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540 CHAPTER 15 Special Functions and Eigenfunction Expansions

We will pause in the development of Bessel functions to look at two applications. The first
is the problem studied by Daniel Bernoulli in perhaps the first appearance of a Bessel function.
Notice that both solutions depend on knowledge of zeros of Bessel functions. These will also
be important for eigenfunction expansions involving Bessel functions, and we will devote some
time to them later.

15.3.4 Displacement of a Hanging Chain
Imagine a heavy flexible chain fixed at its upper end and free at the lower end. We want to
describe oscillations caused by a small displacement of the lower end.
First, we need to model the problem. Assume that each particle of the chain oscillates in a
horizontal straight line. Let m be the mass of the chain per unit length, assumed constant, L its
length, and y(x,t) the horizontal displacement at time t of the particle of chain at distance x from
the top end of the chain. To derive an equation for y, consider an element of chain of length x.
Let the forces acting on the ends of this segment have magnitudes T and T + T . The horizon-
tal component of Newton’s first law of motion (force equals mass times acceleration) requires
that
2
∂ y ∂
∂y
m x = T x.
∂t 2 ∂x ∂x
Then
2
∂ y ∂
∂y
m = T .
∂t 2 ∂x ∂x
We will assume at this point that
T = mg(L − x),
which has been found to be a good approximation for small disturbances. The equation for y is
now
2
2
∂ y ∂y ∂ y
=−g + g(L − x) .
∂t 2 ∂x ∂x 2
This is a partial differential equation. To solve it, first change variables by putting
z = L − x and u(z,t) = y(L − z,t).
The partial differential equation transforms to
∂ u ∂u ∂ z
2
2
= g + gz .
∂t 2 ∂z ∂z 2
We will now anticipate the method of separation of variables (Chapter 16). Look for a solution
having the form of a function of z multiplied by a function of t. Not all functions of z and t have
this form (for example, sin(zt) does not). However, for the equation under consideration, some
thought and trial and error suggest that there might be a solution of the form
u(z,t) = f (z)cos(ωt − δ).
The cosine term is suggested by the fact that we expect the motion of the free end of the chain
to exhibit periodic oscillations. Substitute this expression for u(z,t) into the partial differential
equation to obtain
2

−ω f (z)cos(ωt − δ) = gf (z)cos(ωt − δ) + gzf (z)cos(ωt − δ).

Dividing out the common term cos(ωt − δ), we obtain
2

−ω f (z) = gf (z) + gzf (z).

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October 14, 2010 15:20 THM/NEIL Page-540 27410_15_ch15_p505-562