Applications
of
Partial Differential Equations
Wave Equation
Dr. T. Phaneendra
Professor of Mathematics,
School of Advanced Sciences,
VIT University, Vellore632014,Tamil Nadu, India,
email: [email protected]
1
Vibrating Strings
Consider a thin copper string of length
l
cm, stretched between the fixed ends
x
=
0
and
x
=
l
.
Suppose that it is released from the initial position
f
(
x
)
by giving a velocity
g
(
x
)
to it.
We wish to determine the subsequent vertical displacement
u
(
x
,
t
)
at any point
x
of the
string.
The
transverse displacement
u
(
x
,
t
)
in vibrating strings is governed by the homoge
neous partial differential equation of second order:
∂
2
u
∂
t
2
=
a
2
∂
2
u
∂
x
2
,
(1.1)
where
a
2
is a physical constant, called the
velocity constant
of the string.
Since both the ends
x
=
0 and
x
=
l
are
fixed
, there
will not be
any vertical displacement
at the ends at any time
t
.
Thus the
boundary conditions
are:
u
(
0
,
t
) =
0
for all
t
>
0
,
(1.2)
and
u
(
l
,
t
) =
0
for all
t
>
0
.
(1.3)
We assume a trial solution of the form
u
(
x
,
t
)
≡
XT
=
X
(
x
)
T
(
t
)
.
Substituting this expression in the
wave equation
(1.1),
XT
00
=
a
2
X
00
T
.
Now separating the variables, we get
T
00
a
2
T
=
X
00
X
·
Since
x
and
t
are independent variables, each of these fractions reduces to a constant,
say
T
00
a
2
T
=
X
00
X
=

λ
.
1
This results in two ordinary differential equations:
X
00
+
λ
X
=
0
,
(1.4)
T
00
+
λ
a
2
T
=
0
.
(1.5)
We consider three cases:
Case (a):
λ
<
0
,
say
λ
=

μ
2
,
where
μ
6
=
0
.
The general solution of (1.4) will then be given by
X
(
x
) =
Ae
μ
x
+
Be

μ
x
.
This satisfies the boundary conditions (1.2) and (1.3) only if
A
+
B
=
Ae
μ
l
+
Be

μ
l
=
0
,
that is only if
A
=
0
=
B
.
This leads to the
trivial solution
:
X
≡
0 for all
x
∈
[
0
,
l
]
of (1.4).
Case (b):
λ
=
0
Then the general solution of (1.4) is
X
(
x
) =
Ax
+
B
, which satisfies the boundary con
ditions (1.2) and (1.3) only if
A
=
0
=
B
. This again is the trivial solution
X
≡
0 of (1.4).
Case (c):
λ
>
0
,
say
λ
=
μ
2
,
where
μ
6
=
0
.
Vibrations in strings are in general periodic in nature and the general solution of (1.4)
is given by
X
(
x
) =
A
cos
μ
x
+
B
sin
μ
x
.
(1.6)
Applying the first boundary condition (1.2), we see that
X
(
0
) =
A
=
0.
Thus (1.6) reduces to
X
(
x
) =
B
sin
μ
x
.
But then using the second boundary condition (1.3), this implies that
X
(
l
) =
B
sin
μ
l
=
0
.
In order to get a nontrivial solution, we must have
B
6
=
0 so that
sin
μ
l
=
0
⇒
μ
l
=
n
π
or
μ
=
n
π
l
,
n
=
±
1
,
±
2
,
....
(1.7)
Thus all the nontrivial solutions of (1.4) are of the form:
X
n
(
x
) =
B
n
sin
n
π
x
l
n
=
±
1
,
±
2
,...,
where
B
0
n
s
are arbitrary constants depending on
n
.
Since
sin

k
π
x
l
=

sin
k
π
x
l
for all
k
=
1
,
2
,
3
,...,
2
we see that
X

k
is a constant multiple of
X
k
for each
k
.
Therefore, we shall discard the indices
n
=

1
,

2
,

3
,...
so that the linearly indepen
dent solutions are given by
X
n
(
x
) =
B
n
sin
n
π
x
l
,
n
=
1
,
2
,
3
,
....
(1.8)
Now the general solution of (1.5) is
T
(
t
) =
C
cos
μ
at
+
D
sin
μ
at
.
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 Fall '19