Solved: The horizontal displacement u(x, t) of a heavy chain of length L oscillating in

Chapter 14, Problem 15

(choose chapter or problem)

The horizontal displacement u(x, t) of a heavy chain of length L oscillating in a vertical plane satisfies the partial differential equation

\(g \frac{\partial}{\partial x}\left(x \frac{\partial u}{\partial x}\right)=\frac{\partial^{2} u}{\partial t^{2}}\),  0<x<L, t>0

See FIGURE 14.2.8.

(a) Using \(-\lambda\) as a separation constant, show that the ordinary differential equation in the spatial variable x is \(x X^{\prime \prime}+X^{\prime}+\lambda X=0\). Solve this equation by means of the substitution \(x=\tau^{2} / 4\).

(b) Use the result of part (a) to solve the given partial differential equation subject to

u(L, t)=0,  t>0

u(x, 0)=f(x),  \(\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0\),  0<x<L

[Hint: Assume the oscillations at the free end x = 0 are finite.]