A string of length is held under tension with both ends fixed

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Ordinary differential equations

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

Problem

A string of length

L=1.00\,\mathrm{m}

is held under tension

T=50\,\mathrm{N}

with both ends fixed. Its density falls linearly from

\rho(0)=3.0\,\mathrm{g\,m^{-1}}

\rho(1)=1.0\,\mathrm{g\,m^{-1}}

, so

\rho(x)=(3.0-2.0x)\,\mathrm{g\,m^{-1}}

with

x

in metres. Its transverse modes

y_n(x)

satisfy

-(Ty_n')'=\omega_n^2\rho(x)y_n

with

y_n(0)=y_n(1)=0

. A numerical solution gives the two lowest modes, each scaled so that

\int_0^1 y_n^2\,\mathrm{d}x=1

and

y_n'(0)>0

, with

f_1=78.6\,\mathrm{Hz}

and

f_2=159.1\,\mathrm{Hz}

; the single interior node of

y_2

sits at

x=0.439\,\mathrm{m}

. Direct numerical integration of these two modes gives

\int_0^1 y_1(x)y_2(x)\,\mathrm{d}x = -0.173, \qquad \int_0^1 \rho(x)\,y_1(x)y_2(x)\,\mathrm{d}x = 0.

Someone argues: "The operator

-(T\,\mathrm{d}/\mathrm{d}x)(\mathrm{d}/\mathrm{d}x)

with

y(0)=y(1)=0

is a regular Sturm-Liouville problem, so it is self-adjoint, and the Sturm-Liouville theorem guarantees its eigenfunctions are orthogonal. Therefore

\int_0^1 y_1y_2\,\mathrm{d}x

must equal zero; the reported value

-0.173

is just numerical round-off in an otherwise unremarkable calculation. And because

y_1

and

y_2

are orthogonal, they form a complete set for this string, so any initial displacement shape

f(x)

[0,1]

with

f(0)=f(1)=0

can be written exactly as

f(x)=c_1y_1(x)+c_2y_2(x)

for suitable constants

c_1,c_2

." ((a)) Identify the two separate errors in this argument. ((b)) State, with reference to the two integrals given, what the Sturm-Liouville theorem in fact guarantees for this string.

Related theory

Ordinary differential equations

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