Math 124A – October 07, 2010
Viktor Grigoryan
4
Vibrations and heat flow
In this lecture we will derive the wave and heat equations from physical principles. These are second
order constant coefficient linear PDEs, which we will study in detail for the rest of the quarter.
4.1
Vibrating string
Consider a thin string of length
l
, which undergoes relatively small transverse vibrations (think of a
string of a musical instrument, say a violin string). Let
ρ
be the linear density of the string, measured
in units of mass per unit of length. We will assume that the string is made of homogeneous material
and its density is constant along the entire length of the string. The displacement of the string from its
equilibrium state at time
t
and position
x
will be denoted by
u
(
t, x
). We will position the string on the
x
axis with its left endpoint coinciding with the origin of the
xu
coordinate system.
Considering the motion of a small portion of the string sitting atop the interval [
a, b
], which has mass
ρ
(
b

a
), and acceleration
u
tt
, we can write Newton’s second law of motion (balance of forces) as follows
ρ
(
b

a
)
u
tt
= Total force
.
(1)
Having a thin string with negligible mass, we can ignore the effect of gravity on the string, as well as
air resistance, and other external forces. The only force acting on the string is then the tension force.
Assuming that the string is perfectly flexible, the tension force will have the direction of the tangent
vector along the string. At a fixed time
t
the position of the string is given by the parametric equations
x
=
x,
u
=
u
(
x, t
)
,
where
x
plays the role of a parameter. The tangent vector is then (1
, u
x
), and the unit tangent vector
will be
1
√
1+
u
2
x
,
u
x
√
1+
u
2
x
. Thus, we can write the tension force as
T
(
x, t
) =
T
(
x, t
)
1
p
1 +
u
2
x
,
u
x
p
1 +
u
2
x
!
=
1
p
1 +
u
2
x
(
T, Tu
x
)
,
(2)
where
T
(
t, x
) is the magnitude of the tension.
Since we consider only small vibrations, it is safe to
assume that
u
x
is small, and the following approximation via the Taylor’s expansion can be used
p
1 +
u
2
x
= 1 +
1
2
u
2
x
+
o
(
u
4
x
)
≈
1
.
Substituting this approximation into (2), we arrive at the following form of the tension force
T
= (
T, Tu
x
)
.
With our previous assumption that the motion is transverse, i.e. there is no longitudinal displacement
(along the
x
axis), we arrive at the following identities for the balance of forces (1) in the
x
, respectively
u
directions
0 =
T
(
b, t
)

T
(
a, t
)
ρ
(
b

a
)
u
tt
=
T
(
b, t
)
u
x
(
b, t
)

T
(
a, t
)
u
x
(
a, t
)
.
The first equation above merely states that in the
x
direction the tensions from the two edges of the small
portion of the string balance each other out (no longitudinal motion). From this we can also see that the
tension force is independent of the position on the string. Then the second equation can be rewritten as
ρu
tt
=
T
u
x
(
b, t
)

u
x
(
a, t
)
b

a
.