vibrating string with variable density

The unidimensional wave’s problem may be stated as


with initial conditionsMathworldPlanetmath


and boundary conditions


which may be specialized to a string’s motion if we physically interpret c2(x)=T0/ρ(x) as the ratio between the string’s initial tension and its linear density. We will discuss the free string’s vibrations (i.e. f(x,t)0) given by the string’s problem

2ut2-(1+x)22ux2=0,x(0,1),t>0, (1)

initial conditions

{u(x,0)=f(x),initial string’s form,ut(x,0)=0,string starts from the rest,

and boundary conditions

{u(0,t)=0,string’s left end fixed,u(1,t)=0,string’s right end fixed.

Without loss of generality, we assume unitary the natural undeformed string’s length. The solution of this problem approaches to a string’s motion whose linear density is proportional to (1+x)-2. The method of separation of variablesMathworldPlanetmath (i.e. u(x,t)=X(x)T(t)) gives the equations

X′′+λ(1+x)2X=0, (2)

with boundary conditions X(0)=X(1)=0, and

T′′+λT=0, (3)

with initial conditions T(0)=1, T(0)=0. In these equations, λ is a constant parameter.
In (2), we are dealing with a Sturm-Liouville problem. To find out the eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath, one searches the solution of (2) on the form X(x)=(1+x)a, as we realize that (2) outcomes the associated characteristic equationMathworldPlanetmathPlanetmathPlanetmath

a(a-1)+λ=0.  That is,a=12(1±1-4λ).

In order to satisfying X(0)=0, let us choose


Thus, the boundary condition X(1)=0 becomes


We next study all the possible cases for the eigenvalue λ in the last above equation.

  1. 1.

    λ<1/4. Then 1-4λ is real, and the equation does not have solution.

  2. 2.

    λ=1/4. Then the pair of solutions, above indicated, will not be independent. Indeed the functions (1+x)1/2 and (1+x)1/2log(1+x) are linearly independentMathworldPlanetmath solutions of (2), in (0,1). Nevertheless, although the last one satisfies the boundary condition at x=0, it does not vanish at x=1. Hence, λ=1/4 is not an eigenvalue.

  3. 3.

    λ>1/4. Then 1-4λ is imaginary. We may even get two solutions by setting (i=-1)


    being the real and imaginary partsMathworldPlanetmath of Z(x) two linearly independent solutions. For satisfying X(0)=0 one sets


    then the boundary condition X(1)=0 gives


    Therefore, λ-1/4log2 must be an integral multipleMathworldPlanetmath of π, i.e. λ-1/4log2=nπ, or


    To these eigenvalues correspond the eigenfunctions


and these form a complete system, whenever we impose to f(x) certain conditions that we shall see later. Moreover, f(x) may be expanded in Fourier series




The completeness above mentioned and the Fourier series converges absolutely and uniformly to f(x) in (0,1), only if f(x)𝒫𝒞1(0,1), f(0)=f(1)=0 and 01f(x)2dx is finite. 11A result due to Green-Parseval-Schwarz (GPS) and Bessel’s inequalityMathworldPlanetmath.
On the other hand, for satisfying (3) and its initial conditions, we choose the eigenfunction


Thus, a solution of (1) is given by


and the general solution of (1) may be determined as a linear (infinite) combination of these eigenfunctions, that is


So that, the string’s problem (1) has the formal solution

u(x,t)=n=1cn(1+x)12sin(nπlog(1+x)log2)cos(n2π2log22+14t). (4)

This series converges uniformly, and hence satisfies the initial and boundary conditions, as the series for f(x) converges uniformly. However, in order to assure continuousMathworldPlanetmath derivatives and the partial differential equationMathworldPlanetmath (1) to be satisfied, we need suppose that the series for f′′(x) converges uniformly, i.e. we must suppose that f(x) to be regular 22i.e. f(x)𝒞2(0,1)., that f(0)=f(1)=f′′(0)=f′′(1)=0, and that 01f′′′(x)2dx to be finite. 33By GPS, again.

Title vibrating string with variable density
Canonical name VibratingStringWithVariableDensity
Date of creation 2013-03-22 17:26:42
Last modified on 2013-03-22 17:26:42
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 9
Author perucho (2192)
Entry type Example
Classification msc 35L05