A uniform heavy chain of length hangs vertically

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Ordinary differential equations · Ordinary differential equations

Conditions of Validity, Ch Sturm-Liouville theory: the spectral theorem for a differential operator

Problem

A uniform heavy chain of length LL hangs vertically, fixed at the top (x=Lx=L) and free at the bottom (x=0x=0), where xx is measured upward from the free end. Its small transverse normal modes Y(x)Y(x) satisfy ddx ⁣(x dYdx)+λY=0,0<x≤L,\frac{\mathrm{d}}{\mathrm{d}x}\!\left(x\,\frac{\mathrm{d}Y}{\mathrm{d}x}\right)+\lambda Y=0,\qquad 0<x\le L, where λ=ω2/g\lambda=\omega^2/g, ω\omega is the mode's angular frequency and gg is the gravitational acceleration. Near the free end, the equation has two independent solution behaviours: a bounded one, Y1(x)→cY_1(x)\to c (a finite constant, with Y1′(x)Y_1'(x) tending to a finite limit) as x→0x\to0, and an unbounded one, Y2(x)∼Aln⁡xY_2(x)\sim A\ln x (with Y2′(x)∼A/xY_2'(x)\sim A/x) as x→0x\to0, for constants c≠0c\ne0, A≠0A\ne0. ((a)) Write the chain's equation in Sturm-Liouville form and identify p(x)p(x), q(x)q(x) and w(x)w(x). Show that x=0x=0 is a singular endpoint while x=Lx=L is regular. ((b)) State the boundary condition to impose at the regular end x=Lx=L (fixed top) and explain why no condition of the form αY(0)+βY′(0)=0\alpha Y(0)+\beta Y'(0)=0 is imposed at x=0x=0. ((c)) Evaluate p(x)Y1′(x)p(x)Y_1'(x) and p(x)Y2′(x)p(x)Y_2'(x) as x→0x\to0, and hence show that admitting Y2Y_2 into the domain makes the boundary form p(Y1Y2′−Y1′Y2)p\big(Y_1Y_2'-Y_1'Y_2\big) fail to vanish at x=0x=0, while retaining only bounded multiples of Y1Y_1 makes it vanish for every pair in the domain. State which solution describes a physical hanging chain and why.

Related theory

Browse the problem library