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x = 0 x = l

Figure 4.2: Vibrating String

4.2 Derivation of Equation (4.1)

R b
The amount of pollutant in the interval [0, b] at the time t is M = u(x, t)dx, in grams,
0
say. At the later time t + h, the same molecules of pollutant have moved to the right by
c · h centimeters. Hence
Z Z
b b+ch
M = u(x, t)dx = u(x, t + h)dx.
0 ch
Differentiating with respect to b, we get
u(b, t) = u(b + ch, t + h).

Differentiating with respect to h and putting h = 0, we get 0 = cu x (b, t)+u t (b, t), which
is equation (4.1).

4.3 Vibrating String

Consider a flexible, elastic homogeneous string or thread of length l, which undergoes
relatively small transverse vibrations. For instance, it could be a guitar string or a plucked
violin string. At a given instant t, the string might look as shown in Figure 4.2. Assume
that it remains in a plane. Let u(x, t) be its displacement from equilibrium position at time
t and position x. Because the string is perfectly flexible, the tension (force) is directed
tangentially along the string (Figure 4.3). Let T(x, t) be the magnitude of this tension
vector. Let ρ be the density (mass per unit length) of the string. It is a constant because
the string is homogeneous. We shall write down Newton’s law for the part of the string
between any two points at x = x 0 and x = x 1 . The slope of the string at x 1 is u x (x, t).

T(x 1 , t)

T(x 0 , t) p 1 + u 2 x
u x
1
x 0 x 1

Figure 4.3: Magnitude of tension vector

Newton’s law F = ma in its longitudinal (x) and transverse (u) components is
T x 1
= 0 (longitudinal)
p
1 + u 2 x 0
x

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