ) at t =0
=g( x ), 0 ≤ x ≤ l ....................... (5)

Where f(x) and g(x) are functions such that f(0) = f(l) = 0 and g(0) = g(l) = 0.

To determine y(x,t)
∂2 y 1 ∂ 2 y
One dimensional wave equation is 2 = 2 2 ................... (1)
∂ x c ∂t

Subject to the conditions y(0,t) =0 for all t .................. (2)

y(l,t) = 0 for all t ................... (3) end conditions

y(x,0) =f(x), 0 ≤ x ≤ l ............... (4)

and ( ∂∂ yt ) at t =0
=g( x ), 0 ≤ x ≤ l ....................... (5) Initial conditions.

By the method of separation of variables

y = X(x).T(t) as solution of equation (1), then

2
∂y 1 ∂ y 11
= X ( x ) T (t ) , = X ( x )T (t)
∂x ∂x
2

2
∂y 1 ∂ y
= X ( x ) T ( t ) , 2 = X ( x ) T 11 ( t )
∂t ∂t

From equation (1), we have

11 11
11
X ( x ) T ( t )=
1
X ( x ) T
11
( t ) X ( x ) 1 T (t )
2 ⇨ =
c X ( x ) c2 T ( t )
Since the left hand side is a function of x and the right hand side is a function
of t, the equality is possible if and only if each side is equal to the same
constant λ
11 11
X ( x ) 1 T (t )
⸫ = 2 =λ
X ( x ) c T (t )

Let us take λ to be real. Then three cases are possible λ >0, λ = 0 and λ <0

Case (i): Let λ > 0, Then λ = p2 (p>0)

11 11
X ( x ) 1 T (t ) 2
⸫ = 2 =p
X ( x ) c T (t )
11
X (x) 2 11 2
= p ⇨ X ( x )= p X ( x )
X(x)

i.e., X 11 ( x ) −p 2 X ( x )=0
2 2
d X 2 d 2 d
i.e., 2 −p X =0 Let D= dx
,D = 2
dx dx

i.e., D 2 X− p2 X =0 ⇨ ( D2− p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 −p 2=0

⇨m2= p2 ⇨m=± p

⸫ X ( x )= A1 e px + B1 e−px
11
1 T (t ) 2 11 2 2 11 2 2
and also 2 =p ⇨ T ( t )= p . c . T ( t ) ⇨ T ( t )− p .c .T ( t ) =0
c T (t )

d2 T 2 2
( ) d 2 d2
i.e., 2 − p . c .T t =0 Let D= dt , D = 2
dt dt

i.e., D 2 T −p 2 c 2 T =0 ⇨ ( D2−c 2 p 2 ) T =0

⸫ The auxiliary equation is f ( m )=m2 −c 2 p2=0

⇨m2=c 2 p 2 ⇨m=± c . p

⸫T ( t )=C 1 ecpt + D1 e−cpt

Hence in this case, a typical solution is like

y ( x , t )=[ A1 e + B1 e ][ C1 e cpt + D1 e−cpt ] ........................ (S.1)

px −px

Where A1,B1,C1,D1 are arbitrary constants.

Case (ii): Let λ = 0, then

11 11
X ( x ) 1 T (t )
= 2 =0
X ( x ) c T (t )
11
X (x) 11 2
i.e., =0 ⇨ X ( x ) =0 ⇨ D X=0
X(x)

⸫ The auxiliary equation is f ( m )=m2 =0

⇨m=0,0

⸫ X ( x ) = A2 + B 2 x
11
1 T (t ) 11 2
and also =0 ⇨ T ( t ) =0 ⇨ D T =0
c T (t )
2

⸫ The auxiliary equation is f ( m )=m2 =0

⇨m=0,0

⸫T ( t )=C 2+ D2 t

Hence in this case, a typical solution is like

⸫ y ( x , t )=[ A 2 + B2 x ] [C 2 + D2 t ] .................... (S.2)

Where A2,B2,C2,D2 are arbitrary constants.

Case (iii): Let λ<0, then we can write λ=-p2 where p>0, then
11 11
X ( x ) 1 T (t ) 2
= 2 =−p
X ( x ) c T (t )
11
X (x) 2 11 2
=− p ⇨ X ( x )=− p X ( x )
X(x)
i.e., X 11 ( x ) + p2 X ( x )=0
2 2
d X 2 d 2 d
i.e., 2 + p X=0 Let D= dx , D = 2
dx dx

i.e., D 2 X + p2 X=0 ⇨ ( D2+ p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 + p2=0

⇨m2=− p2 ⇨m=± √ −p 2 ⇨ m=± pi

⸫ X ( x )= A3 cos px+ B 3 sin px

11
1 T (t ) 2 11 2 2 11 2 2
and also 2 =−p ⇨ T ( t )=− p . c . T ( t ) ⇨ T ( t ) + p . c .T ( t )=0
c T (t )

d2T 2 2
( ) d 2 d2
i.e., + p . c . T t =0 Let D=
dt
, D =
d t2 d t2

i.e., D 2 T + p2 c 2 T =0 ⇨ ( D 2+ c2 p2 ) T =0

⸫ The auxiliary equation is f ( m )=m2 + c2 p 2=0

⇨m2=−c2 p 2 ⇨ m=± √−c2 p 2 ⇨ m=± cp i

⸫T ( t )=C 3 cos cpt+ D 3 sin cpt

Hence in this case, a typical solution is like

⸫ y ( x , t )=[ A 3 cos px+ B3 sin px ] [ C3 cos cpt + D3 sin cpt ] .............. (S.3)

Where A3,B3,C3,D3 are arbitrary constants.

Thus the possible solution forms of equation (1) are

y ( x , t )=[ A1 e px + B1 e−px ][ C1 e cpt + D1 e−cpt ] ........................ (S.1)

y ( x , t )=[ A 2 + B2 x ] [C 2 + D 2 t ] .................... (S.2)

y ( x , t )=[ A 3 cos px+ B3 sin px ] [ C3 cos cpt + D3 sin cpt ] .............. (S.3)

We have to note that here A’s, B’s, C’s, D’s are arbitrary constants to be
determined.
The constant p is also to be determined. The A’s, B’s, C’s, D’s depend on f(x)
and g(x). The constant p will be determined using conditions (2) and (3).

We now decide which solution is appropriate for the present problem.

Consider S.1

y ( x , t )=[ A1 e + B1 e ][ C1 e cpt + D1 e−cpt ]

px −px

Using condition (2), y(0,t) = 0 for all t

y ( 0 , t )=0=[ A 1 e + B1 e ][ C1 e cpt + D1 e−cpt ]

p .0 − p .0

⇨( A1 + B1 ) ( C 1 e cpt + D1 e−cpt )=0 for all t

⸫ A1 + B 1 = 0

Using condition (3), y(l,t) = 0 for all t

y ( l , t )=0=[ A1 e p . l +B 1 e− p .l ][ C 1 ecpt + D 1 e−cpt ] for all t

⸫ A1 e p .l +B 1 e− p . l=0

Solving A1 + B1 = 0 and A1 e p .l +B 1 e− p . l=0

We get A1=0=B1

Thus y(x,t) = 0

This implies that there is no displacement for any x and y. This is impossible.

Thus (S.1) is not an appropriate solution.

Consider S.2
y ( x , t )=[ A 2 + B2 x ] [ C 2 + D 2 t ]

Using condition (2), y(0,t) = 0 for all t

y ( 0 , t )=0=[ A 2+ B2 .0 ] [C 2 + D2 t ] ⇨ A2 [ C2 + D2 t ] =0 ⇨ A 2=0

Using condition (3), y(l,t) = 0 for all t

y ( l, t )=0=[ A 2 +B 2 . l ][ C 2 + D2 t ] for all t

⸫ B2 l ( C 2+ D2 t )=0 for all t ⸪A2 = 0

Here l ≠0 , C 2+ D 2 t ≠ 0 for all t ⸪B2 = 0

Thus here again y(x,t)=0 for all x and t.

Thus, as before, this solution also is not valid.

Thus (S.2) is not an appropriate solution.

Consider S.3
y ( x , t )=[ A 3 cos px+ B3 sin px ] [ C3 cos cpt + D3 sin cpt ]

Using condition (2), y(0,t) = 0 for all t

y ( 0 , t )=0=[ A3 cos p .0+ B3 sin p .0 ] [ C3 cos cpt + D3 sin cpt ]

A3 ( C 3 cos cpt+ D3 sin cpt )=0 for all t

⸫A3 = 0
y ( l, t )=0=[ A 3 cos p .l+ B3 sin p . l ][ C 3 cos cpt+ D 3 sin cpt ]

⇨ B3 sin p . l ( C 3 cos cpt + D3 sin cpt ) =0

If B3 = 0, y(x,t) = 0 and this is not valid.

Hence sin p .l=0

⸫ pl=nπ where n = 1,2,3,........

nπ
Thus p= l where n = 1,2,3,.......

Thus a typical solution of equation (1) satisfying conditions (2) and (3) is

y ( x , t )=sin
nπx
l [
C n cos
nπct
l
+ Dn sin
nπct
l ]
for n = 1,2,3,...............
( Here we have write Cn, Dn in place of C3, D3)

Thus the most general solution of equation (1) satisfying (2) and (3) is

( )
∞
nπct nπct nπx
y ( x , t )=∑ Cn cos + Dn sin sin .......... (6)
n=1 l l l

Where Cn, Dn are arbitrary constants to be determined by using the conditions

(4) and (5).

By condition (4)

y(x,0) =f(x), 0 ≤ x ≤ l

Thus putting t = 0 in equation (6), we obtain

( )
∞
nπc .0 nπc .0 nπx
y ( x , 0 ) =f ( x )=∑ C n cos + D n sin sin
n=1 l l l
∞

∑ C n sin nπx
l
=f ( x ) , 0 ≤ x ≤l
n =1

( L.H.S is the half range Fourier sine series of f(x) in [0,l])

l
2 nπx
Hence C n= ∫ f ( x ) sin dx n = 1,2,3,........ ................. (7)
l 0 l

Thus Cn’s are all determined.

By condition (5)

( ∂∂ yt )
at t =0
=g(x ), 0 ≤ x ≤ l

[ ( ) ( )]
∞
∂y nπct nπc nπct nπc nπx
Now ∂ t =∑ −C n sin l l
+ Dn cos
l l
sin
l
n=1

( ∂∂ yt )
at t =0
=g( x ) gives

( ) [ ( ) ( )]
∞
∂y nπc .0 nπc nπc .0 nπc nπx
=g ( x )=∑ −C n sin + Dn cos sin
∂t at t =0 n=1 l l l l l
∑ D n( nπc )
∞
nπx
sin =g ( x ) , 0≤ x ≤ l
n =1 l l
l

Hence Dn
nπc 2
l l 0( )
= ∫ g ( x ) sin
nπx
l
dx

l
2 nπx
Thus D n= ∫
nπc 0
g ( x ) sin
l
dx for n = 1,2,3,.......... .................. (8)

Thus Dn’s are all determined.

Hence the displacement y(x,t) at any point x and at any subsequent time t is
given by

( )
∞
nπct nπct nπx
y ( x , t )=∑ Cn cos + Dn sin sin
n=1 l l l
l
2 nπx
Where C n= ∫ f ( x ) sin dx and
l 0 l
l
2 nπx
D n= ∫
nπc 0
g ( x ) sin
l
dx

Problems
1. A tightly stretched string with fixed end points x = 0 and x = l is
3 πx
initially in a position given by y= y 0 sin l . If it is released from rest
from this position, find the displacement y(x,t).

Solution: The displacement y(x,t) is governed by

∂2 y 1 ∂ 2 y
= ................... (1)
∂ x 2 c 2 ∂ t2
Subject to y(0,t) = 0 for all t ........... (2)
y(l,t) = 0 for all t ............ (3)
3 πx
y(x,0) = y 0 sin l , 0 ≤ x ≤ l ............... (4)
 and ( )
∂y
∂t at t =0
=0 , 0 ≤ x ≤l ....................... (5)

( The condition (5) is implied by the phrase “if it is released from rest
from this position”)
The most general solution of equation (1) satisfying equations (2)
and (3) is given by
Let us seek solution of equation (1) in the form u(x,t) = X(x)T(t)
The three solutions of equation (1) are
y ( x , t )=( A 1 e + B1 e )( C 1 e + D1 e ) .............. (S.1)
px − px pct − pct

y ( x , t )=( A2 + B2 x ) ( C 2+ D 2 t ) ...................... (S.2)

y ( x , t )=( A3 cos px+ B3 sin px ) ( C 3 cos cpt+ D3 sin cpt ) ............ (S.3)

The solution appropriate to the present problem is (S.3) as the

required solution has to be periodic in x and t. Hence the required
solution is of the form

y ( x , t )=( A cos px+ B sin px ) ( C cos cpt+ D sin cpt )

Using conditions (2) and (3), we note that
nπ
A = 0, p= l where n = 1,2,3,...........
The most general solution of equation (1) satisfying (2) and (3) is

[ ]
∞
nπct nπct nπx
y ( x , t )=∑ C n cos + Dn sin sin ............. (6)
n=1 l l l

Where Cn and Dn are to be determined using equations (4) and (5).

Equation (6) partially differentiating with respect to ‘t’ we get
∂y
∂t [
=∑ −C n
nπc
l
sin
nπct
l
+ Dn
nπc
l
cos
nπct
l
sin
nπx
l ]
Using condition (5) i.e., ∂ t (∂ y) t =0
=0 , we get

nπc nπx nπx

0=∑ Dn sin ⇨ D n=0 ⸪sin ≠0
l l l
Substituting D n=0 in equation (6), we get
nπct nπx
y ( x , t )=∑ C n cos sin ................ (7)
l l
 3 πx
Using condition (4) i.e., y(x,0) = y 0 sin l in equation (7), we get
nπx 3 πx
∑ C n sin l
= y 0 sin
l
, ( 0 ≤ x ≤l )

We have sin 3 θ=3 sin θ−4 sin3 θ

3 3 1
i.e., sin θ= 4 sinθ− 4 sin 3 θ
nπx
[ 3
⸫∑ C n sin l = y 0 4 sin l − 4 sin l
πx 1 3 πx
]
Comparing the coefficients of like terms (or by using uniqueness theorem
of Fourier series)
3 y0 − y0
C 1= ,C 3= ∧C 2=0 , C4 =0 , C5 =0 ......................
4 4

3 y0 πct πcx y 0 3 πct 3 πcx

Hence y ( x , t )= cos sin − cos sin
4 l l 4 l l

2. A tightly stretched string with fixed end points x = 0 and x = l is

initially in its equilibrium position. If it is set to vibrate by giving each
of its points a velocity λx(l-x), find the displacement of the string at
any distance x from one end at any time t.

Solution: The displacement y(x,t) is governed by

2 2
∂ y 1 ∂ y
2
= 2 2 ................... (1)
∂ x c ∂t
Subject to y(0,t) = 0 for all t ........... (2)
y(l,t) = 0 for all t ............ (3)
y(x,0)=0 for 0 ≤ x ≤ l ................ (4)

and ( ∂∂ yt )
at t =0
=λx (l−x ) , 0 ≤ x ≤ l ....................... (5)

The most general solution of equation (1) satisfying equations (2)

and (3) is given by
Let us seek solution of equation (1) in the form u(x,t) = X(x)T(t)
The three solutions of equation (1) are
y ( x , t )=( A 1 e + B1 e )( C 1 e + D1 e ) .............. (S.1)
px − px pct − pct

y ( x , t )=( A2 + B2 x ) ( C 2+ D 2 t ) ...................... (S.2)

y ( x , t )=( A3 cos px+ B3 sin px ) ( C 3 cos cpt+ D3 sin cpt ) ............ (S.3)

The solution appropriate to the present problem is (S.3) as the

required solution has to be periodic in x and t. Hence the required
solution is of the form

y ( x , t )=( A cos px+ B sin px ) ( C cos cpt+ D sin cpt )

Using conditions (2) and (3), we note that
nπ
A = 0, p= l where n = 1,2,3,...........
The most general solution of equation (1) satisfying (2) and (3) is

[ ]
∞
nπct nπct nπx
y ( x , t )=∑ C n cos + Dn sin sin ............. (6)
n=1 l l l

Using condition (4)

nπx
∑ C n sin l
=0,0 ≤ x ≤ l

Hence Cn = 0 for all n.

l
2 nπx
D n= ∫
nπc 0
λx ( l−x ) sin
l
dx

[ ]
l
nπx nπx nπx
−cos −sin cos
2λ l l l
D n= x ( l−x ) − (l−2 x ) + (−2 ) 3 3
nπc nπ nπ n π
l l l
3
0

[ ] [
nπl nπl nπl nπ .0
−cos −sin cos −cos −sin
2λ l l l 2λ l
D n= l ( l−l ) −( l−2 l ) + (−2 ) 3 3 − 0. (l−0 ) −( l−2.0 )
nπc nπ nπ n π nπc nπ nπ
l l l
3 l l

[ ]
3 3 3
2 λ −2 l 2l 2 λ 2l
D n= cos nπ + 3 3 ⇨ D n= [ 1−cos nπ ]
nπc n π
3 3
n π nπc n3 π 3
 ⸪cos nπ=(−1 )n
3
4 λl
D n= [ 1−cos nπ ]
n4 π 4 c
If n is even, Dn = 0
If n is odd, Dn =2m + 1
3 3
4 λl 8 λl
D2 m+1 = 4 4
.2=
( 2 m+1 ) π c ( 2 m+1 )4 π 4 c

[ ]
( 2 m+1 ) πct ( 2 m+1 ) πx
3 ∞ sin sin
Hence y ( x , t )= 8 λ l ∑
l l
4 2
π c m=0 ( 2 m+1 )

3. If a string of length ‘l’ is initially at rest in equilibrium position and

3 πx
each of its points is given the velocity V 0 sin l find the displacement
y(x,t).
Solution: The displacement y(x,t) is governed by

∂2 y 1 ∂ 2 y
= ................... (1)
∂ x 2 c 2 ∂ t2
Subject to y(0,t) = 0 for all t ........... (2)
y(l,t) = 0 for all t ............ (3)
y(x,0)=0 for 0 ≤ x ≤ l ................ (4)

and ( ∂∂ yt )
at t =0
=V 0 sin
3 πx
l
, 0 ≤ x ≤l ....................... (5)

The most general solution of equation (1) satisfying equations (2)

and (3) is given by
Let us seek solution of equation (1) in the form u(x,t) = X(x)T(t)
The three solutions of equation (1) are
y ( x , t )=( A 1 e px + B1 e− px )( C 1 e pct + D1 e− pct ) .............. (S.1)
y ( x , t )=( A2 + B2 x ) ( C 2+ D 2 t ) ...................... (S.2)
y ( x , t )=( A3 cos px+ B3 sin px ) ( C 3 cos cpt+ D3 sin cpt ) ............ (S.3)
 The solution appropriate to the present problem is (S.3) as the
required solution has to be periodic in x and t. Hence the required
solution is of the form

y ( x , t )=( A cos px+ B sin px ) ( C cos cpt+ D sin cpt )

Using conditions (2) and (3), we note that
nπ
A = 0, p= l where n = 1,2,3,...........
The most general solution of equation (1) satisfying (2) and (3) is

[ ]
∞
nπct nπct nπx
y ( x , t )=∑ C n cos + Dn sin sin ............. (6)
n=1 l l l
Using equation (4), we get
nπx
∑ C n sin l
=0,0 ≤ x ≤ l

Hence Cn = 0 for all n.

Also using equation (5), we get

∑ Dn
nπc
l
sin
npx
l
=V 0 sin
3 πx
l
⇨ ∑ Dn
nπc
l
sin
npx
l
3 πx 1
=V 0 sin − sin
4 l 4 [
3 πx
l ]
We have sin 3 θ=3 sin θ−4 sin3 θ
3 3 1
i.e., sin θ= 4 sinθ− 4 sin 3 θ
πc 3 3l
⇨ D 1 l = 4 V 0 ⇨ D 1= 4 πc V 0
3 πc −V 0 −l V 0
And D3 = ⇨ D 3=
l 4 12 πc
3l πct πx lV0 3 πct 3 πx
Hence y ( x , t )= V 0 sin sin −¿ sin sin ¿
4 πc l l 12 πc l l

4. If a string of length ‘l’ is initially at rest in equilibrium position and

each of its points is given, the velocity ∂ t =b sin
t =0
(∂ y) 3
( πxl ) find the
displacement y(x,t).
Solution: The displacement y(x,t) is governed by
 2 2
∂ y 1 ∂ y
2
= 2 2 ................... (1)
∂ x c ∂t
Subject to y(0,t) = 0 for all t ........... (2)
y(l,t) = 0 for all t ............ (3)
y(x,0)=0 for 0 ≤ x ≤ l ................ (4)

and ( ∂∂ yt )
at t =0
=b sin
3
( πxl ) ,0 ≤ x ≤ l ....................... (5)
The most general solution of equation (1) satisfying equations (2)
and (3) is given by
Let us seek solution of equation (1) in the form u(x,t) = X(x)T(t)
The three solutions of equation (1) are
y ( x , t )=( A 1 e px + B1 e− px )( C 1 e pct + D1 e− pct ) .............. (S.1)
y ( x , t )=( A2 + B2 x ) ( C 2+ D 2 t ) ...................... (S.2)
y ( x , t )=( A3 cos px+ B3 sin px ) ( C 3 cos cpt+ D3 sin cpt ) ............ (S.3)

The solution appropriate to the present problem is (S.3) as the

required solution has to be periodic in x and t. Hence the required
solution is of the form

y ( x , t )=( A cos px+ B sin px ) ( C cos cpt+ D sin cpt )

Using conditions (2) and (3), we note that
nπ
A = 0, p= l where n = 1,2,3,...........
The most general solution of equation (1) satisfying (2) and (3) is

[ ]
∞
nπct nπct nπx
y ( x , t )=∑ C n cos + Dn sin sin ............. (6)
n=1 l l l
Using equation (4), we get
nπx
∑ C n sin l
=0,0 ≤ x ≤ l

Hence Cn = 0 for all n.

∞
nπct nπx
⸫ y ( x , t )=∑ Dn sin sin .............. (7)
n=1 l l
Equation (7) partially differentiating with respect to ‘t’, we get
∞
∂y nπc nπct nπx
=∑ D n cos sin
∂ t n=1 l l l
 To determine D n use the initial condition

( ) ( )
∞
∂y πx nπc nπc .0 nπx
=∑
3
=b sin Dn cos sin
∂t at t =0 l n=1 l l l

( ∂∂ yt ) =b sin ( πxl )=∑ nπcl D sin nπxl

∞
3
n
at t =0 n=1

Or 4 ( 3sin l −sin l )=D l sin l + D l sin l + D l sin l + ..

b πx 3 πx πc πx 2 πc 2 πx 3 πc 3 πx
1 2 3

We have sin 3 θ=3 sin θ−4 sin3 θ

3 3 1
i.e., sin θ= 4 sinθ− 4 sin 3 θ
πx 2 πx 3 πx
Equating the coefficients of sin l , sin l , sin l , ................. on both
sides, we get
3b πc 2 πc 3 πc −b
4
=D1
l
, D 2
l
= 0, D3
l
=
4
, D4 = 0, D5 = 0 ....................
3 lb −bl
⇨ D 1=
4 πc
, D 2 = 0, D 3=
12 πc
, D4 = 0, D5 = 0 .................... .......... (8)

From equations (7) and (8), we have

y ( x , t )=
bl
12 πc [
9 sin
πct
l
πx
sin −sin
l
3 πct
l
sin
3 πx
l ]
5. Solve the boundary value problem
2
utt =a uxx ; 0< x <l ; t >0 with u ( 0 , t ) =0 ; u ( l , t ) =0∧u ( x , 0 )=0 ; ut ( x , 0 )=u 0 x (1−x ) .

Solution: Given equation is utt =a2 uxx

∂ 2 u 2 ∂2 u
i.e. , 2
=a 2 ................... (1)
∂t ∂x
The boundary conditions are
(i) u(0,t)=0 for all t
(ii) u(l,t) = 0 for all t
(iii) u(x,0) = 0 for 0<x<l
(iv) ut ( x , 0 )=u 0 x ( 1−x )
The required solution of equation (1) is of the form
u ( x , t )=( c 1 cos px+ c2 sin px ) ( c3 cos pat + c 4 sin pat ) ............ (2)

Using conditions (i) and (ii), we have

nπ
c 1=0∧ p= where n = 1,2,3,...........
l

⸫ The general solution of equation (1) satisfying (i) and (ii) is

u ( x , t )=c 2 sin
nπx
l (
c 3 cos
nπat
l
+ c 4 sin
nπat
l )
................ (3), n = 1,2,3,.....

Now using condition (iii) i.e., u(x,0) = 0, we have

0=c 2 sin
nπx
l
c 3 cos (
nπa .0
l
+c 4 sin
nπa.0
l )
nπx nπx
⇨c 2 sin
l
( c 3 cos 0+c 4 sin 0 ) =0 ⇨ c 2 sin l ( c 3 +0 )=0

[
⇨ c 3=0 ⸪ c 2 ≠ 0∧sin
nπx
l
≠0
]
Putting c3 = 0 in equation (3), we have
u ( x , t )=c 2 sin
nπx
l
0. cos (nπat
l
+c 4 sin
nπat
l )
nπx nπat
⇨ u ( x ,t )=c2 c 4 sin sin
l l
nπx nπat
u ( x , t )=c n sin sin since cn =c2 c4
l l
⸫ The most general solution of equation (1) is
∞
nπx nπat
u ( x , t )=∑ c n sin sin ................... (4)
n=1 l l

Equation (4) partially differentiating with respect to ‘t’ , we get

∞
∂u nπa nπx nπat
=∑ c n sin cos
∂ t n=1 l l l

( )
∞
∂u nπa nπx nπa.0
=∑ c n sin cos
∂t t =0 n=1 l l l
( )
∞
∂u nπa nπx
=∑ c n sin
∂t t =0 n=1 l l

Now using condition (iv), we get

∞
nπa nπx
u0 x ( 1− x ) =∑ c n sin ......... (5) ⸪ut ( x , 0 )=u 0 x ( 1−x )
n=1 l l

To find cn, expand u0 x ( 1− x ) as a half – range sine series in (0,l)

∞
nπx
Let u0 x ( 1− x ) =∑ bn sin ................... (6)
n=1 l

From equations (5) and (6), we have

∞ ∞

∑ c n nπa
l
sin
nπx
l
=∑ bn sin
nπx
l
n =1 n=1

Comparing like coefficients, we have

nπa lb
cn =bn ⇨c n= n .................. (7)
l nπa

l
2
But b n= ∫ f ( x ) sin nπx
l 0 l
dx

2u0 l
b n= ∫ ( lx−x 2 ) sin nπx dx
l 0 l

[ ( ) ( ) ( )]
l
nπx nπx nπx
−cos −sin cos
2u0 l l l
b n= ( lx− x2 ) −( l−2 x ) 2 2
+ (−2 ) 3 3
l nπ n π n π
l l 2
l3 0

[ ( ) ( ) ( )]
nπl nπl nπl
−cos −sin cos
2u0 l l l
b n= ( l .l−l2 ) − (l −2l ) −2
l nπ 2 2
n π 3 3
n π
l l
2
l
3
 [ ( ) ( ) ( )]
nπ .0 nπ .0 nπ .0
−cos −sin cos
−2u 0 l l l
( l .0−0 2 ) −( l−2.0 ) 2 2
+ (−2 ) 3 3
l nπ n π n π
l l 2
l3

[ ] [ ]
2u0 cos nπ 2 u0 1
b n= 0+l .0−2 3 3 − 0−l .0−2. 3 3
l n π l n π
3 3
l l

[ ]
2u0 cos nπ 2 2
b n= −2 3 3 + 3 3 4 u0 l
l n π ⇨b n= 3 3 [ 1−(−1 ) ]
n
n π
3 3 n π
l l

{
0 ,if n is even
⸫ bn= 8u 0 l 2
3 3
,if n is odd
n π

Substituting these values of bn, in equation (7), we get

2
l 8 u0 l
c n= . , where n is odd
nπa n3 π 3

8 u0 l 3
c n= , n = 1,2,3,..........
n4 π 4 a

Substituting this value of cn in equation (4), we get

∞
8 u0 l 3 nπx nπat
u ( x , t )= ∑ 4
n π a4
sin
l
sin
l
n=1,3,5, …..

6. Solve the boundary value problem

2
utt =a uxx ; 0< x <l ; t>0 with u ( 0 , t ) =0 ; u ( l , t ) =0∧u ( x , 0 )=0 ;ut ( x , 0 )=sin
3
( πxl ) .
Solution: Given equation is utt =a2 uxx
 2 2
∂ u 2∂ u
i.e. , 2
=a 2 ................... (1)
∂t ∂x
The boundary conditions are
(v) u(0,t)=0 for all t
(vi) u(l,t) = 0 for all t
(vii) u(x,0) = 0 for 0<x<l
(viii) ut ( x , 0 )=sin
3
( πxl )
The required solution of equation (1) is of the form
u ( x , t )=( c 1 cos px+ c2 sin px ) ( c3 cos pat+ c 4 sin pat ) ............ (2)

Using conditions (i) and (ii), we have

nπ
c 1=0∧ p= where n = 1,2,3,...........
l

⸫ The general solution of equation (1) satisfying (i) and (ii) is

u ( x , t )=c 2 sin
nπx
l (
c 3 cos
nπat
l
+ c 4 sin
nπat
l )
................ (3), n = 1,2,3,.....

Now using condition (iii) i.e., u(x,0) = 0, we have

0=c 2 sin
nπx
l (
c 3 cos
nπa .0
l
+c 4 sin
nπa.0
l )
nπx nπx
⇨c 2 sin
l
( c 3 cos 0+c 4 sin 0 ) =0 ⇨ c 2 sin
l
( c 3 +0 )=0

[
⇨ c 3=0 ⸪ c 2 ≠ 0∧sin
nπx
l
≠0 ]
Putting c3 = 0 in equation (3), we have
u ( x , t )=c 2 sin
nπx
l
0. cos (nπat
l
+c 4 sin
nπat
l )
nπx nπat
⇨ u ( x ,t )=c2 c 4 sin sin
l l
nπx nπat
u ( x , t )=c n sin sin since cn =c2 c4
l l
⸫ The most general solution of equation (1) is
∞
nπx nπat
u ( x , t )=∑ c n sin sin ................... (4)
n=1 l l
 Equation (4) partially differentiating with respect to ‘t’ , we get
∞
∂u nπa nπx nπat
=∑ c sin cos
∂ t n=1 n l l l

( )
∞
∂u nπa nπx nπa.0
=∑ c n sin cos
∂t t =0 n=1 l l l

( )
∞
∂u nπa nπx
=∑ c n sin
∂t t =0 n=1 l l

Now using condition (iv), we get

( ) ( )
∞
πx nπa nπx 3 πx
=∑ cn ......... (5) ⸪ut ( x , 0 )=sin
3
sin sin
l n=1 l l l

∞
3 πx 1 3 πx nπa nπx
i .e . , sin − sin =¿ ∑ c n sin ¿
4 l 4 l n=1 l l

We have sin 3 θ=3 sin θ−4 sin3 θ

3 3 1
i.e., sin θ= 4 sinθ− 4 sin 3 θ
3 πx 1 3 πx πa πx 2 πa 2 πx 3 πa 3 πx
⇨ 4 sin l − 4 sin l =¿ c1 l sin l +c 2 l sin l +c 3 l sin l +¿ ¿ ¿ .....
Comparing the coefficients of like terms,

πa 3 3l 3 πa −1 −l
c1 = ⇨c 1=
l 4
, c =0 , c 3 l = 4 ⇨ c 3= 12 πa , c 4 =0
4 πa 2

c 5=0 ...................

substituting these values∈equation ( 4 ) , we get

3l πx πat l 3 πx 3 πat
u ( x , t )= sin sin −¿ sin sin ¿
4 πa l l 12 πa l l

7. A tightly stretched string of length ‘l’ has its ends fastened at x = 0,

x = l. The mid – point of the string is then taken to height ‘h’ and then
released from rest in that position. Find the lateral displacement of a
point of the string at time ‘t’ from the instant of release.

Solution: Let y(x,t) denote the displacement of the string.

The initial displacement is given by OAB.
h−0
y−0= ( x−0 )
Equation Of OA is l
−0
2
2h
⇨ y= l x

Equation of AB is
y−h=
0−h
l−
l
x−( )
l
2
2
2
( )
l
⇨ y−h=(−h ) l x− 2

⇨ y=h− l ( x− 2 )⇨ y =h [ 1− l ( x− 2 )]
2h l 2 l

⇨ y=h [ 1− l x+ 1 ]⇨ y=h ( 2− l x )
2 2

⇨ y=2h ( 1− l )⇨ y = l ( l−x )
x 2h

Thus the problem is to solve the one-dimensional wave equation

∂2 y 2
2∂ y
=a ................... (1)
∂ t2 ∂ x2

With boundary conditions y(0,t) = 0, y(l,t) = 0 and with initial

displacement

{
2h l
x if 0 ≤ x ≤
y ( x , 0 ) =f ( x )=
2h (
l
l
2
l −x ) if ≤ x ≤ l
and
∂y
∂t ( ) t =0
=0
l 2
The solution of equation (1) satisfying the above boundary
conditions and initial conditions is given by

( nπxl ) cos( nπatl ) ................... (2)

∞
y ( x , t )=∑ An sin
n=1
 l
2
Where An = ∫ f ( x ) sin
l 0
nπx
l
dx ( )

[ ]
l
2 l
An =
2
l 0
2h nπx
l
2h
( ) nπx
∫ l x sin l dx +∫ l (l−x ) sin l dx ( )
2

[{ ( ) }{ ( ) ( )} ]
l l

2 2h
An = . x
−cos
nπx
l
−1.
−sin
nπx
l ( ) 2

+ ( l−x )
−cos
nπx
l
− (−1 )
sin
nπx
l
l l nπ 2 2
n π nπ 2 2
n π
l l
2 l l
2 l
0 2

[{ ( )} { ( )} ]
l l

An =
4 h −l
l 2
nπ
x cos
nπx
l
+
l2
2 2
n π ( )
sin
nπx
l 0
2
+
−l
nπ ( )
( l−x ) cos
nπx
l
l2
− 2 2 sin
n π
nπx
l l
2

An =
4h
l2 [{ −l 2
2 nπ
nπ l2
cos + 2 2 sin
2 n π
nπ
2 }{
+ ( 0+0 )+
l2
2 nπ
nπ l2
cos + 2 2 sin
2 n π
nπ
2 }]
[ ( )] ( )
2
4h 2l nπ 8h nπ
An = 2 2 2
sin = 2 2 sin
l n π 2 n π 2

8h nπ
Thus An = 2 2 sin 2
n π ( )
Substituting the values of An in equation (2) we get

( ) ( ) ( )
∞
8h nπ nπx nπat
y ( x , t )=∑ sin sin cos
n=1
2 2
n π 2 l l

y ( x , t )=
8h 1
2
π 1 2[ ( ) ( )
sin
πx
l
cos
πat
l
1
− 2 sin
3
3 πx
l
cos ( ) (
3 πat
l
+…………. ) ]
8. The points of trisection of a tightly stretched string of length ‘l’ with
fixed ends are pulled aside through a distance ‘d’ on opposite sides of
the position of equilibrium, and the string is released from rest.
Obtain an expression for the displacement of the string at any
subsequent time and show that the midpoint of the string is always
at rest.
Solution: Let the points of trisection of the string OA be B and C,
where O and A are the fixed ends of the string. Let the two points of
trisection are displaced by ‘h’. The initial position of the string is as
shown in diagram.

y− y1 x−x 1
Equation of OP is y − y = x −x
2 1 2 1

y−0 x−0
= y 3x 3 hx l
i.e., h−0 l −0 ⇨ h = l ⇨ y = l , 0 ≤ x ≤ 3
3
l
x−
y−h 3
Equation of PQ is −h−h = 2 l l
−
3 3
l
x−
⇨
y−h
−2 h
=
l
3
⇨ y−h=
−6 h
l ( )
l
x− ⇨ y =h−
3
6h
l( )
x−
l
3
3

[ ( )]
⇨ y =h 1−
6
l
x−
l
3
⇨ y =h 1−
l [ ( )]
6 3 x−l
3

[ 2
]
⇨ y =h 1− ( 3 x−l ) ⇨ y=h
l [ ]
l−6 x +2 l
l [
⇨ y =h
3l−6 x
l ]
 3h l 2l
⇨ y= ( l−2 x ) , ≤ x ≤
l 3 3

3h 2l
Similarly, equation of QA is y= l ( x−l ) , 3 ≤ x ≤ l

{
3 hx l
, 0≤ x ≤
l 3
3 h l 2l
⸫ f ( x )= l ( l−2 x ) , 3 ≤ x ≤ 3 ................... (1)
3h 2l
( x−l ) , ≤ x ≤l
l 3
represents the initial position of the string.
The displacement y(x,t) at any point of the string is given by

∂2 y 2
2∂ y
=a ................... (2)
∂ t2 ∂ x2

The boundary conditions are

(i) y(0,t) = 0, for all t
(ii) y(l,t) = 0, for all t
(iii) ( ∂∂ yt )
t =0
=0 , for all t

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

Where f(x) is given by the equation (1).

The solution of the equation (2) consistent with these conditions is

y ( x , t )=( c 1 cos px +c 2 sin px ) ( c 3 cos pat +c 4 sin pat ) ................ (3)

Using the condition (i), we obtain

0= y ( 0 ,t )=( c1 cos p .0+c 2 sin p .0 ) ( c 3 cos pat + c 4 sin pat )

⇨c 1 ( c 3 cos pat +c 4 sin pat )=0⇨ c 1=0

⸫ Equation (3) reduces to

y ( x , t )=( c 2 sin px ) ( c 3 cos pat +c 4 sin pat ) .................. (4)
Using the condition (ii), we obtain
0= y (l , t )=( c 1 cos p . l+c 2 sin p . l )( c 3 cos pat+ c 4 sin pat )

0=( c 2 sin p . l )( c 3 cos pat + c 4 sin pat ) ⸪ c1 = 0

nπ
⇨ sin pl=0∨ pl=nπ ⇨ p=
l

Substituting this value of ‘p’ in equation (4), we get

y ( x , t )=c 2 sin
nπx
l (
c 3 cos
nπat
l
+ c 4 sin
nπat
l )
................. (5)

Equation (5) partially differentiating with respect to ‘t’ , we get

∂y
∂t
=c 2 sin
nπx
l
−c 3 (
nπa
l
sin
nπat
l
+c 4
nπa
l
cos
nπat
l )
Using the condition (iii), we obtain

( ∂∂ yt )t =0
=0=c 2 sin
nπx
l (
−c 3
nπa
l
sin
nπa .0
l
+ c4
nπa
l
cos
nπa .0
l )
0=c 2 sin
nπx
l (
0+c 4
nπa
l )
⇨0=c 2 c 4
nπa
l
sin
nπx
l

(
⇨ c 4=0 ⸪ c 2 ≠ 0 , sin
nπx
l
≠0 )
Hence equation (5) becomes
nπx nπat
y ( x , t )=c 2 c3 sin cos
l l

Putting c2 c3 = cn , we get
nπx nπat
y ( x , t )=c n sin cos
l l

The general solution is obtained by adding all such solutions so that

∞
nπx nπat
y ( x , t )=∑ c n sin cos ................. (6)
n=1 l l
Using the condition (iv) in equation (6), we obtain
∞
nπx nπa .0
y ( x , 0 ) =∑ c n sin cos
n=1 l l
∞
nπx
y ( x , 0 ) =∑ c n sin
n=1 l
∞
nπx
i .e . , f ( x ) =∑ c n sin .................... (7)
n=1 l

To find cn , expand f(x) as a half – range sine series in (0,l)

∞
nπx
Let f ( x )=∑ bn sin ................... (8)
n=1 l

From equations (7) and (8), we have

∞ ∞

∑ c n sin nπx
l
=∑ bn sin
nπx
l
n =1 n=1

Comparing like coefficients,

Here cn = bn .................. (9)

l
2
Where b n= ∫ f ( x ) sin
l 0
nπx
l
dx ( )

[ ]
l 2l
3 3 l
b n=
2
l 0
3 hx nπx
l
3h
( ) nπx
2l
3h nπx
∫ l sin l dx+∫ l ( l−2 x ) sin l dx+∫ l ( x −l ) sin l dx( ) ( )
3 3

{[ ( ) ( ( ) ) ] [ ( ) ( ( ) )] [ (
l 2l
nπx nπx
( ) nπx nπx
(
3 3
−cos −sin −cos −sin −cos
6h l l l l
b n= 2 x −1. + ( l−2 x ) −(−2 ) + ( x−l )
l nπ nπ
2
nπ nπ
2
nπ
l l 0
l l
l
3
l

b n=
6h
l
2
−
{[ (
l2
3 nπ
cos
nπ
3
+
l 2
nπ )
sin
nπ
3
+ −
l2
( )
3 nπ
cos
2 nπ
3
−2
nπ ] [( ( )
l 2
sin
2nπ
3
− −
l2
3 nπ
nπ
cos −2
3 ( ) )] [ ( )
 [( ) ( ) ] [ ]
2 2 2
6h l nπ l 2 nπ 18 h l nπ 2 nπ
b n= 2
3 sin −3 sin = 2 2 2 sin −sin
l nπ 3 nπ 3 l n π 3 3

[ ]
2
18 h l nπ n nπ
b n= sin −(−1 ) sin
2 2 2
l n π 3 3

⸪ sin
2 nπ
3
=sin nπ −
nπ
3 (
n
=(−1 ) sin
nπ
3 )
18 h nπ
⸫ bn= 2 2
sin
3
[ 1−(−1 )n ]
n π

{
0 ,if nis odd
b =
Thus n 36 h sin nπ , if n is even
n2 π2 3

Substituting this value of bn in equation (9), we get

36 h nπ
c n= sin ( n=1,2,3 ,… … . )
2 2
n π 3

Substituting this value of cn in equation (6), we get required solution

∞
36 h nπ nπx nπat
y ( x , t )= ∑ 2 2
sin
3
sin
l
cos
l
n=2,4,6… … n π

∞
36 h 2 nπ 2nπx 2 nπat
y ( x , t )= ∑ sin sin cos
n=1
2 2
n π 3 l l
∞
9h 1 2 nπ 2 nπx 2 nπat
y ( x , t )= 2 ∑ 2 sin sin cos ............... (10)
π n=1 n 3 l l

To obtain the displacement at the mid – point, put x = l/2 in equation

(10)

( )
∞
l 9h 1 2 nπ 2 nπat
⸫ y ,t = 2 ∑ 2 sin sin nπ cos =0 ⸪sin nπ =0
2 π n=1 n 3 l

Hence the mid – point of the string is at rest.